Download Unit 8 Chemical Equilibrium Focusing on Acid

Unit 8
Chemical Equilibrium Focusing on
Acid-Base Systems
Unit 8 - Ch 15 Chem30
unit
11/1/06
1:17 PM
Page 670
8
Chemical Equilibrium
Focusing on Acid–Base
Systems
Equilibrium describes any condition or situation of balance. We recognize equilibrium in a chemical reaction system, oddly enough, by noticing nothing—we see no
change in any property of the system. The easiest conclusion to draw would be that
nothing is happening, but closer study reveals that, at the molecular level, a lot of
change is going on.
Chemical reaction equilibrium is always a dynamic balance between two opposing
changes, which are balanced because they are occurring at equal rates, within a closed
system. What we observe directly is the net effect—neither an increase nor a decrease
in any measurable property.
Chemistry involves the study of change in chemical substances. To predict and control chemical change, we must better understand the nature of the system at the
molecular level. For instance, to understand why and how bubbles of gas form or
dissolve in a liquid, we must take into account the nature of the gas, the nature of
the liquid, and the actions of their invisible entities—all at the same time.
This unit explores the nature of dynamic equilibrium in chemical systems. It
explains much more thoroughly and completely many of the chemical change concepts you have already learned. You will examine some very important reactions—
those involving acids and bases in solution—at a higher conceptual level. This
knowledge will allow you to describe, explain, and predict many new chemical systems and situations. The continual exploration and improvement of concepts such
as these is a critical part of the nature of science.
As you progress through the unit, think about these focusing questions:
• What is happening in a system at equilibrium?
• How do scientists predict shifts in the equilibrium position of a system?
• How do Brønsted–Lowry acids and bases illustrate equilibrium?
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Unit 8
GENERAL OUTCOMES
In this unit, you will
NEL
•
explain that there is a dynamic balance
of opposing reactions in chemical
systems at equilibrium
•
determine quantitative relationships in
acid–base ionization and other simple
systems at equilibrium
Chemical Equilibrium Focusing on Acid–Base Systems
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ARE YOU READY?
Unit 8
Chemical
Equilibrium
Focusing on
Acid–Base
Systems
Prerequisites
Concepts
•
•
•
•
•
•
•
•
•
1:17 PM
collision–reaction theory
dissociation and ionization
amount concentration
ion concentration
These questions will help you find out what you already know, and what you need to
review, before you continue with this unit.
Knowledge
1. A large number of chemical reactions (most notably redox reactions) will only
occur in aqueous solution, at least at a rate great enough to be observable in a
laboratory. In terms of collision–reaction theory (Figure 1), state
(a) why a given reaction might only occur in solution
(b) why collisions between entities do not necessarily result in reaction
(c) the two effects that an increase in temperature has on collisions between
entities involved in a reaction
(d) the effect that an increase in concentration of one kind of entity has on
collisions between all entities involved in a reaction
An Ineffective Collision
percent reaction
stoichiometric calculation
net ionic equations
acids and bases
indicators
Skills
•
•
laboratory safety
scientific problem solving
You can review prerequisite
concepts and skills in the
appendices and on the
Nelson Web site.
A Unit Pre-Test is also
available online.
www.science.nelson.com
672
Unit 8
GO
Figure 1
Collision–reaction theory considers the numbers, speeds, and orientation of colliding entities to
explain the progress of chemical reactions.
2. Assume that each of the following substances is placed in water. Rewrite the
formula and the physical state to indicate whether it is very soluble or only
slightly soluble in water at SATP. For those substances predicted to produce ions
in solution, write symbols for all aqueous ions present after the substance
dissociates upon dissolving.
(a) MgSO4·7H2O(s)
(b) CH3COCH3(l)
(c) CH2CH2(g)
(d) CaCO3(s)
(e) PbCl2(s)
(f) FeCl3(s)
(g) C3H5(OH)3(l)
3. Each of the following substances is mixed with water to form an aqueous
solution. Write an equation (using the modified Arrhenius theory) to explain
the acidity or basicity of the solution in terms of hydronium or hydroxide ions.
Your equations should, when necessary, represent the “ionizing” of the substance
as a reaction with water.
(a) HCl(g)
(b) CH3COOH(l)
(c) H2SO4(l)
(d) HNO3(l)
(e) NaCl(s)
(f) NH3(g)
(g) NaOH(s)
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Unit 8
4. The concentration of chemical substances in solution can vary widely
(Figure 2). Concentration affects solution properties such as colour,
conductivity, freezing point, and viscosity. Concentration also affects the
frequency of particle (entity) collisions; and, thus, will usually affect the
observed rate of a reaction. Complete Table 1.
dilute
solution
Table 1 Concentration of Entities and Quantities of Reagents in Solution
Reagent
Mass
dissolved (g)
Solution
volume (L)
NaOH(s)
1.74
0.500
OH–(aq)
Al2(SO4)3(s)
2.00
SO42–(aq)
0.100
Al2(SO4)3(s)
2.00
Al3+(aq)
0.100
Cl–(aq)
0.00440
CaCl2(s)
Concentration
(mol/L)
Entity
1.00
5. Chemical substances may also have widely varying concentrations in the
gaseous state. Calculate the amount concentration of
(a) 24.0 g of hydrogen in a 2.00 L container
(b) 500 kPa of oxygen in a 10.0 L container at 0 °C
(c) 4.40 mol of carbon dioxide at SATP
(d) 0.227 mol of methane at STP
concentrated
solution
Figure 2
Concentration affects the rate of
chemical reaction as well as physical
properties.
6. Understanding chemical equilibrium theory often involves using a net ionic
equation. For each of the following combinations of reagents, predict the
product(s), and write a net ionic reaction equation. Balance the equation with
simplest integer coefficients, and include physical states for all substances.
(a) Copper(II) chloride and potassium carbonate solutions are mixed.
(b) Ethene (ethylene) reacts with hydrogen chloride to form chloroethane.
(c) Aluminium foil reacts with hydrochloric acid.
(d) Ammonia undergoes simple decomposition.
(e) Magnesium metal is placed in an aqueous solution of gold(III) chloride.
(f) Mixing sulfurous acid solution with sodium hypochlorite solution results in
a spontaneous redox reaction.
7. The acidity of solutions often has a considerable effect on the type, rate, and
extent of the reactions they will undergo. Acids may be classed as strong or
weak, depending on the extent of their reaction with water. Complete Table 2,
identifying the acids as strong or weak.
Table 2 Solution Acidity and Basicity
Acid
HCl(aq)
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Strength
S/W
[Acid]
(mol/L)
% Reaction
in/with water
0.016
99
HBr(aq)
99
HNO3(aq)
99
CH3COOH(aq)
0.100
HCN(aq)
0.200
HNO2(aq)
0.010
pH
[H3O(aq)]
(mol/L)
0.024
4.0
1.3
5.0
7.0
Chemical Equilibrium Focusing on Acid–Base Systems
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chapter
15
Equilibrium Systems
In this chapter
Exploration: Shakin’ the
Blues
Investigation 15.1: The
Extent of a Chemical
Reaction
Web Activity: Equilibrium
State
Mini Investigation:
Modelling Dynamic
Equilibrium
Lab Exercise 15.A: The
Synthesis of an
Equilibrium Law
Web Activity: Paul
Kebarle
Lab Exercise 15.B:
Determining an
Equilibrium Constant
Web Activity: Writing
Equilibrium Expressions
Investigation 15.2:
Equilibrium Shifts
(Demonstration)
Equilibrium in Chemical Systems
The simplest equilibrium systems are static: Nothing is moving or changing to create
the balance. A textbook sitting on a level desktop is an example of static equilibrium. It
stays motionless because two equal and opposite forces act on it simultaneously. The
downward pull of Earth (gravity) on the book is exactly balanced by the upward push
by the desktop. A laboratory balance is a common technology that uses this kind of
equilibrium.
Chemical equilibrium is also a balance between two opposing agents of change, but
always in a dynamic system. An expert juggler in performance (Figure 1) is similar to a
chemical system at equilibrium. The juggler’s act is a dynamic equilibrium, with some balls
moving upward and some moving downward at any given moment. There is no net
change because the rates of upward movement and downward movement are equal at
any given moment.
Chemical systems at equilibrium have constant observable properties. Nothing appears
to be happening because the internal movement involves entities that are too small to see.
A critical task of chemical engineers is to disturb (unbalance) chemical equilibria in
industrial reactions. Production of specific desired products is controlled by manipulating
the conditions under which reactions occur. Some general concepts apply to all chemical equilibrium systems; these concepts are the focus of this chapter.
STARTING Points
Answer these questions as best you can with your current knowledge. Then, using the
concepts and skills you have learned, you will revise your answers at the end of the
chapter.
Biology Connection:
CO2 Transport
1. What, precisely, is happening to the chemical entities involved in a reaction while
Investigation 15.3: Testing
Le Châtelier’s Principle
2. Is anything happening to the chemical entities involved in a reaction when
observation shows that products are being formed?
observation shows the reaction appears to have stopped, with no more products
being formed?
Case Study: Urea
Production in Alberta
3. Why do some reactions seem to occur partially, and apparently stop while some of all
Lab Exercise 15.C: The
Nitrogen Dioxide–
Dinitrogen Tetroxide
Equilibrium
4. Can the chemical amount of product be predicted successfully for reactions that are
Investigation 15.4:
Studying a Chemical
Equilibrium System
Web Activity: Poison
Afloat
674 Chapter 15
of the reactants are still present, while in other reactions all of the limiting reagent
appears to be consumed?
not quantitative?
Career Connection:
Food Science Technologist; Chemical Process Engineer
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Figure 1
How is a juggler similar to a
chemical system in equilibrium?
Exploration
Shakin’ the Blues
When small pieces of zinc are added to a dilute solution of
excess hydrochloric acid in an open beaker (Figure 2), a
vigorous reaction occurs with lots of gas and heat given off. The
zinc continues to react and, when it is completely consumed,
the visible signs of reaction come to an end. After seeing many
reactions such as this one in your science studies, you may
have come to think that all chemical reactions only go one way:
from reactants to products. But do they always?
Materials: lab apron; eye protection;
400 mL flask and stopper; 250 mL water; 5.0 g potassium
hydroxide (KOH(s)); 3.0 g glucose or dextrose; 2% methylene
blue; stirring rod
•
•
•
•
•
•
•
NEL
Pour 250 mL of water into the flask.
Add 6 drops of methylene blue and all of the potassium
hydroxide and glucose to the flask.
Stir the mixture with the stirring rod until the solids have
dissolved.
Stopper the flask and set it on the bench. Observe the
colour of the solution.
Shake the solution vigorously and note any changes
(Figure 3).
Set the flask on the table and leave it standing until another
change is noticed.
Repeat the previous two steps many times. Make
observations each time.
(a) Describe the reaction in the flask in relation to the
discussion at the beginning of this activity.
(b) What evidence do you have to substantiate your answer to
question (a)?
(c) Predict whether the colour changes will continue forever.
•
Test your prediction over a reasonable period of time.
(d) Evaluate your prediction.
Potassium hydroxide is poisonous and corrosive.
Keep potassium hydroxide away from skin and
eyes. Wear eye protection.
Figure 2
Zinc reacts rapidly
and quantitatively with
hydrochloric acid.
Figure 3
How many times does this reaction
happen?
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15.1
Page 676
Explaining Equilibrium Systems
Scientists describe chemical systems in terms of empirical properties such as temperature, pressure, volume, and amounts of substances present. Chemical systems are simpler to study when separated from their surroundings by a definite boundary. This
separation gives an experimenter control over the system so that no matter can enter
or leave. Such a physical arrangement is called a closed system. A reaction in solution
in a test tube or a beaker can be considered a closed system, as long as no gas is used or
produced in the reaction. Systems involving gases must be closed on all sides by a solid
container. Separating a chemical reaction system from its surroundings makes studying
its properties, conditions, and changes much simpler. The use of controlled systems is
an integral part of scientific study.
Closed Systems at Equilibrium
Figure 1
When the pressure on this
equilibrium system changes, the
equilibrium is disturbed.
One example of a chemical system at equilibrium is a soft drink in a closed bottle—a closed
system in equilibrium. Nothing appears to change, until the bottle is opened. Removing
the bottle cap and reducing the pressure alters the equilibrium state, as the carbon
dioxide is allowed to leave the system (Figure 1). Carbonated drinks that have gone
“flat” because of the decomposition of carbonic acid can be carbonated again by the
addition of pressurized carbon dioxide to the solution to reverse the reaction, and then
capping the container to restore the original equilibrium.
Collision–reaction theory is fundamental to the study of chemical systems. As originally introduced in this textbook to provide a basis for stoichiometric calculations, this
theory required us to initially assume, for simplicity, that reactions are always spontaneous,
rapid, quantitative, and stoichiometric. Common experience, however, shows that this
assumption is not always true. Not all reactions are rapid; for example, corrosion of a car
body may take years. A study of oxidation–reduction reactions soon provides evidence
that many reactions are not spontaneous. This chapter will examine and test the assumption of quantitative reaction in detail to significantly increase your understanding of
chemical systems.
INVESTIGATION 15.1 Introduction
The Extent of a Chemical Reaction
In Chapter 8, you performed experiments that produced evidence
that reactions are quantitative. In a quantitative reaction, the
limiting reagent is completely consumed. To identify the limiting
reagent, you can test the final reaction mixture for the presence of
the original reactants. For example, in a diagnostic test, you might
try to precipitate ions from the final reaction mixture that were
present in the original reactants.
Purpose
The purpose of this investigation is to test the validity of the
assumption that chemical reactions are quantitative.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1, 2, 3)
Problem
What are the limiting and excess reagents in the chemical
reaction of selected quantities of aqueous sodium sulfate and
aqueous calcium chloride?
Design
Samples of sodium sulfate solution and calcium chloride solution
are mixed in different proportions and the final mixture is filtered.
Samples of the filtrate are tested for the presence of excess
reagents, using diagnostic tests.
To perform this investigation, turn to page 700.
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Section 15.1
DID YOU KNOW
1. The evidence gathered in Investigation 15.1 may be classified as an anomaly—an
unexpected result that contradicts previous rules or experience.
(a) Write the balanced equation for the double replacement reaction of sodium
sulfate and calcium chloride solutions.
(b) Write a statement describing the anomaly that occurred, using chemical names
from the equation.
(c) Write the net ionic equation for the reaction.
(d) Use chemical names from the net ionic equation to write a statement about the
anomaly.
(e) Which of the previous statements more accurately describes the chemical system,
according to collision–reaction theory?
2. When scientists first encounter an apparent anomaly, they carefully evaluate the
design, procedure, and technological skills involved in an investigation. One
important consideration is the reproducibility of the evidence. Compare your evidence
in Investigation 15.1 with the evidence collected by other groups. Is there support for
the reproducibility of this evidence?
Evidence obtained from many reactions contradicts the assumption that reactions
are always quantitative. In Investigation 15.1, there is direct evidence for the presence of
both reactants after the reaction appears to have stopped. This apparent anomaly can be
explained, in terms of collision–reaction theory, by the idea that a reverse reaction can
occur: the products, calcium sulfate and sodium chloride, can react to re-form the original reactants. The final state of this chemical system can be explained as a competition
between collisions of reactants to form products and collisions of products to re-form
reactants.
Na2SO4(aq) CaCl2(aq)
Anomalies—Signals for
Change
Anomalies, or discrepant events,
are important as scientists acquire
and develop scientific knowledge.
Sometimes these events have
been ignored, discredited, or
elaborately explained away by
scientists who do not wish to
question or reconsider accepted
laws and theories. Investigating
anomalies sometimes leads to the
restriction, revision, or replacement
of scientific laws and theories.
H2O(g)
H2O(l)
forward
0 CaSO4(s) 2 NaCl(aq)
reverse
This competition requires that the system be closed so that reactants and products
cannot escape from the reaction container. The chemical system in Investigation 15.1 can
be considered a closed system, bounded by the volume of the liquid phase.
We assume that any closed chemical system with constant macroscopic properties
(no observable change occurring) is in a state of equilibrium, usually classified, for convenience, as one of three types. Phase equilibrium involves a single chemical substance
existing in more than one phase in a closed system. Water placed in a sealed container
evaporates until the water vapour pressure (concentration of water in the gas phase)
rises to a maximum value, and then remains constant (Figure 2). Solubility equilibrium involves a single chemical solute interacting with a solvent substance, where excess
solute is in contact with the saturated solution (Figure 3). A chemical reaction equilibrium involves several substances: the reactants and products of a chemical reaction.
All three types of equilibrium are explained by a theory of dynamic equilibrium—a
balance between two opposite processes occurring at the same rate.
The terms forward and reverse are used to identify which process is being referred to,
and are specific to a written equilibrium equation. When any equation is written with
arrows to show that the change occurs both ways, the left-to-right change is called the
forward reaction, and the right-to-left change is called the reverse reaction.
NEL
?
0
Practice
Figure 2
According to the theory of dynamic
equilibrium, as long as the container
remains closed at constant
temperature, the rate at which
molecules in the liquid state
evaporate is equal to the rate at
which molecules in the gas state
condense.
H2O(l) 0 H2O(g)
Figure 3
For excess solid copper(II) sulfate in
equilibrium with its saturated
aqueous solution, the rates of
dissolving and crystallization are
equal.
CuSO4(s) 0 Cu2(aq) SO42–(aq)
Equilibrium Systems 677
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WEB Activity
Simulation—Equilibrium State
This simulation illustrates the establishment of a simple dynamic equilibrium by showing how
the concentrations of entities change over time, beginning from an initial condition. A textbook
can only represent this process as a sequence of static (still) diagrams (such as those in
Figure 4) or graphs (such as that in Figure 5).
www.science.nelson.com
mini Investigation
GO
Modelling Dynamic Equilibrium
This activity models the progress of a chemical reaction to
equilibrium, representing concentrations of reactants and
products as volumes of water. The establishment of equilibrium
is graphed as volume versus number of transfers. This graph
then represents a typical concentration–time graph of a
chemical reaction to equilibrium.
•
After each transfer, measure and record the volume of water
in each cylinder to 0.1 mL. Also record the number of the
transfer.
•
Repeat the transfer step until no significant change in water
volumes has occurred for at least three transfers.
•
Graph the volume of water in each cylinder as a dependent
(responding) variable against the number of transfers as the
independent (manipulated) variable.
•
Repeat the activity, switching the straws used in each
cylinder. Plot the values on the same axes as for the first
trial.
Materials: two drinking straws of different diameters; two
25 mL graduated cylinders; graph paper; water; meniscus finder
•
Label one 25 mL graduated cylinder R (for reactants), and
fill with water to the 25.0 mL mark.
•
Label the other cylinder P (for products) and leave it empty.
•
Holding a straw in each hand, place one straw in each
graduated cylinder so that each straw rests on the bottom
of its cylinder. Use the larger straw in the R cylinder.
•
Transfer water simultaneously from each cylinder to the
other by placing an index finger over the open end of each
straw (to seal it). Then lift the straws out of their original
cylinders, move the bottom of each straw over the other
cylinder, and lift your index fingers to allow the water in
each straw to drain into the cylinder below. Replace the
straws in their original cylinders.
(a) How does the graph change, and how is it the same, when
the smaller straw is initially in the R cylinder?
(b) How would doing the transfers more slowly affect the final
volumes in each cylinder?
(c) How would replacing the larger straw with an even larger
one affect the final volumes in each cylinder?
(d) Express the equilibria of the water transfer trials as percent
yields; that is, the percentage of “reactant” water that is
converted to “product”.
(e) Predict the graph’s shape if both of the straws chosen had
the same diameter.
Chemical Reaction Equilibrium
Chemical reaction equilibria are more complex than phase or solubility equilibria, due
to the variety of possible chemical reactions and the greater number of substances
involved. To explain chemical equilibrium systems, we need to combine ideas from
atomic theory, kinetic molecular theory, collision–reaction theory, and the concepts of
reversibility and dynamic equilibrium. Although this synthesis is successful as a description, and also as an explanation, it has only limited application in predicting quantitative properties of an equilibrium system.
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Section 15.1
The Hydrogen–Iodine Reaction System
Chemists have studied the reaction of hydrogen gas and iodine gas extensively, because
the molecules are simple in structure and the reaction takes place in the gas phase. Once
hydrogen and iodine are mixed, the reaction proceeds rapidly at first. The initial
dark purple colour of the iodine vapour gradually fades, and then remains constant
(Figure 4).
I2(g)
I2(g)
I2(g)
H2(g)
H2(g)
H2(g)
HI(g)
HI(g)
Initially, hydrogen (in excess)
and iodine are added to the
flask. The colour of the iodine
vapour is the only easily
observable property.
Early in the reaction, hydrogen
and iodine form hydrogen
iodide faster than hydrogen
iodide forms hydrogen and
iodine. Overall, the amount of
iodine decreases, so the colour
of the flask contents appears
to lighten. Both hydrogen and
hydrogen iodide are colourless.
Figure 4
When hydrogen and iodine are
added to the flask, the colour of the
iodine vapour is the only easily
observable (empirical) property. An
equilibrium equation describes this
evidence theoretically.
H2(g) I2(g) 0 2 HI(g), t 448 °C
At equilibrium, analysis shows
that the flask contains all
three substances. The purple
colour shows that some iodine
remains. The constancy of the
colour is evidence that
equilibrium exists. Forward
and reverse reactions are
occurring at equal rates.
Reaction of Hydrogen
and Iodine
t = 448 °C
Table 1 The Hydrogen–Iodine System at 448 °C
System
Initial system concentrations
(mmol/L)
Equilibrium system concentrations
(mmol/L)
H2(g)
I2(g)
HI(g)
H2(g)
I2(g)
HI(g)
1
5.00
5.00
0
1.10
1.10
7.80
2
0.50
0.50
1.70
0.30
0.30
2.10
3
0
0
3.20
0.35
0.35
2.5
NEL
Quantity
HI
Table 1 contains data from three experiments with the hydrogen–iodine system: one
in which hydrogen and iodine are mixed; one in which hydrogen, iodine, and hydrogen
iodide are mixed; and one in which only hydrogen iodide is present initially. At a temperature of 448 °C, the system quickly reaches an observable equilibrium each time.
Chemists use evidence such as that in Table 1 to describe a state of equilibrium in two
ways: in terms of percent reaction and in terms of an equilibrium constant. Percent
reaction describes the equilibrium for one specific system example only, whereas an
equilibrium constant describes all systems of the same reaction at a given temperature.
Alternatively, you can draw a graph of the reaction progress, plotting quantity (or concentration) of the reagents versus time (Figure 5).
H2
I2
Time (reaction progress)
Figure 5
The graph of the quantity of each
substance against time shows that
the rate of reaction of the reactants
decreases as the number of reactant
molecules decreases, and the rate
at which the product changes back
to reactants increases as the
number of product molecules
increases. These two rates must
become equal at some point, after
which the quantity of each
substance present will not change.
Equilibrium Systems 679
Unit 8 - Ch 15 Chem30
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DID YOU KNOW
1:18 PM
?
Lavoisier and Closed Systems
Antoine Laurent Lavoisier
(1743–1794) is recognized as the
father of modern chemistry for many
reasons. He created the basic
nomenclature system we still use for
compounds, demonstrated the law
of conservation of mass, and
explained and clarified the theory of
combustion. Most importantly,
perhaps, his successes convinced
the chemical community of the
critical importance of his methods of
careful measurement, and of
carrying out experiments in closed
systems—carefully accounting for,
and preventing, invisible reactants
and products (notably, gases) from
escaping. When chemists began to
understand the importance of
maintaining closed systems in order
to draw correct conclusions about
reactions, chemistry could finally
move forward as a true science.
Page 680
A percent yield is defined as the yield of product measured at equilibrium compared
with the maximum possible yield of product. In other words, percent yield can be useful
for communicating the position of an equilibrium. The maximum possible yield of
product is calculated using the method of stoichiometry, assuming a quantitative forward
reaction with no reverse reaction. Percent yield provides an easily understood way to
refer to quantities of chemicals present in equilibrium systems. For example, analysis
of the evidence in System 1, Table 1, shows that, at 448 °C, this particular hydrogen–iodine
system reaches an equilibrium with a percent yield of 78.0% (Table 2).
Table 2 Percent Yield of the Hydrogen–Iodine System at 448 °C
System
Equilibrium [HI]*
(mmol/L)
Maximum possible [HI]*
(mmol/L)
Percent yield
(%)
7.80
10.0
78.0
1
2
2.10
2.70
77.8
3
2.50
3.20
78.1
*Square brackets [ ] indicate amount concentration.
Equilibrium arrows (0) communicate that an equilibrium exists. To communicate the
extent of a reaction, a percent yield may be written above the equilibrium arrows in a
chemical equation. The following equation describes the position of a hydrogen–iodine
equilibrium in System 1, Table 2, at 448 °C.
78%
H2(g) I2(g)
Learning Tip
When a reaction is shown to be
quantitative (as written in the
equation), it means that the
reverse reaction happens so
little that it can be ignored for
all normal purposes. Another
way to think of a quantitative
reaction is that if the products
shown (as written) were mixed
together as reactants, there
would be no apparent reaction.
A reaction that is quantitative in
the forward direction is
necessarily nonspontaneous in
the reverse direction. In this
unit, we will (arbitrarily) assume
that "quantitative" specifies a
reaction that, at equilibrium, is
more than 99.9% complete.
Another way to think of it is that
less than one part per thousand
(0.1%) of an original reactant
remains unreacted, at
equilibrium, in a quantitative
reaction.
680
Chapter 15
0 2 HI(g)
t 448 °C
Scientists now think of all chemical reactions as occurring in both forward and reverse
directions. Any reaction falls loosely into one of four categories. Reactions that favour
reactants very strongly, that is, reactions that normally have a percent yield of much less
than 1%, are simply observed as being nonspontaneous. In these reactions, mixing reactants has no observable result. Reactions producing observable equilibrium conditions
may react less or more than 50%, favouring reactants or products respectively. Significant
amounts of both reactants and products are always present. Finally, reactions that favour
products very strongly, much more than 99%, are observed to be complete (quantitative).
The chemical equations for quantitative reactions are generally written with a single
arrow to indicate that the effect of the reverse reaction is negligible. Table 3 shows how
percent yield may be used to classify equilibrium systems and how the classification may
be communicated in reaction equations.
Table 3 Classes of Chemical Reaction Equilibria
Percent yield
Description of equilibrium
Position of equilibrium
negligible
nonspontaneous
(no apparent reaction)
50%
reactants favoured
50%
50%
products favoured
50%
99.9%
quantitative
0
0
→
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 681
Section 15.1
When considering equilibrium systems, we cannot use the simple assumption of
quantitative reaction. When there is no limiting reagent for a reaction, and when we
cannot assume complete reaction, stoichiometric calculations require a little more
thought. Such calculations may conveniently be set up as an ICE table, meaning that
the initial, change, and equilibrium values are arranged in tabular form.
SAMPLE problem 15.1
Consider the reaction equation for the formation of hydrogen iodide at 448 °C. Assume the
reaction is begun with 1.00 mmol/L concentrations of both H2(g) and I2(g). Construct an
ICE table to determine equilibrium concentrations of the reagents. The equilibrium
concentration of I2(g) (determined by colour intensity) is 0.22 mmol/L.
Set up the ICE table as follows:
Table 4 The H2(g) I2(g)
Concentration
Initial
0
2 HI(g) Equilibrium
[H2(g)]
(mmol/L)
[I2(g)]
(mmol/L)
[HI(g)]
(mmol/L)
1.00
1.00
0.00
DID YOU KNOW
?
Diabetes: Blood Sugar
Equilibrium
For people with diabetes, reaction
equilibrium established by sugar in
the human body is critically
important. Recent advances in the
technology allow testing of blood
sugar concentration with personal
devices such as the One Touch®
SureStep® blood-glucose meter
(Figure 6). The meter displays the
concentration in mmol/L after
analyzing a single drop of blood
extracted from a fingertip. Multiple
readings, including date and time,
are stored electronically and can
be displayed at any time. The data
can be downloaded to a computer.
Change
Equilibrium
0.22
Begin by calculating the change (decrease) in concentration of iodine.
(0.22 1.00) mmol/L 0.78 mmol/L
The changes of the other concentrations may be calculated directly from this value, using
stoichiometric ratios from the balanced equation.
For hydrogen, the change is also a decrease (negative value) of
1
0.78 mmol/L 0.78 mmol/L
1
For hydrogen iodide, the change is an increase (positive value) of
2
0.78 mmol/L 1.6 mmol/L
1
Complete the ICE table, using these values to enter the concentrations at equilibrium.
Table 5 The H2(g) I2(g)
Concentration
Initial
Change
Equilibrium
0
2 HI(g) Equilibrium
[H2(g)]
(mmol/L)
[I2(g)]
(mmol/L)
[HI(g)]
(mmol/L)
1.00
1.00
0.00
0.78
0.78
1.6
0.22
0.22
1.6
In Sample Problem 15.1, notice that every substance in the reaction is a gas. Therefore,
all of the stoichiometric calculations can use concentrations directly, rather than chemical amounts, because the volume must be the same for every gaseous substance in a
closed container. The volume is a common factor in the calculation step that uses the
stoichiometric ratio. This same reasoning means that concentrations can also be used
directly for stoichiometric calculation whenever every substance in a reaction is an
aqueous entity dissolved in the same volume of solvent.
NEL
Figure 6
This device helps diabetics monitor
their blood glucose levels.
CAREER CONNECTION
Food Science Technologist
The development and analysis of
food products for individuals with
diabetes is crucial. Food science
technologists carefully measure
and conduct tests on
carbohydrates so that patients can
control their blood sugar
equilibrium.
Learn more about the many
food industries that employ food
science specialists.
www.science.nelson.com
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Equilibrium Systems 681
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 682
Practice
3. For a chemical system at equilibrium:
(a) What are the observable characteristics?
(b) Why is the equilibrium considered “dynamic”?
(c) What is considered “equal” about the system?
4. In a gaseous reaction system, 2.00 mol of methane, CH4(g), is initially added to
10.00 mol of chlorine, Cl2(g). At equilibrium the system contains 1.40 mol of
chloromethane, CH3Cl(g), and some hydrogen chloride, HCl(g).
(a) Write a balanced reaction equation for this equilibrium and calculate the
maximum possible yield of chloromethane product.
(b) Calculate the percent yield at this equilibrium and state whether products or
reactants are favoured.
5. Combustion reactions, such as the burning of methane, often favour products so
strongly that they are written with a single arrow. Assuming the forward reaction has
a very low activation energy (Chapter 12) and the reverse reaction has a very high
activation energy (to account for the difference in the tendency to occur), sketch a
possible potential energy diagram representing the progress of such a reaction.
Chemists often refer to
“homogeneous” reaction
systems. This term means that
every entity involved in the
reaction exists in the same
physical state. The reaction
from Sample Problem 15.1,
where all reactants and
products are gases, is a typical
case. The other common case
involves reactions where all
entities involved are in aqueous
solution. A system that has
more than one phase, such as
solid copper reacting in
aqueous silver nitrate solution,
is a “heterogeneous” reaction
system.
6. After 4.0 mol of C2H4(g) and 2.50 mol of Br2(g) are placed in a sealed container, the
reaction
C2H4(g) Br2(g)
0
C2H4Br2(g)
establishes an equilibrium. Figure 7 shows the concentration of C2H4(g) as it
changes over time at a fixed high temperature until equilibrium is reached.
Reaction of Ethene and Bromine
Concentration (mol/L)
Learning Tip
5.0
4.0
3.0
C2H4
2.0
1.0
0.0
Time
Figure 7
A graph of the reaction of ethene with bromine
(a) Sketch this graph. Draw lines on your copy to show how the concentration of
each of the other two substances changes.
(b) Create an ICE table, using reagent amount concentrations.
(c) What is the volume of the container?
(d) Calculate the percent yield of dibromoethane.
7. Write a balanced net ionic equation for each of the following described reactions,
showing appropriate use of equilibrium "arrow" symbols where appropriate, and
indicating (with symbols) whether products or reactants are favoured.
(a) An excess of solid copper reacts with virtually all of the silver ions in a sample
solution.
(b) When a solution containing calcium ions is mixed with a solution containing a
large excess of sulfate ions, a precipitate forms, but tests indicate that a small
quantity of calcium ions remains in solution.
(c) When acetic acid is dissolved in water, the acetic acid molecules react with water
molecules to form hydronium and acetate ions. Careful pH testing shows that
about 980 of every 1000 acetic acid molecules remain in their molecular form, at
equilibrium.
682
Chapter 15
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 683
Section 15.1
LAB EXERCISE 15.A
Report Checklist
Purpose
Problem
Hypothesis
Prediction
The Synthesis of an Equilibrium Law
The following chemical equation represents a chemical
equilibrium:
Fe3(aq) SCN(aq)
0
FeSCN2(aq)
Design
Materials
Procedure
Evidence
Analysis
Evaluation
Design
The purpose of this investigation is the synthesis of an equilibrium
law. Complete the Analysis of the investigation report.
Reactions are performed using various initial concentrations of
iron(III) nitrate and potassium thiocyanate solutions. The
equilibrium concentrations of the reactants and the product are
determined from the measurement and analysis of the colour
intensity using a spectrophotometer. Possible mathematical
relationships among the concentrations are tried and analyzed to
determine if the mathematical formula gives a constant value.
Problem
Evidence
What mathematical formula, using equilibrium concentrations
of reactants and products, gives a constant for the iron(III)–
thiocyanate reaction system?
Table 6 Iron(III)–Thiocyanate Equilibrium at SATP
This equilibrium is convenient to study because the colour of the
system characterizes the equilibrium position of the system
(Figure 8).
Purpose
Trial
[Fe3(aq)]
(mol/L)
[SCN(aq)]
(mol/L)
[FeSCN2(aq)]
(mol/L)
1
3.91 102
8.02 105
9.22 104
2
1.48 102
1.91 104
8.28 104
3
6.27 103
3.65 104
6.58 104
4
2.14 103
5.41 104
3.55 104
3
4
3.23 104
1.78 10
5
SCN–(aq)
3+(aq)
Fe
2+(aq)
FeSCN
Figure 8
The two reactants
combine to form a dark
red equilibrium mixture.
The red colour of the
solution is due to the
aqueous thiocyanate–
iron(III) product,
FeSCN2(aq).
6.13 10
Analysis
Test the following mathematical relationships for constancy:
1. [Fe3(aq)][SCN(aq)][FeSCN2(aq)]
2. [Fe3(aq)] [SCN(aq)] [FeSCN2(aq)]
[FeSCN2(aq)]
3. 3
[Fe (aq)][SCN(aq)]
[Fe3(aq)]
4. [FeSCN2(aq)]
[SCN(aq)]
5. [FeSCN2(aq)]
WEB Activity
DID YOU KNOW
Computers
Canadian Achievers—Paul Kebarle
Paul Kebarle (Figure 9) pioneered the measurements of gas-phase
ion-molecule equilibria. Kebarle’s findings, now significantly expanded
by other workers, constitute a central database that is of fundamental
importance in many diverse fields of scientific research.
1.
What fundamental data did Kebarle and his co-workers obtain
from their research?
2.
List three fields of research that Kebarle’s work has aided.
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GO
?
Figure 9
Paul Kebarle
Scientists often use computers to
analyze numerical evidence in
order to establish mathematical
relationships among experimental
variables. The mathematical
formulas derived are useful in
understanding chemical processes
and in applying these processes to
technology.
Equilibrium Systems 683
Unit 8 - Ch 15 Chem30
11/1/06
DID YOU KNOW
1:18 PM
?
Related Interests
Two Norwegian chemists, Cato
Maximilian Guldberg and Peter
Waage, conducted detailed empirical
studies of many equilibrium systems
in the mid-1800s (Figure 10). By
1864, they had proposed a
mathematical description of the
equilibrium condition that they called
the “law of mass action.” Analyzing
the results of their experiments,
Guldberg and Waage noticed that,
when they arranged the equilibrium
concentrations into a specific form of
ratio, the resulting value was the
same no matter what combinations
of initial concentrations were mixed.
Page 684
The Equilibrium Constant, Kc
Analysis of the evidence from many experiments such as those in Lab Exercise 15.A
(page 683) reveals a mathematical relationship that provides a constant value for a chemical system over a range of amount concentrations. This constant value is called the
equilibrium constant, Kc , for the reaction system. Evidence and analysis of many equilibrium systems have resulted in the following equilibrium law.
For the reaction a A b B
0 c C d D,
[C]c[D]d
the equilibrium law expression is Kc [A]a[B]b
In this mathematical expression, A, B, C, and D represent chemical entity formulas and
a, b, c, and d represent their coefficients in the balanced chemical equation. The relationship holds only when amount concentrations are observed to remain constant, in a
closed system, at a given temperature.
COMMUNICATION example 1
Write the equilibrium law expression for the reaction of nitrogen monoxide gas with
oxygen gas to form nitrogen dioxide gas.
Solution
2 NO(g) O2(g)
0
2 NO2(g)
2
[NO2(g)]
Kc [NO(g)]2[O2(g)]
Figure 10
Cato Maximilian Guldberg
(1836–1902) and Peter Waage
(1833–1900) were related by more
than their interest in chemistry:
They were brothers-in-law!
Learning Tip
It is common practice
(convention) to ignore units
and list only the numerical
value for an (amount
concentration) equilibrium
constant. The expression of
units is often very complex for
Kc relationships. But, because
each entity concentration is
always entered with mol/L
units, any entity concentration
we calculate from a Kc value
will always give an answer
having mol/L units, so you need
only memorize this (simplifying)
rule.
684
Chapter 15
We use a balanced chemical equation with whole-number coefficients to write the mathematical expression of the equilibrium law. The coefficients of the balanced equation
become the exponents of the amount concentrations. If the equation were to be written
in reverse, the equilibrium law expression would simply be the reciprocal of the expression above, and the equilibrium constant would be the reciprocal of the one for the
reaction as written here. Using the products over reactants convention results in a relationship between the numerical value of Kc and the forward extent of the equilibrium that
is easier to visualize for the equation as written. The higher the numerical value of the
equilibrium constant, the greater the tendency of the system to favour the forward direction; that is, the greater the equilibrium constant, the more the products are favoured at
equilibrium.
COMMUNICATION example 2
The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at temperature t.
Determine the value of Kc for the decomposition of HI(g) at the same temperature.
Solution
2 Hl(g)
0
H2(g) I2(g)
[H2(g)][I2(g)]
1
= 0.025
Kc [Hl(g)]2
40
Note that the decomposition reaction equation is the reverse of the formation reaction
equation, and the value of Kc for decomposition is the reciprocal of the Kc for formation.
NEL
Unit 8 - Ch 15 Chem30
11/3/06
9:42 AM
Page 685
Section 15.1
Experiments have shown that the value of the equilibrium constant depends on temperature. The value is also affected by very large changes in the equilibrium concentration of a reactant or a product. A moderate change in the concentration of any one of
the reactants or products results in a change in the other concentrations, so that the
equilibrium constant remains the same. The equilibrium constant provides only a
measure of the equilibrium position of the reaction; it does not provide any information
on the rate of the reaction. Because they hold for a significant range of different concentrations, equilibrium constant expressions have been found to be very useful, and
Kc values for reactions are in common use throughout the scientific community.
Equilibrium constants are adjusted to reflect the fact that pure substances in solid or
liquid (condensed) states have concentrations that are essentially fixed—the chemical
amount (number of moles) per unit volume is a constant value. For example, a litre of
liquid water at SATP has a mass of 1.00 kg (a chemical amount of 55.5 mol) and, thus,
a fixed amount concentration of 55.5 mol/L. The concentration of condensed states is
not included in a Kc expression—we assume that these constant values become part of
the expressed equilibrium constant. Substances in a gaseous or dissolved state have variable concentrations, and must always be shown in an equilibrium law expression.
COMMUNICATION example 3
Write the equilibrium law expression for the decomposition of solid ammonium chloride to
gaseous ammonia and gaseous hydrogen chloride.
Solution
NH4Cl(s)
0
NH3(g) HCl(g)
Kc [NH3(g)][HCl(g)]
The concentration of solid NH4Cl(s) is omitted from the equilibrium law expression.
The role of temperature in equilibrium constant expressions is critical, although the
temperature is not written in the expression directly. The value of the equilibrium constant, Kc , always depends on the temperature. Any stated numerical value for an equilibrium constant, or any calculation using an equilibrium constant expression, must
specify the reaction temperature at equilibrium.
Since equilibrium depends on the concentrations of reacting substances, these substances must be represented in the expression as they actually exist—meaning that ions
in solution must be represented as individual entities. Equilibrium constant expressions
are always written from the net ionic form of reaction equations, balanced with simplest
whole-number (integral) coefficient values unless otherwise specified.
DID YOU KNOW
?
Using Constant
Relationships
You are already familiar with the
usefulness of some other constant
mathematical relationships about
real phenomena. Finding a
relationship that is constant for
equilibrium concentrations is just
another example.
If you examine circles of
different sizes carefully, you
discover that the distance around
any circle divided by the distance
across it (at the widest part)
always gives the same (constant)
answer, no matter how big or small
the circle. This constant value,
3.14159…, is so useful that it has
been given its own symbol, the
Greek letter pi, . This relationship
is most usefully expressed as
C d because measuring a
diameter is much easier than
measuring a circumference.
Many other relationship
expressions produce this same
constant, including those usually
used to calculate the area of a
circle and the time period of a
pendulum’s swing.
+ EXTENSION
The Meaning of the
Equilibrium Constant
Try this simulation to deepen your
understanding of the equilibrium
constant.
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COMMUNICATION example 4
Write the equilibrium law expression for the reaction of zinc in copper(II) chloride solution.
Solution
Zn(s) Cu2(aq)
0
Cu(s) Zn2(aq)
[Zn2(aq)]
Kc [Cu2(aq)]
Again, note that the solids, as well as the spectator ions (the chloride ions in this example),
are omitted from the equilibrium law expression.
NEL
Equilibrium Systems 685
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 686
LAB EXERCISE 15.B
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Determining an Equilibrium Constant
Determining the equilibrium constant for a reaction at specific
conditions is often essential for an industrial chemist. This
knowledge is necessary before adjusting the reaction conditions
in order to optimize production of the desired substances.
Complete the Analysis of the investigation report.
Analysis
Evaluation
Problem
What is the value of the equilibrium constant for the
decomposition of phosphorus pentachloride gas to phosphorus
trichloride gas and chlorine gas, at a temperature of 200 °C?
Purpose
The purpose of this investigation is to use the equilibrium law to
determine the equilibrium constant at 200 °C for the
decomposition reaction of the molecular compound phosphorus
pentachloride.
SUMMARY
Design
Materials
Procedure
Evidence
Evidence
equilibrium temperature 200 °C
equilibrium concentrations:
[PCl3(g)] [Cl2(g)] 0.014 mol/L
[PCl5(g)] 4.3 104 mol/L
Writing Equilibrium Law Expressions
Write an equilibrium law expression based on a balanced equation for the reaction
system. Use single whole-number coefficients, written in net ionic form, and ignore
concentrations of pure solid or liquid phases:
If: aA bB 0 cC dD
[C]c[D]d
then: Kc [A]a[B]b
An equilibrium constant value
• always depends on the system temperature
• is independent of the reagent concentrations
• is independent of any catalyst present
• is independent of the time taken to reach equilibrium
• is normally stated as a numerical value, ignoring any units
• is greater, the more the system favours the formation of products
Predicting Final Equilibrium Concentrations
For simple homogeneous systems, it is possible to algebraically predict reagent concentrations at equilibrium using a known value for Kc and initial reactant concentration
values. For more complex systems, the calculation becomes more difficult—such systems
are left for more advanced chemistry courses.
SAMPLE problem 15.2
In a 500 mL stainless steel reaction vessel at 900 °C, carbon monoxide and water vapour
react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction
establishes an equilibrium with only partial conversion of reactants to products. Initially,
2.00 mol of each reactant is placed in the vessel. Kc for this reaction is 4.20 at 900 °C.
What amount concentration of each substance will be present at equilibrium?
Write a balanced equation for the reaction equilibrium.
CO(g) H2O(g)
686
Chapter 15
0
CO2(g) H2(g)
Kc 4.20 at 900 °C
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 687
Section 15.1
Use the balanced equation to write the equilibrium law expression.
[CO2(g)][H2(g)]
Kc 4.20 [CO(g)][H2O(g)]
The initial amount concentrations of the CO(g) and the H2O(g) are the same:
2.00 mol
c 4.00 mol/L [CO(g)] [H2O(g)]
0.500 L
An ICE table makes it easier to keep track of amount concentration changes that occur
during a reaction, and to find the amount concentrations at equilibrium.
At equilibrium, let the final amount concentration of the product H2(g) be x mol/L (any
convenient symbol could be used). Then [CO2(g)] must also be x mol/L, since the
stoichiometric ratio of the reaction is 1:1:1:1. By this same reasoning, at equilibrium, the
initial concentrations of CO(g) and H2O(g) must have decreased by x mol/L; so
[CO(g)] [H2O(g)] (4.00 – x) mol/L. When you enter values for “Change” into the ICE
table, you must show increases as “”, and decreases as “”.
Table 7 The CO(g) H2O(g)
0
CO2(g) H2(g) Equilibrium
[CO(g)]
(mol/L)
[H2O(g)]
(mol/L)
[CO2(g)]
(mol/L)
[H2(g)]
(mol/L)
Initial
4.00
4.00
0
0
Change
x
x
x
x
(4.00 x)
(4.00 x)
x
x
Concentration
Equilibrium
Substitute equilibrium concentrations in the equilibrium law expression.
[CO2(g)][H2(g)]
x2
4.20 2
[CO(g)][H2O(g)]
(4.00 x)
+ EXTENSION
Since the right side of the equation is a perfect square, solving for x is quite
straightforward.
4.20
Equilibrium Constant and
Reaction Quotient
x2
2
(4.00 x)
Is there any way of knowing
whether or not a reaction is at
equilibrium? If we know the
concentrations of reactants and
products, and the equilibrium
constant at the appropriate
temperature, we can use a
concept called the "reaction
quotient" to predict which way the
reaction will proceed, or if it is
already at equilibrium.
x
2.05 4.00 x
x (2.05)(4.00 x)
8.20 – 2.05x
3.05x 8.20
x 2.69
Assign positive and negative signs and complete the ICE table.
Table 8 The CO(g) H2O(g)
Concentration
Initial
Change
Equilibrium
At equilibrium, at 900 °C,
0
CO2(g) H2(g) Equilibrium
[CO(g)]
(mol/L)
[H2O(g)]
(mol/L)
[CO2(g)]
(mol/L)
[H2(g)]
(mol/L)
0
0
4.00
4.00
2.69
2.69
2.69
2.69
1.31
1.31
2.69
2.69
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[CO(g)] [H2O(g)] 1.31 mol/L and
[CO2(g)] [H2(g)] 2.69 mol/L
Note that, if both initial reactant concentrations are not the same, solving the equation
for x is more complicated, requiring use of the quadratic formula. Questions in this text
are restricted to examples that do not require the quadratic formula for solution.
NEL
Equilibrium Systems 687
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 688
WEB Activity
Simulation—Writing Equilibrium Expressions
This simulation allows you to select a reaction type and the initial reactant concentrations,
which the program uses to plot the resulting equilibrium graph. You will then be guided
through a series of questions.
www.science.nelson.com
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Section 15.1 Questions
expression of the equilibrium law for each of the following
reaction systems at fixed temperature.
(a) Hydrogen gas reacts with chlorine gas to produce
hydrogen chloride gas in the industrial process that
eventually produces hydrochloric acid.
(b) In the Haber process (Chapter 8), nitrogen reacts with
hydrogen to produce ammonia gas.
(c) At some time in the future, industry and consumers
may make more extensive use of the combustion of
hydrogen as an energy source.
(d) When aqueous ammonia is added to an aqueous
nickel(II) ion solution, the Ni(NH3)62(aq) complex ion is
formed (Figure 11).
Figure 11
A Ni2(aq)
solution is green.
Ammonia reacts
with the nickel(II)
ion to form the
intensely blue
hexaamminenickel(II) ion,
Ni(NH3)62(aq).
Ni2+(aq)
Ni(NH3)62+(aq)
3. Interpret the graph in Figure 12 to answer the questions
about the reaction. Hydrogen and iodine were placed in a
reaction vessel, which was then sealed, and heated to
450 °C.
Reaction of Hydrogen and Iodine t = 448 °C
Concentration (mol/L)
1. Write a balanced equation with integer coefficients and the
8.0
7.2
HI
6.0
4.0
H2
2.4
2.0
I2
0.4
Time
Figure 12
The progress of a hydrogen–iodine reaction
(a) All three substances are gases. If the container has a
volume of 2.00 L, what chemical amount of each
substance was present initially?
(b) What chemical amount of hydrogen iodide had formed
at equilibrium? (Create an ICE table.)
(c) Describe the rate at which hydrogen is reacting from
the moment the reactants are mixed to the time when
equilibrium has been established, in terms of
collision–reaction theory.
4. For each of the following, write the chemical reaction
(e) In the Solvay process for making washing soda
(Chapter 7), one reaction involves heating solid calcium
carbonate (limestone) to produce solid calcium oxide
(quicklime) and carbon dioxide.
(f) In Investigation 15.1, aqueous solutions of sodium
sulfate and calcium chloride are mixed. (Remember to
use a net ionic equation.)
(g) In a sealed can of soda, carbonic acid, H2CO3(aq),
decomposes to liquid water and carbon dioxide gas.
2. You can apply the empirical and theoretical concepts of
equilibrium to many different chemical reaction systems.
Use the generalizations from your study of organic
chemistry to predict the position of equilibrium for bromine
placed in a reaction container with ethylene at a high
temperature.
688
Chapter 15
equation with appropriate equilibrium arrow, as shown
in Table 3 (page 680).
(a) The Haber process is used to manufacture ammonia
fertilizer from hydrogen and nitrogen gases. Under lessthan-desirable conditions, only an 11% yield of
ammonia is obtained at equilibrium.
(b) A mixture of carbon monoxide and hydrogen, known as
water gas, is used as a supplementary fuel in many
large industries. At high temperatures, the reaction of
coke and steam forms an equilibrium mixture in which
the products (carbon monoxide and hydrogen gases)
are favoured. (Assume that coke is pure carbon.)
(c) Because of the cost of silver, many high school science
departments recover silver metal from waste solutions
containing silver compounds or silver ions. A
quantitative reaction of waste silver ion solutions with
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:18 PM
Page 689
Section 15.1
copper metal results in the production of silver metal
and copper(II) ions.
(d) One step in the industrial process used to manufacture
sulfuric acid is the production of sulfur trioxide from
sulfur dioxide and oxygen gases. Under certain
conditions, the reaction produces a 65% yield of
products.
5. Write the expression of the equilibrium law for the
hydrogen–iodine–hydrogen iodide system at 448 °C. Using
the evidence for System 1 as reported in Table 1 on page
679, calculate the value of the equilibrium constant.
6. In the Haber process for synthesizing ammonia gas from
nitrogen and hydrogen, the value of Kc is 6.0 102 for the
reaction at 500 °C. In a sealed container at equilibrium at
500 °C, the concentrations of H2(g) and of N2(g) are
measured to be 0.50 mol/L and 1.50 mol/L, respectively.
Write the equilibrium law expression and calculate the
equilibrium concentration of NH3(g).
7. At a certain constant (very high) temperature, 1.00 mol of
HBr(g) is introduced into a 2.00 L container. Decomposition
of this gas to hydrogen and bromine gases quickly
establishes an equilibrium, at which point the amount
concentration of HBr(g)is measured to be 0.100 mol/L.
(a) Write a balanced equation for the reaction.
(b) Write the equilibrium law expression.
(c) Calculate the chemical amount of HBr(g) present at
equilibrium.
(d) Calculate the chemical amount of HBr(g) that has
reacted to form H2(g) and Br2(g) products when
equilibrium is established.
(e) Calculate the chemical amounts of H2(g) and Br2(g)
that have been produced, and, thus, are present, when
equilibrium is established.
(f) Calculate the amount concentration of all substances
present at equilibrium.
(g) Calculate Kc for this reaction at this temperature.
8. To a heated reaction vessel with a volume of 1.00 L, a lab
technician adds 6.23 mmol H2(g), 4.14 mmol of I2(g), and
22.40 mmol of HI(g). At equilibrium, a spectrophotometer is
used to determine that the concentration of iodine vapour
is 2.58 mmol/L. Construct an ICE table and find Kc for the
reaction system
H2(g) I2(g) 0 2 HI(g).
NEL
9. Consider the system
CO2(g) H2(g)
0
CO(g) H2O(g)
Initially, 0.25 mol of water and 0.20 mol of carbon monoxide
are placed in the reaction vessel. At equilibrium,
spectroscopic evidence shows that 0.10 mol of carbon
dioxide is present. Construct an ICE table and find Kc for
this system.
10. Consider the system
2 HBr(g)
0
H2(g) Br2(g)
Initially, 0.25 mol of hydrogen and 0.25 mol of bromine are
placed into a 500 mL electrically heated reaction vessel.
Kc for the reaction at the temperature used is 0.020.
(a) Find the concentrations of the substances at
equilibrium.
(b) Calculate the chemical amount of each substance
present at equilibrium.
11. Explain briefly how atomic theory, kinetic molecular theory,
collision–reaction theory, and the concepts of reaction rate
and reversible reactions are all necessary to explain
chemical reaction equilibrium observations.
Extension
12. In a very long-term sense, Earth may be considered a
closed system. One equilibrium of concern to scientists is
the same one involved in carbonation of soft drinks, on a
vastly larger scale. Scientists believe that over time, the
carbon dioxide gas in the atmosphere should be in
equilibrium with carbon dioxide dissolved in the oceans.
They also know that the concentration of CO2(g) in the
atmosphere has been increased significantly (by about
20%) in the last century, which, they believe, is mostly due
to the burning of fossil fuels. Concerns about the
consequences of global warming make it imperative that
scientists improve their theories about the various cycles,
processes, and equilibria involving this greenhouse gas.
Research and summarize currently accepted theory about
carbon dioxide dissolved in the oceans, and list some other
cycles and systems involving reaction or production of
CO2(g).
www.science.nelson.com
GO
Equilibrium Systems 689
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
15.2
Figure 1
Henri Louis Le Châtelier
(1850–1936), French chemist and
engineer, worked in chemical
industries. To maximize the yield of
products, Le Châtelier used
systematic trial and error. After
measuring properties of equilibrium
states in chemical systems, he
discovered a pattern and stated it as
a generalization. This generalization
has been supported extensively by
evidence and is now considered a
scientific law. By convention, it is
known as Le Châtelier’s principle.
Figure 2
Fe3(aq) SCN(aq)
Chapter 15
Qualitative Change in Equilibrium
Systems
Observing the effects of varying system properties on the equilibrium of systems contributes greatly to our understanding of the equilibrium state. From a technological
perspective, controlling the extent of equilibrium by manipulating properties is very
desirable because control leads to more efficient and economic processes. From a scientific
perspective, observing systems at equilibrium leads to improved theories that describe,
explain, and predict the nature of equilibrium, thus increasing our understanding.
Equilibrium is an area of study where, historically, technology has led science. Reactions
were first manipulated in response to some human need, although the reactions’ responses
were not explained until much later by successive theories of increasing validity. This
section of the chapter will examine equilibrium manipulation in the same way, with
empirical descriptions of equilibrium manipulation given first, followed by theoretical
explanations of the observed results.
According to Le Châtelier’s principle, when a chemical system at equilibrium is disturbed by a change in a property of the system, the system always appears to react in
the direction that opposes the change, until a new equilibrium is reached (Figures 1 to
3). The application of Le Châtelier’s principle involves a three-stage process: an initial equilibrium state, a shifting non-equilibrium state, and a new equilibrium state.
Le Châtelier’s principle provides a method of predicting the response of a chemical
system to an imposed change. Using this simple and completely empirical approach,
chemical engineers could produce more of the desired products, making technological
processes more efficient and more economical. For example, Fritz Haber used
Le Châtelier’s principle to devise a process for the economical production of ammonia
from atmospheric nitrogen. (See the Haber process, Chapter 8, page 325.)
0
FeSCN2(aq)
The test tube on the left is at
equilibrium, as shown by the
constant colour of the FeSCN2(aq)
ion. The equilibrium is disturbed by
the addition of Fe3(aq) ions. The
system shifts and some of the
additional Fe3+(aq) reacts to
produce more FeSCN2(aq), thus
establishing a new equilibrium state.
When the shift is complete, the
concentration of Fe3+(aq) is higher
than before (only some of the added
ions react), the concentration of
SCN(aq) is lower, and the
concentration of FeSCN2(aq) is
higher. The higher concentration of
FeSCN2(aq) is evident from the
more intense colour of the solution
observed in the test tube on the
right.
690
Page 690
Fe3(aq) SCN(aq)
0 FeSCN2(aq)
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 691
Section 15.2
INVESTIGATION 15.2 Introduction
Equilibrium Shifts (Demonstration)
In this investigation, you will be looking at two equilibrium
systems:
N2O4(g) energy 0 2 NO2(g)
colourless
reddish brown
CO2(g) H2O(l)
0
H (aq) HCO3(aq)
The second equilibrium system, produced by the reaction of
carbon dioxide gas and water, is commonly found in the human
body and in carbonated drinks. A diagnostic test is necessary to
detect shifts in this equilibrium. Bromothymol blue, an acid–base
indicator, can detect an increase or decrease in the hydrogen ion
concentration in this system. Bromothymol blue turns blue when
the hydrogen ion concentration decreases, and yellow when the
hydrogen ion concentration increases.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Purpose
The purpose of this demonstration is to test Le Châtelier’s
principle by studying two chemical equilibrium systems: the
equilibrium between two oxides of nitrogen, and the equilibrium
of carbon dioxide gas and carbonic acid.
Problem
How does a change in temperature affect the nitrogen dioxide–
dinitrogen tetroxide equilibrium system? How does a change in
pressure affect the carbon dioxide–carbonic acid equilibrium
system?
To perform this investigation, turn to page 700.
Le Châtelier’s Principle and Concentration Changes
CCl4(l) 2 HF(g)
0 CCl2F2(g) 2 HCl(g)
CCI4(I) + 2 HF(g) 0
CCI2F2(g) + 2 HCI(g)
Concentration
(mol/L)
Le Châtelier’s principle predicts that if the addition of a reactant to a system at equilibrium
increases the concentration of that substance, then that system will undergo an equilibrium shift forward (to the right). The effect of the shift is that, temporarily, we observe the
reactant concentration decreasing, as some of the added reactant changes to products.
This period of change ends with the establishment of a new equilibrium state where, once
again, there are no observable changes. The system has changed in such a way as to oppose
the change introduced. For example, the production of freon-12, a CFC refrigerant,
involves the following equilibrium reaction taking place at a fixed temperature:
[HF]
[HCI]
[CCI2F2]
freon-12
Time
To improve the yield of the primary product, freon-12, more hydrogen fluoride is added
to the initial equilibrium system. The additional concentration of reactant disturbs the
equilibrium state and the system shifts to the right, consuming some of the added
hydrogen fluoride by reaction with carbon tetrachloride. As a result, more freon-12 is produced and a new equilibrium state is reached. In chemical reaction equilibrium shifts,
an imposed concentration change is normally only partially counteracted, and the final
equilibrium state concentrations of the reactants and products are usually different from
the values at the original equilibrium state. See Figure 3 for a graphic interpretation of
the freon-12 equilibrium shift.
Note that adding more carbon tetrachloride, CCl4(l), would have no effect on the
equilibrium state in the container. This reactant is (and stays) in liquid form, so its concentration is constant and would not be increased by increasing the amount of CCl4(l)
present.
Adjusting an equilibrium state by adding and/or removing a substance is by far the most
common application of Le Châtelier’s principle. For industrial chemical reactions, engineers strive to design processes where reactants are added continuously and products are
continuously removed, so that an equilibrium is never allowed to establish. If the reaction is always shifting forward, the process is always making product (and, presumably,
the industry is always making money).
NEL
Figure 3
The reaction establishes an
equilibrium that is disturbed (at the
time indicated by the vertical dotted
line) by the addition of HF(g). Some
of the added HF reacts, decreasing
in concentration, while the
concentration of both products
increases until a new equilibrium is
established and concentrations
become constant again. Note that
the concentration of HF(g) at the
new equilibrium is greater than at
the original equilibrium, so the
imposed change is only partly
counteracted. The initial Kc value
and final Kc value are the same.
Equilibrium Systems 691
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Concentration
(mol/L)
CCI4(I) + 2 HF(g) 0
CCI2F2(g) + 2 HCI(g)
[HF]
[HCI]
[CCI2F2]
Time
Figure 4
The reaction establishes an
equilibrium that is disturbed (at the
time indicated by the vertical dotted
line) by the removal of HCl(g). The
equilibrium shifts forward,
increasing the concentration of both
products while decreasing HF(g)
concentration, until a new
equilibrium is established. The initial
Kc value and the final Kc value are
the same.
air
O2
air-filled
O2
O2
O2
O2
O2 sacs in
lungs
heart
O2
capillary blood vessels
and body cells
Figure 5
Oxygenated blood from the lungs is
pumped by the heart to body
tissues. The deoxygenated blood
returns to the heart and is pumped
to the lungs. Shifts in equilibrium
occur over and over again as oxygen
is picked up in the lungs and
released throughout the body.
692
Chapter 15
Page 692
The removal of a product (if the removal decreases concentration as well as chemical
amount) will also shift an equilibrium forward, producing more product to counteract
the change imposed. The freon-12 reaction can be shifted forward by removing either
gaseous product, since decreasing the amount of a gas lowers its concentration in any reaction container of fixed size (Figure 4).
The following equation represents the final step in the production of nitric acid:
3 NO2(g) H2O(l)
0 2 HNO3(aq) NO(g)
In this industrial process, nitrogen monoxide gas is removed from the chemical system
by a reaction with oxygen gas. The removal of the nitrogen monoxide causes the system
to shift to the right—some nitrogen dioxide and water react, replacing some of the
removed nitrogen monoxide. As the system shifts, more of the desired product, nitric acid,
is produced.
Although equilibrium systems are important in industrial chemical production, they
are even more vital in biological systems. A particularly important biological equilibrium
is that of hemoglobin (a protein in red blood cells), oxygen, and oxygenated hemoglobin.
Hb O2
0 HbO2
As blood circulates to the lungs, the high concentration of oxygen shifts the equilibrium to the right and the blood becomes oxygenated (Figure 5). As the blood circulates
throughout the body, cell reactions consume oxygen. This removal of oxygen shifts the
equilibrium to the left and more oxygen is released.
Collision–Reaction Theory and Concentration Changes
Collision–reaction theory provides a simple explanation of the equilibrium shift that
occurs when a reactant concentration is increased. We assume that the number of reactant entities per unit volume suddenly increases, so that collisions are suddenly much more
frequent for the forward reaction. The forward reaction rate, therefore, increases significantly. Since the reverse reaction rate is not changed, the opposing rates are no longer
equal, and, for a time, the difference in rates results in an observed increase of
products.
Of course, as the concentration of products increases, so does the reverse reaction
rate. At the same time, the new forward rate decreases as reactant is consumed, until
eventually the two rates become equal to each other again. The rates at the new equilibrium
are faster than those at the original equilibrium, because the system now contains a
larger number of particles (and, therefore, a higher concentration) in dynamic equilibrium. If a substance is removed, causing an equilibrium shift, the explanation is similar
except that the initial effect is to suddenly decrease either the forward or the reverse rate
by decreasing the concentration.
Addition or removal of a reagent present in pure solid or pure liquid state does not
change the concentration of that substance. The reaction of condensed phases (solids
and liquids) takes place only at an exposed surface—and if the surface area exposed
is changed, it is always exactly the same change in available area for both forward
and reverse reaction collisions. The forward and reverse rates change by exactly the same
amount if they change at all, so equilibrium is not disturbed and no shift occurs.
NEL
Unit 8 - Ch 15 Chem30
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1:19 PM
Page 693
Section 15.2
Le Châtelier’s Principle and Temperature Changes
BIOLOGY CONNECTION
The heat energy in a chemical equilibrium equation is treated as though it were a reactant or a product.
CO2 Transport
reactants
0 products
(endothermic in the forward direction)
0 products energy
(exothermic in the forward direction)
Heating or cooling a system adds or removes heat energy from the system. In either situation, the equilibrium shifts to minimize the change. If the system is cooled, the equilibrium shifts so that more heat energy is produced. If the system is heated, the equilibrium
shifts in the direction in which heat energy is absorbed.
For example, in the salt–sulfuric acid process used to produce hydrochloric acid, the
system is heated in order to increase the percent yield of hydrogen chloride gas.
2 NaCl(s) H2SO4(l) energy
0 2 HCl(g) + Na2SO4(s)
Adding heat energy shifts the system to the right, absorbing some of the added energy.
In the production of sulfuric acid, the key reaction step is the equilibrium represented
by the following equation. Percent yield of the product is increased at low temperature.
2 SO2(g) O2(g)
0 2 SO3(g) energy
Removing heat energy causes the system to shift to the right. This shift yields more sulfur
trioxide while partially replacing the heat energy that was removed.
Collision–Reaction Theory and Energy Changes
Collision–reaction theory explains the equilibrium shift (that occurs when the heat
energy of a system at equilibrium is changed) as the result of an imbalance of reaction
rates. Consider the previously mentioned reaction equation—a typical exothermic reaction. The reaction energy is shown this time in standard ∆r H notation.
2 SO2(g) O2(g)
0 2 SO3(g)
∆rH 198 kJ
We explain the result of cooling the system by assuming that both forward and reverse
reaction rates are slower at lower temperatures, because the particles move more slowly
and collide less frequently. The reverse rate decreases more than the forward rate, however. While the rates remain unequal, the observed result is the production of more
product and the release of more heat energy. The shift causes concentration changes
that will increase the reverse rate and decrease the forward rate until they become equal
again, at a new, lower temperature (Figure 6).
Note that industrial exothermic equilibrium reactions are often carried out at high temperatures, even though adding heat energy shifts the equilibrium toward reactants (lowers
the percent yield). The Haber process (Case Study, Section 8.3) is a good example. Heat
energy is added because the forward and reverse reaction rates are too slow at lower
temperatures to allow the reaction to reach equilibrium in a reasonable time. Making large
quantities of the marketable product in a short time is much more important to a manufacturer than creating a small increase in the yield of each batch. Whenever possible,
chemical engineers try to design a continuous process for an industrial reaction—one
that shifts the reaction forward by constantly adding reactants, and constantly removing
products. This system is no longer a closed system, so the reaction never establishes
equilibrium. Such an industrial reaction may run continuously for months or even years.
NEL
www.science.nelson.com
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2 SO2(g) + O2(g) 0 2 SO3(g)
Concentration
(mol/L)
reactants energy
Many biological processes depend
on equilibria. For example, the
transportation of carbon dioxide
depends on the CO2 0 H2CO3
equilibrium system. In the tissues
of the body where carbon dioxide
is produced, the equilibrium shifts
so that more carbonic acid is
formed. In the lungs, the shift is in
the reverse direction, as carbon
dioxide is released. You will
discover other examples of
equilibria in a biology textbook.
[SO2]
[O2]
[SO3]
Time
Figure 6
The reaction establishes an equilibrium that is disturbed (at the time
indicated by the vertical dotted line)
by a decrease in temperature. The
equilibrium shifts forward, increasing
the concentration of SO3 product
while decreasing the concentration
of both reactants, until a new equilibrium is established. Because the
temperature is changed, the final Kc
value for this example is greater
than the initial Kc value because the
shift favours the forward reaction.
CAREER CONNECTION
Chemical Process Engineer
What role do chemical engineers
play in designing systems and
sequences of reactions for industrial
chemical production? Research the
education requirements, job
prospects, and work assignments of
people in this profession.
www.science.nelson.com
GO
Equilibrium Systems 693
Unit 8 - Ch 15 Chem30
11/3/06
9:43 AM
Page 694
Le Châtelier’s Principle and Gas Volume Changes
According to Boyle’s law, the amount concentration of a gas in a container is inversely
proportional to the volume of the container. Since the amount concentration of a gas is
directly proportional to its pressure, we can predict the possible effect of container
volume change on the equilibrium position of homogeneous gaseous systems. Decreasing
the volume by half doubles the concentration of every gas in the container. To predict
whether a change in pressure will affect a system’s equilibrium, you must consider the
total chemical amount of gas reactants and the total chemical amount of gas products.
For example, in the equilibrium reaction of sulfur dioxide and oxygen, three moles of
gaseous reactants produce two moles of gaseous products.
2 SO2(g) O2(g)
Concentration
(mol/L)
2 SO2(g) + O2(g) 0 2 SO3(g)
[SO2]
[O2]
[SO3]
Time
Figure 7
The reaction equilibrium is
disturbed by a decrease in container
volume (at the time indicated by the
vertical dotted line). The equilibrium
shifts forward, increasing the
concentration of SO3 while
decreasing the concentration of
reactants, until a new equilibrium is
established. The initial Kc value and
the final Kc value are the same.
Learning Tip
Since increasing the
temperature of any exothermic
reaction at equilibrium always
shifts the equilibrium left
(toward reactants), an increase
in temperature must decrease
the value of Kc for such a
reaction. Similarly, increasing
the temperature will increase
the value of Kc for any
endothermic reaction. You have
already learned that the value
of Kc is temperature dependent.
No other change imposed on a
system at equilibrium changes
the numerical value of the
equilibrium constant.
694
Chapter 15
0 2 SO3(g)
If the volume is decreased, the overall pressure is increased. Increased pressure causes a
shift to the right, which decreases the total number of gas molecules (three moles to
two moles) and, thus, reduces the pressure. If the volume is increased, the pressure is
decreased, and the shift is in the opposite direction. A system with equal numbers of
gas molecules on each side of the equation, such as the equilibrium reaction between
hydrogen and iodine (page 679), is not affected by a change in volume. Similarly, systems
involving only liquids or solids are not affected by changes in pressure. Note that adding
a gas that is not involved in the equilibrium (such as an inert gas) to the container will
increase the overall pressure in the container, but will not cause a shift in equilibrium.
Adding or removing gaseous substances not involved in the reaction does not change the
concentrations of the reactant and product gases.
Collision–Reaction Theory and Gas Volume Changes
When a system involving gaseous reactants and products is changed in volume, the
resulting equilibrium shift is again explained as an imbalance of reaction rates.
2 SO2(g) O2(g)
0 2 SO3(g) 198 kJ
Collision–reaction theory explains the result of decreasing the volume of this system by
assuming that both forward and reverse reaction rates become faster because the concentrations of reactants and products both increase. For this example, however, the forward rate increases more than the reverse rate because there are more particles involved
in the forward reaction. Consequently, the increase in the total number of collisions is
greater for the forward reaction process. Again, while the rates remain unequal, the
observed result is the production of more product. The shift causes concentration changes
that gradually increase the reverse rate and decrease the forward rate until they become
equal again (Figure 7).
Catalysts and Equilibrium Systems
Catalysts are used in most industrial chemical systems. A catalyst decreases the time
required to reach an equilibrium position, but does not affect the final position of equilibrium. The presence of a catalyst in a chemical reaction system lowers the activation
energy for both forward and reverse reactions by an equal amount (Chapter 12), so the
equilibrium establishes much more rapidly but at the same position as it would without
the catalyst present. Forward and reverse rates increase equally. The final equilibrium concentrations are reached in a shorter time compared with the same, but uncatalyzed,
reaction. The value of catalysts in industrial processes is to decrease the time required for
equilibrium shifts created by manipulating other variables, allowing a more rapid overall
production of the desired product.
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Unit 8 - Ch 15 Chem30
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1:19 PM
Page 695
Section 15.2
SUMMARY
Variables Affecting Chemical Equilibria
Variables
Imposed Change
Response of System
concentration
increase
shifts to consume some of the added
reactant or product
decrease
shifts to replace some of the removed
reactant or product
increase
shifts to absorb some of the added
heat energy
decrease
shifts to replace some of the removed
heat energy
increase
(decrease in pressure)
shifts toward the side with the larger
total chemical amount of gaseous
entities
decrease
(increase in pressure)
shifts toward the side with the smaller
total chemical amount of gaseous entities
temperature
volume
(gaseous systems
only)
Practice
1. What three types of changes shift the position of a chemical equilibrium?
2. For each of the following chemical systems at equilibrium, use Le Châtelier’s principle
to predict the effect of the change imposed on the chemical system. Indicate the
direction in which the equilibrium is expected to shift. For each example, sketch the
graph of concentrations versus time, plotted from just before the change to the
established new equilibrium.
(a) H2O(l) energy 0 H2O(g)
The container is heated.
(b) H2O(l) 0 H(aq) OH(aq)
A few crystals of NaOH(s) are added to the container.
(c) CaCO3(s) energy 0 CaO(s) CO2(g)
CO2(g) is removed from the container.
(d) CH3COOH(aq) 0 H(aq) CH3COO(aq)
A few drops of pure CH3COOH(l) are added to the system.
3. Much methanol is produced industrially by the exothermic reaction
CO(g) 2 H2(g) 0 CH3OH(l), carried out at high pressure (5–10 MPa) and
temperature (250 °C) in the presence of several catalyst substances. Methanol is less
flammable than gasoline, and so it is a safer fuel. It is the fuel used in open-wheel
Champ Car racing, and also in the Indianapolis 500.
(a) State, in terms of forward and reverse reaction rates, why using a very high
pressure of the reactant gases is economically desirable for the manufacturer .
(b) State in which direction a high temperature will shift this reaction equilibrium.
(c) Explain why using a high temperature is desirable, in terms of the time required
for the reaction to reach equilibrium.
(d) Explain, in terms of equilibrium position and equilibrium shift, why this reaction is
done in an open system, where reactants are continually added to the pressure
vessel and liquid product is continually removed.
NEL
Equilibrium Systems 695
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 696
INVESTIGATION 15.3 Introduction
Testing Le Châtelier’s Principle
The equilibria chosen for this investigation involve chemicals that
provide coloured solutions. The investigation tests predictions
about equilibrium shifts (made using Le Châtelier’s principle) by
observing colour changes.
In order to complete the Prediction section of the report, you
must read the Design, Materials, and Procedure carefully. Then
make a Prediction about the result of each change made in the
Procedure.
Purpose
The purpose of this investigation is to test Le Châtelier’s principle
by applying stress to four different chemical equilibria.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Problem
How does applying changes to conditions of particular chemical
equilibria affect the systems?
Design
Stresses are applied to four chemical equilibrium systems and
evidence is gathered to test predictions made using Le Châtelier’s
principle. Control samples are used in all cases.
To perform this investigation, turn to page 701.
Case Study
Urea Production in Alberta
Canada has a vast wealth of natural resources that we can use
to produce many chemicals with an incredible variety of uses.
Because plants require nitrogen for growth, the primary
purpose of some of these chemicals is for agricultural use,
such as nitrogen fertilizers. You have already learned that
Alberta produces large amounts of ammonia, which can be
used directly as a fertilizer. Ammonia is stored as a liquid at
high pressure, and injected directly into the soil (see Section
8.3, Figure 7), but it is toxic and corrosive to human tissue,
which makes it dangerous to use. Special equipment and care
are required. Many food producers prefer to use a highnitrogen, nontoxic, solid compound.
Urea (Figure 8) is another simple molecular chemical—also
a nitrogen fertilizer—that is inexpensive, simple to produce,
easy to transport, and extraordinarily useful. This chemical is
used in the millions of tonnes, for applications as diverse as
•
•
•
•
•
•
•
•
wastewater plants (for treating effluent)
air transportation (for de-icing runways)
forestry and agriculture (for fertilizer)
livestock feeding (for a protein supplement)
woodworking (for making glues and resins)
construction supplies (for making insulation)
furniture (for making particle board and chipboard)
clothing (for making certain dyes)
Until the 1900s, the common source of this chemical was stale
animal urine. The body forms this compound as its principal
means of removing excess nitrogen. In fact, until the early
1800s, it was thought that this, or any other “organic” chemical
extracted from living things, always had to be produced by a
living organism. In 1828, however, Friedrich Wöhler, in a
famous classic experiment, synthesized urea in his laboratory.
His work forever changed the “organic” concept of chemistry.
Early in the last century, an efficient industrial method was
696
Chapter 15
H
H
N
C
H
O
N
H
urea
Figure 8
Urea is a small and simple molecule, but a critically important
nitrogen-containing compound. It has a very high nitrogen
percentage by mass, and is very soluble in water—both very
important points for a compound to be useful as a plant
fertilizer. Three different representations of the urea molecule
are shown here.
developed, and has since been used for mass production of
this versatile and valuable substance.
Urea is produced through the reaction of ammonia, NH3(g),
with carbon dioxide, CO2(g), at high temperature and
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 697
Section 15.2
pressure. Most of the reactants exist in liquid form at the
pressures used. A hot concentrated solution, containing about
80% urea by mass, results from this reaction. This hot solution
is further concentrated and cooled through evaporation of the
water content, to form either granules (angular crystals) or
prills (small round pellets) of white solid urea (Figure 9). The
overall reaction may be written as
CO2(l) 2 NH3(l)
0
NH2CONH2(aq) H2O(l)
t 150–200 °C
P 12–20 MPa
The science and technology for urea production developed
in response to a strong demand for this compound. The
process depends on an understanding of reaction equilibrium,
and of the effect (on equilibrium) of high temperature and
pressure conditions. Also key was the design and construction
of reaction containers (vessels) able to withstand high
pressures and temperatures.
Sometimes, as well as building on existing scientific
knowledge, new technology results in scientific advances. A
technology, such as equipment to create very high pressures,
may allow great leaps forward in science, such as the
synthesis of completely new forms of crystalline solids. New
observations then lead to new theories and sometimes even
new laws.
Figure 9
Prills form when sprayed droplets of very hot urea solution are
made to fall through air in a huge tower, cooling and
evaporating to dryness on the way down.
Case Study Questions
1. There are two phase equilibria that shift during the
reaction of carbon dioxide and ammonia. Write
equilibrium equations to represent these phase changes,
and use Le Châtelier’s principle to explain how they are
continuously being shifted, both by pressure, and also by
the effect of the chemical reaction that is occurring in the
vessel.
2. Crystallization of urea from aqueous solution can be
expressed as a solubility equilibrium, according to the
equation
NH2CONH2(aq)
0
NH2CONH2(s)
(a) On which side of this equation would heat energy be
written? Explain your reasoning.
(b) Based on molecular structure and bonding theory,
explain why you would expect urea to be a very highly
soluble compound.
(c) Urea granules that are bagged and sold for fertilizer
use are quite uniform in size (Figure 10). Explain
what physical process would likely be used to
separate these granules from any larger or smaller
granules that form during crystallization.
Figure 10
Urea fertilizer is a nitrogen source
for crops.
Extension
3. Research Friedrich Wöhler’s classic experiment of 1828.
Write the balanced equation for the reaction he
performed.
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4. Find out how much urea is produced annually in Alberta,
where it is produced, and its current price per tonne.
Assemble your findings into an attractive presentation.
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NEL
GO
GO
Equilibrium Systems 697
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 698
LAB EXERCISE 15.C
Report Checklist
Purpose
Problem
Hypothesis
Prediction
The Nitrogen Dioxide–Dinitrogen
Tetroxide Equilibrium
Complete the Prediction, Analysis, and Evaluation sections of the
report.
Purpose
The purpose of this problem is to use Le Châtelier’s principle to
predict the response of an equilibrium to an introduced change in
conditions.
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1, 2, 3)
Design
A sample of nitrogen dioxide gas is compressed in a syringe and
the intensity of the colour is used as evidence to test the
Prediction.
Evidence
The orange-brown nitrogen dioxide gas colour increases in
intensity when the plunger on the syringe is depressed, and then
decreases in intensity (Figure 11). The final colour is slightly more
intense than the original colour (before moving the plunger).
Problem
How does increasing the pressure affect the nitrogen
dioxide–dinitrogen tetroxide equilibrium?
Figure 11
2 NO2(g) 0 N2O4(g)
An increase in pressure on the nitrogen dioxide–dinitrogen tetroxide equilibrium in the closed system results initially in a more intense
colour followed by a decrease in colour intensity.
INVESTIGATION 15.4 Introduction
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Studying a Chemical Equilibrium System
Figure 2, page 690, shows the colours of aqueous solutions of
iron(III), thiocyanate, and iron(III) thiocyanate ions. Use your
knowledge of Le Châtelier’s principle to write a Problem
statement, and then design and carry out a simple investigation
to determine whether the reaction as written is exothermic or
endothermic.
Fe3(aq)
almost colourless
Design
Materials
Procedure
Evidence
SCN(aq)
colourless
Analysis
Evaluation (1, 2, 3)
0
FeSCN2(aq)
red
Purpose
The purpose of this investigation is to use Le Châtelier’s principle
to solve a problem concerning the effect of an energy change on
the following equilibrium system.
To perform this investigation, turn to page 703.
WEB Activity
Web Quest—Poison Afloat
Have you ever considered becoming a crime scene investigator? In this Web Quest, you are
promoted to Chief Chem Crime Investigator. You are presented with a body and a series of
clues… The detecting is up to you. You will have a chance to use your knowledge of chemistry
to solve this puzzle and gather the evidence to unravel what happened.
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698
Chapter 15
GO
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 699
Section 15.2
Section 15.2 Questions
1. The following equation represents part of the industrial
production of nitric acid. Predict the direction of the
equilibrium shift for each of the following changes. Explain
any shift in terms of the changes in forward and reverse
reaction rates.
4 NH3(g) 5 O2(g)
(a)
(b)
(c)
(d)
0
4 NO(g) 6 H2O(g) energy
O2(g) is added to the system.
The temperature of the system is increased.
NO(g) is removed from the system.
The pressure of the system is increased by decreasing
the volume.
2. The following chemical equilibrium system is part of the
Haber process for the production of ammonia.
N2(g) 3 H2(g)
0
2 NH3(g) energy
Suppose you are a chemical process engineer. Use
Le Châtelier’s principle to predict five specific changes that
you might impose on the equilibrium system to increase
the yield of ammonia.
3. In a solution of copper(II) chloride, the following
equilibrium exists:
CuCl42(aq) 4 H2O(l)
dark green
0
Cu(H2O)42(aq) 4 Cl(aq)
blue
For the following stresses put on the equilibrium, predict
the shift in the equilibrium and draw a graph of
concentration versus time to communicate the shift.
(a) Concentrated hydrochloric acid is added.
(b) Saturated aqueous silver nitrate is added, causing a
precipitation reaction.
4. Identify the nature of the changes imposed on the following
equilibrium system at the four times indicated by
coordinates A, B, C, and D (Figure 12).
Concentration
(mol/L)
C2H4(g) + H2(g) 0 C2H6(g) + energy
C 2H 6
C 2H 4
H2
A
B
C
D
6. Chloromethane (methyl chloride) is manufactured by
“chlorinating” methane. For this reaction system at
equilibrium, explain the effect of each of the imposed
changes on the position of reaction equilibrium.
CH4(g) Cl2(g)
(a)
(b)
(c)
(d)
0
CH3Cl(g) HCl(g)
r H is negative
More methane is injected into the reaction vessel.
The container volume is increased.
The temperature is lowered.
A catalyst is introduced into the system.
7. Ethyne (acetylene) is manufactured by a high-temperature
combustion of methane, using a large excess of methane.
For this endothermic reaction system at equilibrium, explain
the effect of each of the imposed changes on the value of
the equilibrium constant.
6 CH4(g) O2(g)
(a)
(b)
(c)
(d)
0
2 C2H2(g) 10 H2(g) 2 CO(g)
More methane is injected into the reaction vessel.
The container volume is decreased.
The temperature is lowered.
A catalyst is introduced into the system.
Extension
8. In a deep lake or in the ocean, the pressure that a human
considers “normal” has doubled by the time a diver reaches
a depth of 10 m, and increases by about one atmosphere
for every extra 10 m, to a maximum of about 100 MPa at the
deepest points in Earth’s oceans. This fact is of major
concern to scuba divers for several reasons. Pressure inside
a scuba diver’s lungs must constantly be adjusted to equal
outside water pressure, otherwise the lungs could collapse
upon diving, and could explode upon rising. In fact, the
development of the pressure regulator (Figure 13) was the
technology that made scuba diving possible.
The gases in a diver’s lungs are dissolved to some extent
in the blood that passes through. Pressure changes can
change several solubility equilibria, with some serious
effects. Research the gases dissolved in human blood, and
what equilibrium shifts cause the diving conditions called
nitrogen narcosis, and the “bends.”
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GO
Time
Figure 12
Graph showing four disturbances to an equilibrium system
5. In which of the following cases would an increase in
temperature increase the percent yield at equilibrium?
(a) H2O(l) 0 H2O(g)
(b) N2(g) 3 H2(g) 0 2 NH3(g)
r H 91 kJ
(c) KOH(s) 0 K(aq) OH(aq) heat
(d) 2 C(s) 2 H2(g) 0 C2H4(g)
r H 53 kJ
NEL
Figure 13
The mouthpiece pressure
from a diver’s tank
automatically adjusts for
changes in depth.
Equilibrium Systems 699
Unit 8 - Ch 15 Chem30
11/3/06
Chapter 15
9:43 AM
Page 700
INVESTIGATIONS
INVESTIGATION 15.1
The Extent of a Chemical Reaction
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1, 2, 3)
In Chapter 8, you performed experiments that produced
evidence that reactions are quantitative. In a quantitative
reaction, the limiting reagent is completely consumed. To
identify the limiting reagent, you can test the final reaction
mixture for the presence of the original reactants. For example,
in a diagnostic test, you might try to precipitate ions from
the final reaction mixture that were present in the original
reactants.
• If a few drops of Ba(NO3)2(aq) are added to the filtrate
and a precipitate forms, then sulfate ions are present.
Ba2(aq) SO42(aq) → BaSO4(s)
• If a few drops of Na2CO3(aq) are added to the filtrate and
a precipitate forms, then calcium ions are present.
Ca2(aq) CO32(aq) → CaCO3(s)
Purpose
Materials
The purpose of this investigation is to test the validity of the
assumption that chemical reactions are quantitative.
lab apron
eye protection
25 mL of 0.50 mol/L
CaCl2(aq)
25 mL of 0.50 mol/L
Na2SO4(aq)
1.0 mol/L Na2CO3(aq) in
dropper bottle
saturated Ba(NO3)2(aq) in
dropper bottle
Problem
What are the limiting and excess reagents in the chemical
reaction of selected quantities of aqueous sodium sulfate and
aqueous calcium chloride?
Design
Samples of sodium sulfate solution and calcium chloride
solution are mixed in different proportions and the final mixture is filtered. Samples of the filtrate are tested for the presence of excess reagents, using the following diagnostic tests:
two 50 mL or 100 mL
beakers
two small test tubes
10 mL or 25 mL graduated
cylinder
filtration apparatus
filter paper
wash bottle
stirring rod
Soluble barium compounds are toxic. Remember to
wash your hands before leaving the laboratory.
Barium nitrate in solid form is a strong oxidizing
substance.
INVESTIGATION 15.2
Equilibrium Shifts (Demonstration)
In this investigation, you will be looking at two equilibrium
systems:
N2O4(g) energy 0 2 NO2(g)
colourless
CO2(g) H2O(l)
reddish brown
0 H(aq) + HCO3(aq)
The second equilibrium system, produced by the reaction of
carbon dioxide gas and water, is commonly found in the
human body and in carbonated drinks. A diagnostic test is
necessary to detect some shifts in this equilibrium.
Bromothymol blue, an acid–base indicator, can detect an
increase or decrease in the hydrogen ion concentration in
700
Chapter 15
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
this system. Bromothymol blue turns blue when the hydrogen
ion concentration decreases, and yellow when the hydrogen
ion concentration increases.
Purpose
The purpose of this demonstration is to test Le Châtelier’s
principle by studying two chemical equilibrium systems: the
equilibrium between two oxides of nitrogen, and the equilibrium of carbon dioxide gas and carbonic acid.
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 701
Chapter 15
INVESTIGATION 15.2 continued
Problem
How does a change in temperature affect the nitrogen
dioxide–dinitrogen tetroxide equilibrium system? How does
a change in pressure affect the carbon dioxide–carbonic acid
equilibrium system?
Materials
lab apron
eye protection
two NO2(g)/N2O4(g) sealed flasks
carbon dioxide–hydrogen carbonate ion equilibrium
mixture (pH 7)
bromothymol blue indicator in dropper bottle
small syringe with needle removed (5 to 50 mL)
solid rubber stopper to seal end of syringe
beaker of ice–water mixture
beaker of hot water
Be careful with the flasks containing nitrogen
dioxide: this gas is highly toxic. Use in a fume hood
in case of breakage.
INVESTIGATION 15.3
Testing Le Châtelier’s Principle
The equilibria chosen for this investigation involve chemicals that provide coloured solutions. This investigation tests
predictions about equilibrium shifts (made using Le Châtelier’s
principle) by observing colour changes. For example, if the
colour in Reaction 3 becomes more intensely red, the equilibrium has shifted right to produce more FeSCN2(aq) ions,
increasing their concentration. See Table 1.
In order to complete the Prediction section of the report,
you must read the Design, Materials, and Procedure carefully. Then, make a Prediction about the result of each change
made in the Procedure.
Purpose
The purpose of this investigation is to test Le Châtelier’s principle by applying stress to four different chemical equilibria.
Procedure
1. Place the sealed NO2(g)/N2O4(g) flasks in hot and
cold water baths (Figure 1) and record your
observations.
2. Place two or three drops of bromothymol blue
indicator in the carbon dioxide–hydrogen carbonate
ion equilibrium mixture.
3. Draw some of the carbon dioxide–hydrogen
carbonate ion equilibrium mixture into the syringe,
and then block the end with a rubber stopper.
4. Slowly move the syringe plunger and record your
observations.
NO2(g) 0 N2O4(g)
hot water
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Table 1 Solution Colours
Ion
Colour
2
CoCl4 (alc)
2
blue
Co(H2O)6 (alc)
pink
H2Tb(aq)
red
HTb(aq)
yellow
2
Tb (aq)
blue
3
Fe (aq)
pale yellow
SCN(aq)
colourless
FeSCN2(aq)
red
2
pale blue
2
deep blue
Cu(H2O)4 (aq)
Cu(NH3)4 (aq)
NEL
ice water
Figure 1
Each of these flasks
contains an equilibrium
mixture of dinitrogen
tetroxide and nitrogen
dioxide. Shifts in
equilibrium can be seen
when one of the flasks is
heated or cooled.
Equilibrium Systems 701
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 702
INVESTIGATION 15.3 continued
Problem
How does applying stresses to particular chemical equilibria
affect the systems?
Part I
CoCl42(alc) 6 H2O(alc)
0
Co(H2O)62(alc) 4 Cl(alc) energy
hot water bath
cobalt(II) chloride equilibrium mixture in ethanol
dropper bottles containing
0.2 mol/L AgNO3(aq)
thymol blue indicator
0.1 mol/L HCl(aq)
0.1 mol/L NaOH(aq)
iron(III) thiocyanate equilibrium mixture
0.2 mol/L Fe(NO3)3(aq) 0.2 mol/L KSCN(aq)
6.0 mol/L NaOH(aq)
0.1 mol/L CuSO4(aq)
1.0 mol/L HCl(aq)
1.0 mol/L NH3(aq)
Part II
0 H(aq) HTb(aq)
HTb(aq) 0 H(aq) Tb2(aq)
H2Tb(aq)
Part III
Fe3(aq) SCN(aq)
0 FeSCN2(aq)
Part IV
Cu(H2O)42(aq) 4 NH3(aq)
0
Cu(NH3)42(aq) 4 H2O(l)
The chemicals used may be corrosive or
poisonous, and may cause other toxic
effects. Exercise great care when using the
chemicals and avoid skin and eye contact.
Immediately rinse the skin if there is any
contact. If any chemicals get in the eyes,
flush eyes for a minimum of 15 min and
inform the teacher. Ethanol is flammable.
Make sure there are no open flames in the
laboratory when using the ethanol solution
of cobalt(II) chloride.
Design
Stresses are applied to four chemical equilibrium systems and
evidence is gathered to test predictions made using
Le Châtelier’s principle. Control samples are used in all cases.
For example, before adding sodium hydroxide to a new equilibrium solution, split the solution into two samples in order
to have a control sample for colour comparison.
Part I Cobalt(II) Complexes
Water, saturated silver nitrate, and heat are added to, and heat
is removed from, samples of the provided equilibrium mixture. Note: This reaction equilibrium is in solution using an
alcohol solvent, shown as (alc), so the concentration of water
is a variable in this system.
Part II Thymol Blue Indicator
Hydrochloric acid and sodium hydroxide are added to samples of the provided equilibrium mixture.
Part III Iron(III)–Thiocyanate Equilibrium
Iron(III) nitrate, potassium thiocyanate, and sodium
hydroxide are added to samples of the provided equilibrium
system.
Procedure
Part I Cobalt(II) Complexes
1. Obtain 25 mL of the equilibrium mixture with the
cobalt(II) chloride complex ions.
2. Place a small amount of the mixture into each of five
small test tubes. Use the fifth test tube as a control for
comparison purposes.
3. Add drops of water to one test tube until a change is
evident. Record the evidence.
4. Add drops of 0.2 mol/L silver nitrate to another test
tube and record the evidence.
5. Heat another equilibrium mixture in a hot water bath
and record the evidence.
6. Cool an equilibrium mixture in an ice bath and
record the evidence.
Part II Thymol Blue Indicator
7. Add about 5 mL of distilled water to each of two
small test tubes.
Part IV Copper(II) Complexes
Aqueous ammonia and hydrochloric acid are added to samples of the provided equilibrium mixture.
8. Add 1 to 3 drops of thymol blue indicator to the
water in each test tube to obtain a noticeable colour.
Use one test tube of solution as a control.
Materials
9. Add drops of 0.1 mol/L HCl(aq) to the experimental
test tube to test for the predicted colour changes.
lab apron
100 mL beaker
6 to 12 small test tubes
distilled water
702
Chapter 15
eye protection
large waste beaker
test-tube rack
crushed ice
10. Add drops of 0.1 mol/L NaOH(aq) to the same tube
to test for the predicted colour changes.
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 703
Chapter 15
Part IV Copper(II) Complexes
INVESTIGATION 15.3 continued
Part III Iron(III)–Thiocyanate Equilibrium
16. Obtain 2 mL of 0.1 mol/L CuSO4(aq) in a small test
tube.
11. Obtain about 20 mL of the iron(III)–thiocyanate
equilibrium mixture.
17. Add three drops of 1.0 mol/L NH3(aq) to establish
the equilibrium mixture.
12. Place about 5 mL of the equilibrium mixture in each
of three test tubes. Use one test tube as a control.
18. Add more 1.0 mol/L NH3(aq) to the above
equilibrium mixture and record the results.
13. Add drops of Fe(NO3)3(aq) to one test tube until a
change is evident.
19. Add 1.0 mol/L HCl(aq) to the equilibrium mixture
from step 18 and record the results.
14. Add drops of 6.0 mol/L NaOH(aq) to this new
equilibrium mixture until a change occurs. (Iron(III)
hydroxide has very low solubility.)
15. Add drops of KSCN(aq) to another equilibrium
mixture until a change is evident.
INVESTIGATION 15.4
Dispose of the chemicals as directed by your teacher. Identify
each as toxic (to be collected) or nontoxic (disposable in the
sink).
Ensure that all equipment and surfaces are clean and
wash your hands thoroughly before leaving the laboratory.
Report Checklist
Studying a Chemical Equilibrium
System
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1, 2, 3)
Figure 1 shows the colours of aqueous solutions of iron(III),
thiocyanate, and iron(III) thiocyanate ions. Use your knowledge of Le Châtelier’s principle to write a Problem statement,
and then design and carry out a simple investigation to determine whether the reaction as written is exothermic or
endothermic.
Purpose
The purpose of this investigation is to use Le Châtelier’s principle to solve a problem concerning the effect of an energy
change on the following equilibrium system.
Fe3(aq) SCN(aq)
almost colourless
colourless
0 FeSCN2(aq)
red
Iron(III) compounds are irritants. Thiocyanate ion
solutions are toxic. Avoid skin and eye contact. If
there is any skin or eye contact, immediately rinse
with plenty of water. Flush the eyes for at least
15 min and inform the teacher.
Figure 1
The iron–thiocyanate reaction
NEL
Equilibrium Systems 703
Unit 8 - Ch 15 Chem30
11/1/06
Chapter 15
1:19 PM
Page 704
SUMMARY
Outcomes
Key Terms
Knowledge
15.1
•
define equilibrium and state the criteria that apply to a
chemical system in equilibrium (15.1)
•
identify, write, and interpret chemical equations for systems
at equilibrium (15.1, 15.2)
•
predict, qualitatively, using Le Châtelier’s principle, shifts in
equilibrium caused by changes in temperature, pressure,
volume, concentration, or the addition of a catalyst, and
describe how these changes affect the equilibrium constant
(15.2)
•
define Kc and write equilibrium law expressions for given
chemical equations, using lowest whole-number coefficients
(15.1)
•
calculate equilibrium constants and concentrations for
homogeneous systems when concentrations at equilibrium
are known, when initial concentrations and one equilibrium
concentration are known, and when the equilibrium constant
and one equilibrium concentration are known (15.1)
STS
•
state that the goal of science is knowledge about the natural
world (15.1, 15.2)
•
list the characteristics of empirical and theoretical
knowledge (15.2)
•
state that a goal of technology is to solve practical problems
(15.2)
Skills
15.2
Le Châtelier’s principle
equilibrium shift
Key Equations
For the reaction a A b B
0
c C d D,
[C] [D]d
the equilibrium law is Kc [A]a[B]b
c
MAKE a summary
1. Make a concept map, beginning with the word
“Equilibrium” in the centre of a page. Link all of the Key
Terms from this chapter, together with points of your
own, to explain and illustrate how connections among
these terms include the equilibrium law expression, ICE
table format, Le Châtelier’s principle, and points from
the section Summaries.
2. Refer back to your answers to the Starting Points
•
initiating and planning: predict variables that can cause a
shift in equilibrium (15.2); design an experiment to show
equilibrium shifts (15.2); describe procedures for safe
handling, storage, and disposal of materials used in the
laboratory (15.1, 15.2)
•
performing and recording: perform an experiment to test,
qualitatively, predictions of equilibrium shifts (15.2)
•
analyzing and interpreting: write the equilibrium law
expression for a given equation (15.1); analyze, qualitatively,
the changes in concentrations of reactants and products
after an equilibrium shift (15.2); interpret data from a graph
to determine when equilibrium is established, and determine
the cause of a stress on the system (15.2)
•
forward reaction
reverse reaction
ICE table
equilibrium constant, Kc
equilibrium law
closed system
equilibrium
phase equilibrium
solubility equilibrium
chemical reaction equilibrium
dynamic equilibrium
communication and teamwork: work collaboratively in
addressing problems and communicate effectively (15.1, 15.2)
questions at the beginning of this chapter. How has
your thinking changed?
Go To
www.science.nelson.com
GO
The following components are available on the Nelson
Web site. Follow the links for Nelson Chemistry Alberta 20–30.
• an interactive Self Quiz for Chapter 15
• additional Diploma Exam-style Review questions
• Illustrated Glossary
• additional IB-related material
There is more information on the Web site wherever you see
the Go icon in this chapter.
+ EXTENSION
Plague and the Little Ice Age
A Dutch researcher discusses his theory that the atmospheric
carbon dioxide equilibrium shifted when 40% of Europe’s
human population was killed by a plague in the 14th century.
Could this have caused “the little ice age”?
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704
Chapter 15
GO
NEL
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 705
REVIEW
Chapter 15
Chapter 15
Many of these questions are in the style of the Diploma
Exam. You will find guidance for writing Diploma Exams in
Appendix H. Exam study tips and test-taking suggestions
are on the Nelson Web site. Science Directing Words used
in Diploma Exams are in bold type.
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GO
DO NOT WRITE IN THIS TEXTBOOK.
Part 1
1. A “closed” system for a chemical equilibrium means that
A.
B.
C.
D.
the reaction container must be solid and sealed, with a
fixed volume
no substance involved in the equilibrium must be able
to enter or leave
pressure and temperature in the reaction vessel must
be constant
no chemical of any kind may be added to the reaction
vessel
Figure 1
Most sulfuric acid is produced in plants such as this one by the
contact process, which includes two exothermic combustion
reactions. Sulfur reacts with oxygen, forming sulfur dioxide; then,
sulfur dioxide, in contact with a catalyst, reacts with oxygen,
forming sulfur trioxide. Sulfur trioxide and water form sulfuric acid.
2. The possible equilibrium that is not dynamic is the one
between
A. the liquid and gas phases of octane
B. chromate and dichromate ions in aqueous solution
C. the downward force of gravity exerted by Earth on an
object, and the upward force of a balance exerted on
the object
D. the oxygen dissolved in water in a lake and the oxygen
dissolved in nitrogen in the atmosphere
4. Which is the correct form of the equilibrium law for this
reaction, as the equation is written?
A.
[SO2(g)]2[O2(g)]
Kc [SO3(g)]
B.
Kc [SO2(g)]2[O2(g)][SO3(g)]2
C.
[SO2(g)]2[O2(g)]
Kc [SO3(g)]2
D.
[SO3(g)]2
Kc [SO2(g)]2[O2(g)]
3. In a reaction, 4.22 mol of product has formed but there are
NR
no more visible signs of change. Calculations show that as
much as 6.00 mol of product could have formed from the
chemical amounts of reactants used. The percent yield is
__________ %.
Use this information to answer questions 4 to 6.
Sulfuric acid is the most common commercial acid, with
millions of tonnes produced each year (Figure 1). The second
step in the “contact” process for industrial production of
sulfuric acid involves the oxidation of sulfur dioxide gas
catalyzed by contact with V2O5(s) powder. The reaction
equation is
2 SO2(g) O2(g)
NEL
0
2 SO3(g)
∆H° 198 kJ
5. The imposed condition that does not shift equilibrium
toward the product is
A. adding the vanadium pentoxide catalyst
B. decreasing the temperature
C. decreasing the container volume
D. adding oxygen to the system container
6. If the equilibrium constant, Kc , for the above oxidation
reaction at a given temperature is 2.4 103, then the
constant for the reverse reaction (written as the
decomposition of sulfur trioxide) is
A. 2.4 103
B. 4.2 104
C. 2.4 103
D. 4.2 104
Equilibrium Systems 705
Unit 8 - Ch 15 Chem30
11/1/06
1:19 PM
Page 706
7. If excess copper reacts in a solution of silver nitrate, it is
correct to state that
A. the reaction is not considered quantitative
B. the equilibrium constant at SATP will have a numerical
value between 1 and 100
C. water is not written in the equilibrium law expression
because it is a spectator species
D. silver nitrate is not written in the equilibrium law
expression because it is a species with a constant
concentration
Use this information to answer questions 8 to 10.
Consider the following reaction to produce hydrogen, done as
a first step in the industrial process to make ammonia.
Methane (from natural gas) reacts with steam over a nickel
powder catalyst. The reaction equation is
CH4(g) 2 H2O(g) heat energy
0
CO2(g) 4 H2(g)
8. This reaction is done in a laboratory autoclave (a stainless
steel pressure vessel) and allowed to reach equilibrium.
More methane is then injected into the autoclave. We can
predict that, when a new equilibrium is reached (at the
same temperature), the concentration of every reagent in
the equation will have increased, except that of
A. methane
B. water
C. carbon dioxide
D. hydrogen
9. When this gaseous reaction system at equilibrium is
disturbed by heating the autoclave, we theorize that
A. both forward and reverse reaction rates increase, but
the forward rate increases more
B. the forward reaction rate does not change, but the
reverse reaction rate increases
C. the reverse reaction rate does not change, but the
forward reaction rate increases
D. both reaction rates increase equally, so the equilibrium
position is unchanged
10. A test reaction is done starting with only methane and
NR
excess water in the autoclave. If the initial concentration of
methane is 0.110 mol/L, and the methane concentration at
equilibrium is 0.010 mol/L, then the equilibrium
concentration of hydrogen is __________ mol/L.
Part 2
11. Define chemical equilibrium empirically.
12. What main idea explains chemical equilibrium?
13. What phrase is used to describe a reaction equilibrium in
which the proportion of reactants to products is quite high?
706
Chapter 15
14. Describe and explain a situation in which a carbonated
soft drink is in
(a) a non-equilibrium state
(b) an equilibrium state
15. Predict whether adding a catalyst affects a state of
equilibrium. What does the catalyst do?
16. For each of the following descriptions, write a chemical
equation for the system at equilibrium. Communicate the
position of the equilibrium with equilibrium arrows. Then
write a mathematical expression of the equilibrium law for
each chemical system.
(a) A combination of low pressure and high temperature
provides a percent yield of less than 10% for the
formation of ammonia in the Haber process.
(b) At high temperatures, the formation of water vapour
from hydrogen and oxygen is quantitative.
(c) The reaction of carbon monoxide with water vapour to
produce carbon dioxide and hydrogen has a percent
yield of 67% at 500 °C.
17. Scientists and technologists are particularly interested in
the use of hydrogen as a fuel. Interpret this reaction
equation by predicting the relative proportions of reactants
and products in this system at equilibrium.
2 H2(g) O2(g)
0
2 H2O(g)
Kc 1 1080 at SATP
Use this information to answer questions 18 to 24.
The solubility of pure oxygen in contact with liquid water at
SATP is very low: only about 42 ppm, or 42 mg/L. The solubility
equilibrium when water is in contact with air (21% oxygen)
at SATP, is even lower: about 8.7 mg/L. The equilibrium
equation is
O2(g)
0
O2(aq)
18. Write the equilibrium law expression for a saturated
solution of oxygen in water.
19. If the gas in the closed system is pure oxygen, the solubility
is higher than it is if the gas is air. Express the solubility of
pure oxygen in water at SATP as an amount concentration.
20. Express the concentration of pure oxygen gas at SATP as an
amount concentration. (Recall that the molar volume of
gases at SATP is 24.8 L/mol.)
21. Use the answers to the previous two questions to
determine the value of Kc for an equilibrium of pure
oxygen gas in contact with its saturated solution at SATP.
22. Predict whether the value of Kc for this equilibrium will be
different for an equilibrium of air in contact with water at
25 °C. Predict which system condition changes will, and
which will not, change the value of an equilibrium constant.
NEL
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Chapter 15
23. In terms of the equilibrium law, explain why more oxygen
dissolves in water when the gas above it is pure oxygen
than when the gas above it is air.
0 2 HBr(g) Kc 12.0 at t °C
(a) 8.00 mol of hydrogen and 8.00 mol of bromine are
added to a 2.00 L reaction container. Construct an ICE
table and use it to predict the concentrations at
equilibrium.
(b) 12.0 mol of hydrogen and 12.0 mol of bromine are
added to a 2.00 L reaction container. Construct an ICE
table and use it to predict the concentrations at
equilibrium.
27. H2(g) Br2(g)
DE
24. The quality of surface water in lakes and streams (Figure 2)
is of critical importance to society. Dissolved oxygen
content of the surface fresh water in Canada averages
10 ppm. Explain what stress is placed upon fish and other
organisms in streams, lakes, and wetlands if climate change
increases the average temperature of the water.
28. CO(g) H2O(g)
0
CO2 (g) H2(g) Kc 4.00 at 900 °C
In a container, carbon monoxide and water vapour react to
produce carbon dioxide and hydrogen. The equilibrium
concentrations are
[H2O(g)] 2.00 mol/L, [CO2(g)] 4.00 mol/L, and
[H2(g)] 2.00 mol/L.
Determine the equilibrium concentration of carbon
monoxide.
29. Write a statement of Le Châtelier’s principle.
30. What variables are commonly manipulated to shift a
chemical equilibrium system?
31. Describe how a change in volume of a closed system
containing a gaseous reaction at equilibrium affects the
pressure of the system.
Figure 2
The Bow River begins as a cold, glacier-fed mountain
stream with a higher-than-average oxygen content, and is
world famous for its trout fly fishery. It also supplies the city
of Calgary with water for a million people daily, as well as
providing water for agriculture in southern Alberta.
25. In many processes in industry, engineers try to maximize
the yield of a product. Outline how concentration can be
manipulated in order to increase the yield of a product.
26. In a container at high temperature, ethyne (acetylene) and
DE
hydrogen react to produce ethene (ethylene). No ethene is
initially present. Later, at equilibrium, the concentration of
ethene is 0.060 mol/L.
C2H2(g) H2(g)
0
C2H4(g)
The initial concentrations of both acetylene and hydrogen
are 1.00 mol/L.
(a) Use an ICE table to determine the equilibrium
constant.
(b) Sketch a reaction progress graph to show the change
in concentration values over time, from the beginning
of the reaction to equilibrium.
NEL
32. In a sealed container, nitrogen dioxide is in equilibrium with
DE
dinitrogen tetroxide.
2 NO2(g)
0
N2O4(g)
Kc 1.15, t 55 °C
(a) Write the mathematical expression for the equilibrium
law applied to this chemical system.
(b) If the equilibrium concentration of nitrogen dioxide is
0.050 mol/L, predict the concentration of dinitrogen
tetroxide.
(c) Predict the shift in equilibrium that will occur when
the concentration of nitrogen dioxide is increased.
33. Predict the shift in the following equilibrium system
resulting from each of the following changes:
4 HCl(g) O2(g)
0
2 H2O(g) 2 Cl2(g) 113 kJ
(a) an increase in the temperature of the system
(b) a decrease in the system’s total pressure due to an
increase in the volume of the container
(c) an increase in the concentration of oxygen
(d) the addition of a catalyst
34. Chemical engineers use Le Châtelier’s principle to predict
shifts in chemical systems at equilibrium resulting from
changes in the reaction conditions. Predict the changes
necessary to maximize the yield of product in each of the
following industrial chemical systems:
(a) the production of ethene (ethylene)
C2H6(g) energy 0 C2H4(g) H2(g)
(b) the production of methanol
CO(g) 2 H2(g) 0 CH3OH(g) energy
Equilibrium Systems 707
Unit 8 - Ch 15 Chem30
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Page 708
35. Apply Le Châtelier’s principle to predict whether, and in
which direction, the following established equilibrium
would be shifted by the change imposed:
2 CO(g) O2(g)
(a)
(b)
(c)
(d)
(e)
0
2 CO2(g) heat energy
temperature is increased
vessel volume is increased
oxygen is added
platinum catalyst is added
carbon dioxide is removed
36. For each example, predict whether, and in which direction,
an established equilibrium would be shifted by the change
imposed. Explain any shift in terms of changes in forward
and reverse reaction rates.
(a) Cu2(aq) 4 NH3(g) 0 Cu(NH3)42(aq)
CuSO4(s) is added
(b) CaCO3(s) energy 0 CaO(s) CO2(g)
temperature is decreased
(c) Na2CO3(s) energy 0 Na2O(s) CO2(g)
sodium carbonate is added
(d) H2CO3(aq) energy 0 CO2(g) H2O(l)
vessel volume is decreased
(e) KCl(s) 0 K(aq) Cl(aq)
AgNO3(s) is added
(f) CO2(g) NO(g) 0 CO(g) NO2(g)
vessel volume is increased
(g) Fe3(aq) SCN(aq) 0 FeSCN2(aq)
Fe(NO3)3(s) is added
37. Predict in which of the following equilibria a decrease in
temperature favours the forward reaction.
(a) Br2(l) 0 Br2(g)
(b) N2(g) 3 H2(g) 0 2 NH3(g) rH is negative
(c) LiCl(s) 0 Li(aq) Cl(aq) heat
(d) 6 C(s) 3 H2(g) 0 C6H6(l)
rH 49 kJ
(e) CaCO3(s) energy 0 CaO(s) CO2(g)
Use this information to answer questions 38 and 39.
Alberta’s petroleum industry has a chronic problem—all fossil
fuels found in the province contain some sulfur. If not
removed, this sulfur will react upon burning to release SO2(g).
Sulfur dioxide is very irritating to lung tissue and is highly
corrosive; thus, it is a major contributor to air pollution, acid
rain, and respiratory disease. Furthermore, sulfur impurities
may damage the fuel injection and anti-pollution systems of
modern internal combustion engines if not removed from
gasoline and diesel fuels. For these reasons, recent Canadian
legislation requires that sulfur content in diesel fuels sold for
on-road use must be less than 15 ppm (15 mg/kg) by July,
2006.
In a refinery or bitumen upgrader, the sulfur is first removed
from fossil fuel feedstock by cracking and/or hydrogenation,
resulting in reaction of the sulfur to produce (extremely toxic)
H2S(g). Standard industry technology to remove hydrogen
sulfide gas from petrochemical gas stream mixtures involves
the use of an amine scrubber unit, in a two-step process that
depends on two kinds of equilibrium. For simplicity, assume
that an amine scrubber reaction vessel contains a 25%
aqueous solution of diethanolamine (C2H4OH)2NH(aq), which
approximates the actual solution used in the various oil sands
plants in Alberta.
38. The scrubbing operation involves a gas mixture of lowDE
molar-mass hydrocarbons, carbon dioxide, and hydrogen
sulfide. This mixture of gases is injected into the scrubber
unit at high pressure, with the scrubber solution at about
40 °C. For hydrogen sulfide, first the toxic gas dissolves
H2S(g)
0
H2S(aq)
(negative rH)
which is followed immediately by the chemical reaction
(C2H4OH)2NH(aq) + H2S(aq) 0
(C2H4OH)2NH2+(aq) + HS–(aq) (negative rH)
(a) Draw a structural formula for diethanolamine.
(b) Explain, using Le Châtelier’s principle, how the
relatively low temperature and high pressure act to
help the scrubber solution “absorb” toxic hydrogen
sulfide.
39. When the unabsorbed hydrocarbon gases are removed
from the scrubber unit, the sulfur atoms remain behind,
trapped in solution as hydrogen sulfide ions. In the next
process step, the scrubber solution is “regenerated”: the
absorbed toxic compound is now emitted from solution and
removed from the reaction vessel. Use Le Châtelier’s
principle to describe how conditions should be altered in
the scrubber vessel, to shift equilibrium positions to make
this process as efficient as possible.
708
Chapter 15
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Page 709
Chapter 15
Research and describe the role of temperature in the
operation of a halogen lamp. For example, how is it
possible for a halogen lamp to operate with the filament at
2700 °C when the tungsten normally would not last very
long at this high temperature? Why is such a high
temperature desirable?
40. When carbon dioxide gas “dissolves” in water, the process
is more correctly thought of as being an exothermic
chemical reaction with water, to form aqueous carbonic
acid. Soft drink beverages are “carbonated” in this way, at
high pressure. Write and balance an equilibrium equation
for this reaction.
41. Carbonic acid, H2CO3(aq), then reacts exothermically with
basic aqueous diethanolamine in essentially the same way
that hydrosulfuric acid, H2S(aq), does. Write and balance
the equilibrium equation for this reaction, and use it to
explain whether the process conditions for scrubbing
H2S(g) should work to remove CO2(g) as well.
Extension
42. Work cooperatively to research, assemble, and present a
DE
more complete and accurate summary of the operation of a
typical Alberta industrial amine “scrubber” system.
Your presentation should include
•
a description of the role of chemical equilibrium in the
system
•
information on applications of this process throughout
Alberta
•
the use of the best features of any available word
processing or slideshow software
www.science.nelson.com
www.science.nelson.com
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44. When the Olympic Games were held in Mexico in 1968,
DE
many athletes arrived early to train in the higher altitude
(2.3 km) and lower atmospheric pressure of Mexico City.
Exertion at high altitudes, for people who are not
acclimatized, may make them dizzy or “lightheaded” from
lack of oxygen. Explain this observation. Your explanation
should include
•
the theory of dynamic equilibrium
•
Le Châtelier’s principle
•
a description of how people who normally live at high
altitudes are physiologically adapted to their reducedpressure environment
www.science.nelson.com
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GO
43. A halogen light bulb contains a tungsten (wolfram)
filament, W(s), in a mixed atmosphere of a noble gas and a
halogen; for example, Ar(g) and I2(g) (Figure 3). The
operation of a halogen lamp depends, in part, on the
equilibrium system
W(s) I2(g)
0
WI2(g)
Figure 3
Halogen light bulbs are more efficient than ordinary
incandescent bulbs, but they burn so hot they may require
special fixtures, and must be treated with extra care.
NEL
Equilibrium Systems 709
Unit 8 - Ch 16 Chem30
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11:08 AM
Page 710
chapter
16
Equilibrium in
Acid–Base Systems
In this chapter
Exploration: Salty Acid or
Acidic Salt?
Lab Exercise 16.A: The
Chromate–Dichromate
Equilibrium
Web Activity: Edgar
Steacie
Investigation 16.1:
Creating an Acid–Base
Strength Table
Lab Exercise 16.B:
Predicting Acid–Base
Equilibria
Web Activity: Pool
Chemistry
Lab Exercise 16.C:
Aqueous Bicarbonate Ion
Acid–Base Reactions
Investigation 16.2: Testing
Brønsted–Lowry Reaction
Predictions
Lab Exercise 16.D:
Creating an Acid–Base
Table
Case Study: Changing
Ideas on Acids and
Bases—The Evolution of a
Scientific Theory
Web Activity: Titration of
Polyprotic Acids and
Bases
Biology Connection:
Homeostasis
The nature of science involves constant questioning and testing of theories—and so it
is with theories about acids and bases. A great many chemical reaction systems involve
acids in some way, including the one that begins the decomposition (digestion) of the
food you eat, and, thus provides you with energy. Many other types of aqueous reaction systems have rates that are easily controlled by adjusting the level of acidity. Because
such systems are commonly found both in nature and in industry, scientists seek to
work with the most complete and successful acid–base concepts—which in turn means
seeking new ways to test previously accepted theories. Theories that do not describe,
explain, and predict the chemistry of acids and bases well enough will initially be restricted
to only those situations where they work. As a result of further testing, scientists will
either revise the theory, or replace it altogether.
This chapter presents new hypotheses, evidence, and analyses to help you develop a
more comprehensive understanding of both the aqueous reaction environment, and
the activity of acids and bases within that environment. Your knowledge of chemical
equilibrium (Chapter 15) allows you to explore these questions from a new perspective, and, in turn, will allow you to form a more complete and less restrictive theory of
acids and bases. In fact, few topics in chemistry illustrate this scientific principle of
ongoing theory testing and development so well.
These underlying principles—that theories must be supported by evidence, and that
understanding is increased by always questioning and testing existing knowledge—are
the basis of the uniquely productive “way of knowing” about the natural world that we
call science. To these principles, and to the enormous accumulation of knowledge they
have made possible, we owe most aspects of our present technological civilization.
STARTING Points
Answer these questions as best you can with your current knowledge. Then, using
the concepts and skills you have learned, you will revise your answers at the end of
the chapter.
1. How can some substances neutralize both acids and bases?
2. Can acid–base reactions and their products be predicted? Explain.
3. Can pH curves for titrations of weak acids and weak bases predict equivalence points
as strong acid–strong base pH curves do? Explain.
4. How is the buffering of some medications related to stomach fluid acidity?
Web Activity: Preparation
of Buffer Solutions
Web Activity: Maud
Menten
Investigation 16.3: Testing
a Buffer Effect
710
Chapter 16
Career Connection:
Environmental Engineer; Chemistry Researcher; Microbiologist
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Page 711
Figure 1
Testing your concepts is a continual part of “doing” science.
Exploration
Salty Acid or Acidic Salt?
In Chapter 5, you learned about saturated solutions as examples •
of dynamic equilibrium, and prepared a saturated sodium
chloride solution. The strong acid, HCl(aq), shares half of its
•
chemical formula with NaCl(aq). How might the two solutions
interact?
Measure approximately 2 mL of concentrated hydrochloric
acid into the smaller graduated cylinder.
Materials: two 10 mL graduated cylinders; saturated sodium
chloride solution, NaCl(aq); concentrated hydrochloric acid,
HCl(aq)
Dispose of all substances down the sink, using lots of water.
(d) Explain what has apparently happened to the saturated
NaCl(aq) equilibrium.
(e) Explain which ion concentration in the large cylinder
must have changed, whether it was increased or
decreased, and what principle you use to “know” this
answer.
(f) Which initial solution was more concentrated? Use
Le Châtelier’s principle to explain how you “know” this
answer.
(g) Was the initial HCl(aq) a saturated solution? Was it at
equilibrium? Was it at equilibrium before being removed
from its closed storage container? Explain how you can
use the basic principles of equilibrium to “know” these
answers.
(a) Write and balance a chemical reaction equation for any
reaction you can predict when sodium chloride solution and
concentrated hydrochloric acid are mixed.
(b) Write an equation to express the nature of a saturated
sodium chloride aqueous solution as a dynamic equilibrium.
(c) Write the equilibrium law expression for a saturated
aqueous sodium chloride solution. Identify the substance
with a constant concentration.
Hydrochloric acid is corrosive. Wear appropriate
eye protection, lab gloves, and a lab apron.
•
NEL
•
Pour the concentrated hydrochloric acid into the cylinder
containing the saturated sodium chloride solution.
Record your observations.
Measure approximately 8 mL of saturated sodium chloride
solution into the larger graduated cylinder.
Equilibrium in Acid–Base Systems 711
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
16.1
_
CrO42 (aq)
_
Cr2O72 (aq)
Figure 1
The chromate—dichromate aqueous
ion system, in equilibria at high and
low pH
Page 712
Water Ionization and
Acid–Base Strength
In many of the preceding units, you have studied examples of chemical reactions and systems that in some way depend on the nature of acids and bases and/or the pH of solution. For example, aqueous permanganate ions are powerful oxidizing agents, but can act
to oxidize other reagents only in the presence of hydrogen (hydronium) ions. Almost any
soluble R–COOH organic compound will make an aqueous solution with a pH below
7, because the hydrogen atom of such a group is relatively easily removed. The electrolysis of potassium iodide solution produces a solution that is strongly basic. Understanding
these connections involves considering equilibrium effects, as shown by the chromate–dichromate aqueous ion system (Figure 1).
You have already learned several concepts about the strengths and properties of acids
and bases (see Chapter 6 Summary, page 262). If we now combine equilibrium concepts with these acid–base concepts, we can develop a much more comprehensive understanding of acids and bases. This understanding, in turn, will allow you to better explain
and predict how acids and bases behave. For instance, the reaction examined in Lab
Exercise 16.A is an easily explained example of how the acidity of a solution can directly
affect other ion equilibria.
LAB EXERCISE 16.A
The Chromate–Dichromate
Equilibrium
In an aqueous solution, chromate ions are in equilibrium with
dichromate ions (Figure 1).
2 CrO42(aq) 2 H(aq)
0
Cr2O72(aq) H2O(l)
Complete the Prediction and Design (including diagnostic tests)
of the investigation report.
Purpose
The scientific purpose of this investigation is to test a Design for
varying the acidity of an equilibrium.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation
Problem
How does changing the hydrogen ion concentration affect the
chromate–dichromate equilibrium?
Hypothesis
The position of this equilibrium depends on the acidity of the
solution.
For simplicity, a great many chemical reaction equations use H(aq) to represent an
aqueous hydrogen ion. This representation often works very well, as in the redox equations used in Unit 7, or the titration analysis equations used for stoichiometric calculations in Chapter 8. For other purposes, however, it is necessary to represent this ion more
accurately in order to understand and explain the theoretical nature of the reaction. As
you learned in Chapter 6, it is more consistent with evidence to think of this entity as a
hydronium ion, and to represent it as H3O(aq). Acid–base equilibrium theory necessarily
involves collision–reaction theory for a variety of entities. This chapter will, therefore, use
the hydronium ion convention almost exclusively. Before aqueous acidic or basic solution equilibrium can be investigated further, the equilibrium nature of the ions of the solvent (water) must first be examined, understood, and taken into consideration.
712
Chapter 16
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Page 713
Section 16.1
The Water Ionization Constant, Kw
Even highly purified water has a very slight conductivity that is only observable if measurements are made with very sensitive instruments (Figure 2). According to Arrhenius’
theory, conductivity is due to the presence of ions (Figure 3). Therefore, the conductivity observed in pure water must be the result of ions produced by the ionization of some
water molecules into hydronium ions and hydroxide ions. Because the conductivity is so
slight, the equilibrium at SATP must greatly favour the water molecules.
H
H
H
O
H
H
O
O
H
H
H
–
O
H
106 %
0
H2O(l) H2O(l)
O
O
H
H
+
H
H3O(aq)
OH(aq)
Figure 3
Collisions of water molecules very occasionally produce this cation–anion pair.
Figure 2
A sensitive multimeter is required to
detect the electrical conductivity of
the highly purified water that is
typically used in a chemistry
laboratory. (See Appendix C.3 for
guidance on using a multimeter.)
DID YOU KNOW
[H3O (aq)][OH (aq)]
or Kc [H3O(aq)] [OH(aq)]
Kc [H2O(l)][H2O(l)]
Evidence indicates that, at 25 °C, at any given moment, fewer than two of every billion
molecules in pure liquid water exist in ionized form! When we write an equilibrium
constant expression for this ionization, the value of Kc is an extremely small number.
Recall that, because pure liquid water (or the water in any dilute aqueous solution) has
an essentially constant concentration, it does not appear in the expression; it is simply
incorporated into the equilibrium constant value.
The water ionization equilibrium relationship is so important in chemistry that this
particular Kc constant is given its own special symbol and name. This new constant is called
the ion product or ionization constant for water, Kw.
Kw [H3O(aq)][OH(aq)] 1.00 1014 at SATP
The equilibrium equation for the ionization of water shows that hydronium ions and
hydroxide ions form in a 1:1 ratio. Therefore, the concentration of hydronium ions and
hydroxide ions in pure water must be equal. This equality must also be true for any neutral aqueous solution. Using the mathematical expression for Kw, and the value of Kw at
SATP, the concentrations of H3O(aq) and OH–(aq) can be calculated by taking the
square root of the Kw value. Recall (from Chapter 15) that in all entity concentration
calculations from any Kc value, the entity concentration is simply assumed to have units
of mol/L.
[H3O(aq)] [OH(aq)] 1.00 1014 1.00 107 mol/L
The ionization of water is especially important in the empirical and theoretical study
of acidic and basic solutions. Recall from Chapter 6 that, according to the modified
Arrhenius theory, an acid is a substance that reacts with water to produce hydronium ions.
The additional hydronium ions provided by the acid increase the hydronium ion
concentration. Since the hydronium ion concentration is greater than 10–7 mol/L, the solution is acidic. A basic solution is one in which the hydroxide ion concentration is greater
NEL
?
Successful Collisions
A collision that successfully forms
hydronium and hydroxide ions is
very rare. This is not because
collisions are rare—each water
molecule collides with others tens
of trillions of times every second!
But, the chance that any given
collision will both have sufficient
energy, and also be at exactly the
right orientation, is very, very small
indeed. Ordinarily, an equilibrium
so strongly favouring the reverse
reaction would just be ignored—
thought of as not happening at
all—but, in this case, the ions are
uniquely important because of the
effects they have on all reactions in
aqueous solution.
Conversely, you already know
that the reverse reaction
(hydronium ions with hydroxide
ions) is quantitative. Almost every
such ion collision will be effective
because the required energy is
very low, and the attraction of
opposite ion charges acts to orient
the entities correctly.
Modelling water molecules this
way—as space-filling models with
superimposed atomic symbols and
lines to represent bonds—gives an
overall representation of the
process that is logically consistent
with experimental evidence.
Equilibrium in Acid–Base Systems 713
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Learning Tip
Keep in mind that the value for
K w is subject to the same
restrictions as any other
equilibrium constant: one being
that it will change if the
temperature changes. For
example, in pure water at 20 °C,
K w has a numerical value of
6.76 1015, whereas at 30 °C
its value is 1.47 1014.
Kw will also change enough
to be invalid in any aqueous
solution with a very high solute
concentration—because then
the assumption that the
concentration of the water
solvent is at or near a constant
value (55.5 mol/L) no longer
holds true.
CAREER CONNECTION
Environmental Engineer
Monitoring and protecting water
quality is an essential part of the
work of environmental engineers.
They design systems and
treatment processes that control
pH levels, temperature, and
amounts of dissolved oxygen for
industries that use water.
Would you like to become an
environmental engineer? Look into
salaries, employment
opportunities, and educational
programs at at least two different
educational institutions.
www.science.nelson.com
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Page 714
than 107 mol/L. Basic solutions are produced in two ways: either by the complete dissociation (upon dissolving in water) of an ionic hydroxide, or by partial reaction of
some weak base entity (ion or molecule) with water to produce hydroxide ions.
The most important point about Kw is that it applies to pure water, and also to any solution that is mostly water. This means that this ionization equilibrium will be involved in
any other reaction going on in aqueous solution, if that reaction involves hydronium
ions or hydroxide ions in any way. As an example, consider the equilibrium reaction
just studied in Lab Exercise 16.A. Since that reaction involves hydronium (hydrogen)
ions, the chromate–dichromate equilibrium can be controlled (shifted) easily, simply
by adjusting either the hydronium ion or the hydroxide ion concentration; which really
just means deliberately shifting the water ionization equilibrium. A great many chemical reactions are dependent in this way on the water ionization equilibrium. Many of them
(those containing H3O(aq) or OH–(aq) ions) are evident in the Relative Strengths of
Oxidizing and Reducing Agents table (Appendix I).
Note that, since the mathematical relationship is simple, we can easily use Kw to calculate either the hydronium ion amount concentration or the hydroxide ion amount
concentration in an aqueous solution, if the other concentration is known.
Since [H3O(aq)][OH(aq)] Kw
Kw
then [H3O(aq)] [OH(aq)]
Kw
and [OH(aq)] [H3O(aq)]
In ordinary dilute aqueous acidic or basic solutions, the presence of substances other
than water decreases the certainty of the Kw value at 25 °C to two significant digits.
All questions and examples in this text assume temperatures of 25 °C, and aqueous solutions that are not highly concentrated, with a Kw value of 1.0 1014, unless specifically
stated otherwise.
COMMUNICATION example 1
A 0.15 mol/L solution of hydrochloric acid at 25 °C is found to have a hydronium ion
concentration of 0.15 mol/L. Calculate the amount concentration of the hydroxide ions.
Solution
HCl(aq) H2O(l)
Learning Tip
These communication examples
show the application of the
usual units convention for
simplifying calculations from
equilibrium constants. Units for
the constant are ignored, and
other concentration units are
always entered in mol/L. Then,
since the units for the
calculated value are always
mol/L, they are just written in
with the answer.
714
Chapter 16
0
H3O(aq) Cl(aq)
Kw
[OH(aq) [H3O(aq)]
1.0 1014
(entity concentration units assumed to be mol/L)
0.15 mol/L
6.7 1014 mol/L
Using the Kw relationship, the hydronium ion concentration is 6.7 1014 mol/L.
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Page 715
Section 16.1
COMMUNICATION example 2
Calculate the amount concentration of the hydronium ion in a 0.25 mol/L solution of
barium hydroxide.
Solution
Ba(OH)2(s) → Ba2(aq) 2 OH(aq)
[OH(aq)] 2 [Ba(OH)2(aq)]
2 0.25 mol/L
0.50 mol/L
Kw
[H3O(aq)] [OH(aq)]
1.0 1014
0.50 mol/L
2.0 1014 mol/L
Using the Kw relationship, the hydronium ion concentration is 2.0 1014 mol/L .
COMMUNICATION example 3
?
Determine the hydronium ion and hydroxide ion amount concentrations in 500 mL of an
aqueous solution for home soap-making containing 2.6 g of dissolved sodium hydroxide.
DID YOU KNOW
Solution
Many households, in previous
centuries, kept supplies of lye
(sodium hydroxide) on hand for
making soap. Because it looked
like sugar, it was occasionally
swallowed by curious children,
causing terrible injuries to their
throats. A prominent American
physician, Dr. Chevalier Jackson
(1865–1958) realized that warnings
on the packaging would
encourage parents to keep this
dangerous substance out of the
reach of their children. This is one
of the earliest instances of warning
labelling on packaging. Partially
because of Dr. Jackson's efforts,
the United States Congress passed
the Federal Caustic Labelling Act
in 1927. In Canada, we have the
Consumer Chemicals and
Containers Regulations.
1 mol
nNaOH 2.6 g 0.065 mol
40.00 g
0.065 mol
[NaOH(aq)] 0.13 mol/L
0.500 L
NaOH(s) → Na(aq) OH(aq)
[OH(aq)] [NaOH(aq)] 0.13 mol/L
Kw
[H3O(aq)] [OH(aq)]
1.0 1014
0.13 mol/L
7.7 1014 mol/L
Using the Kw relationship, the hydronium ion concentration is 7.7 1014 mol/L, and the
hydroxide ion concentration is 0.13 mol/L.
NEL
Warnings on Packaging
Equilibrium in Acid–Base Systems 715
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Page 716
COMMUNICATION example 4
Calculate the amount concentration of hydronium ions in a 0.100 mol/L aqueous solution
of ammonia that is used in a spray bottle for window cleaning solution (Figure 4). A
reference states that there is 2.1% reaction of dissolved ammonia with water (ionization) at
this concentration, at SATP.
Solution
NH3(aq) H2O(l)
0
NH4(aq) OH–(aq)
2.1
[OH(aq)] 0.100 mol/L
100
Figure 4
Many solutions sold for cleaning
windows contain ammonia.
0.0021 mol/L
Kw
[H3O(aq)] [OH(aq)]
1.0 1014
0.0021 mol/L
4.8 1012 mol/L
Using the Kw relationship, in 0.100 mol/L aqueous ammonia, the hydronium ion
concentration is 4.8 10–12 mol/L.
Practice
1. The hydronium ion concentration in an industrial effluent is 4.40 mmol/L. Determine
the concentration of hydroxide ions in the effluent.
2. The hydroxide ion concentration in a household cleaning solution is 0.299 mmol/L.
Calculate the hydronium ion concentration in the cleaning solution.
3. Calculate the hydroxide ion amount concentration in a solution prepared by
dissolving 0.37 g of hydrogen chloride in 250 mL of water.
4. Calculate the hydronium ion amount concentration in a saturated solution of calcium
hydroxide (limewater) that has a solubility of 6.9 mmol/L.
5. What is the hydronium ion amount concentration in a solution made by dissolving
20.0 g of potassium hydroxide in water to form 500 mL of solution?
6. Calculate the percent ionization of water at SATP. Recall that 1.000 L of water has a
mass of 1000 g.
Communicating Concentrations: pH and pOH
Recall (Chapter 6) that the enormous range of aqueous solution hydronium ion concentrations is more easily expressed using the logarithmic pH scale (Figure 5).
Mathematically,
pH log [H3O(aq)] and, inversely, [H3O(aq)] = 10pH
For basic solutions, it is sometimes more useful to use a scale based on the amount concentration of hydroxide ions. Recall that the definition of pOH follows the same format
and the same certainty rule as pH.
pOH log[OH(aq)] and, inversely, [OH(aq)] 10pOH
716
Chapter 16
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Page 717
Section 16.1
battery
acid
vinegar soft
drink
normal
rain
blood sea
water
antacid household lye
solution ammonia
pH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
pOH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
increasing acidity pH < 7
increasing basicity pH > 7
neutral (pure water)
Figure 5
The pH scale
The mathematics of logarithms allows us to express a simple relationship between pH
and pOH. According to the rules of logarithms,
log(ab) log(a) log(b)
Using the equilibrium law for the ionization of water,
[H3O(aq)][OH(aq)] Kw
log[H3O(aq)] log[OH(aq)] log(Kw)
(pH) (pOH) 14.00
pH pOH 14.00
(at SATP)
This relationship allows for a quick conversion between pH and pOH values.
SUMMARY
Water Ionization Conversions
and Values (at SATP)
Kw [H3O(aq)][OH(aq)] 1.0 1014
pH log[H3O(aq)]
[H3O(aq)] 10pH
pOH log[OH(aq)]
[OH(aq)] 10pOH
pH pOH 14.00
NEL
Equilibrium in Acid–Base Systems 717
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Learning Tip
Recall this convenient “rule of
thumb” for keeping track of
significant digits in
(logarithmic) calculations of pH
or pOH.
The number of digits following
the decimal point in the pH or
pOH (logarithm) value should be
equal to the number of significant digits shown in the amount
concentration of the ion.
For example, a hydronium
ion concentration of
2.7 103 mol/L is expressed
as a pH of 2.57, and a pOH of
4.3 is expressed as a hydroxide
ion concentration of
5 105 mol/L.
Page 718
Practice
7. Food scientists and dieticians measure the pH of foods when they devise recipes and
special diets.
(a) Copy and complete Table 1.
Table 1 Acidity of Foods
Food
oranges
[H3O(aq)]
(mol/L)
[OH(aq)]
(mol/L)
pH
5.5 10
asparagus
olives
pOH
3
5.6
2.0 10
11
blackberries
10.6
(b) Based on pH only, predict which of the foods would taste most sour.
8. To clean a clogged drain, 26 g of sodium hydroxide is added to water to make 150 mL
of solution. What are the pH and pOH values for the solution?
9. What mass of potassium hydroxide is contained in 500 mL of solution that has a pH
of 11.5? Comment on the degree of certainty of your answer.
Acid Strength as an Equilibrium Position
Learning Tip
For any aqueous solution of an
acid, percent reaction of that
acid with water is also
commonly called its percent
ionization in chemistry
references, because the net
effect is the same as if the acid
molecules simply ionize, as was
once (simplistically) assumed in
Arrhenius’ original theory.
You should consider “reaction
with water” and “ionization in
water” to be equivalent terms
for acid–base solution theory,
and may expect to see either
term used routinely in text and
in questions.
718
Chapter 16
In Chapter 6, you learned that acidic solutions of different substances at the same concentration do not possess acid properties to the same degree. The pH of a 1.00 mol/L solution of an acid can vary anywhere from a value of nearly 7 to a value of nearly 0, depending
on the specific acid in the solution. Other properties can also vary. For example, acetic
acid does not conduct an electric current nearly as well as hydrochloric acid of equal
concentration (Figure 6). When we observe chemical reactions of these acids, it is
apparent that acetic acid, although it reacts in the same manner and amount as
hydrochloric acid, does not react as quickly. The concepts of strong and weak acids were
developed to describe and explain these differences in properties of acids.
An acid is described as weak if its characteristic properties are less than those of a
common strong acid, such as hydrochloric acid. Weak acids are weaker electrolytes and
react at a slower rate than strong acids do; the pH of solutions of weak acids is closer to
7 than the pH of strong acids of equal concentration.
In Chapter 6, strong acids were explained as ionizing quantitatively by reacting with
water to form hydronium ions, whereas weak acids were explained as ionizing only partially (usually 50%). The empirical distinction between strong and weak acids can be
explained much more completely now by combining the modified Arrhenius theory
with equilibrium theory. A strong acid is explained as an acid that reacts quantitatively
with water to form hydronium ions. For example, the reaction of dissolved hydrogen
chloride (hydrochloric acid) with water is virtually complete. Even though the equation could be written with double equilibrium arrows and the extent shown with a
99.9% note, it is simpler, and much more common, to just use a single arrow to show
that the reaction is quantitative.
HCl(aq) H2O(l) → H3O(aq) Cl(aq)
A weak acid is an acid that reacts partially with water to form hydronium ions.
Measurements of pH indicate that most weak acids react less than 50%. For example, acetic
acid reacts only 1.3% in solution at 25 °C and 0.10 mol/L concentration.
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:08 AM
Page 719
Section 16.1
Recall, from Chapter 15, that any equilibrium position depends on concentration(s)
as well as on temperature; so, this 1.3% ionization value for acetic acid is only valid for
a 0.10 mol/L solution at 25 °C. Laboratory pH experiments show that the higher the
concentration of a weak acid solution, the lower its percent ionization becomes.
Figure 6
In solutions of equal concentration,
a weak acid such as acetic acid
conducts electricity to a lesser
extent than does a strong acid such
as hydrochloric acid.
1.3%
CH3COOH(aq) H2O(l)
0 H3O(aq) CH3COO(aq)
The hydronium ion concentration of any acid solution can be calculated by multiplying the percent reaction by the initial amount concentration of the acid solute. For
example, in HCl(aq) solution, virtually 100% of the HCl molecules react with water
molecules at equilibrium.
HCl(aq) H2O(l) → H3O(aq) Cl(aq)
100
[H3O(aq)] 0.10 mol/L
100
0.10 mol/L
There are six acids ordinarily classed as “strong”: hydrochloric, nitric, sulfuric, hydrobromic, hydroiodic, and perchloric acid solutions. Only the first three are common.
For any aqueous strong acid, we can simply assume that the concentration of hydronium ions in solution is equal to the initial concentration of the acid dissolved. For weak
acids (the majority of examples), we always need to calculate the hydronium ion concentration in solution, because only a small proportion of the initial acid concentration
will be converted to ions at equilibrium.
NEL
Equilibrium in Acid–Base Systems 719
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Page 720
COMMUNICATION example 5
In a 0.10 mol/L solution of acetic acid, only 1.3% of the CH3COOH molecules have reacted
at equilibrium to form hydronium ions. Calculate the hydronium ion amount
concentration.
Solution
1.3%
CH3COOH(aq) H2O(l)
0
H3O(aq) CH3COO(aq)
1.3
[H3O(aq)] 0.10 mol/L
100
1.3 103 mol/L
The hydronium ion concentration in 0.10 mol/L acetic acid is 1.3 103 mol/L.
As explained in Chapter 6, we can easily compare the strengths of different acids by
comparing the measured pH values for aqueous solutions of equal concentration. The
lower the pH, the higher the hydronium ion concentration, the greater the percent reaction, and thus the stronger the acid. Furthermore, we can find the percent reaction for
ionization of any weak acid solution from the measured pH of a solution of known initial concentration.
Learning Tip
Percent ionization:
[H3O(aq)]
p 100
[HA(aq)]
Rearrange the equation to solve
for hydronium ion
concentration:
p
[H3O(aq)] [HA(aq)]
100
where p percent ionization
and
[HA(aq)] initial concentration
of weak acid
COMMUNICATION example 6
The pH of a 0.10 mol/L methanoic acid solution is 2.38. Calculate the percent reaction for
ionization of methanoic acid.
Solution
[H3O(aq)] 10pH
102.38 mol/L
4.2 103 mol/L
p
[H3O(aq)] 100 [HCOOH(aq)]
[H3O(aq)]
p 100
[HCOOH(aq)]
4.2 103 mol/L
100
0.10 mol/L
4.2%
The percent ionization of 0.10 mol/L aqueous methanoic acid is 4.2%.
720
Chapter 16
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Page 721
Section 16.1
WEB Activity
Canadian Achievers—Edgar Steacie
Edgar Steacie (Figure 7) was an internationally acclaimed research scientist and a senior
administrator of the National Research Council.
1. What was Steacie’s main area of research?
2. Why was Steacie known as a statesman of science for Canada?
3. What is a Steacie Fellowship?
www.science.nelson.com
GO
Figure 7
Edgar Steacie (1900–1962)
Section 16.1 Questions
1. How does the hydronium ion concentration compare with
the hydroxide ion concentration if a solution is
(a) neutral?
(b) acidic?
(c) basic?
2. What two diagnostic tests can distinguish a weak acid from
a strong acid?
3. According to Arrhenius’ original theory, what do all bases
have in common?
4. Hydrocyanic acid is a very weak acid.
(a) Write an equilibrium reaction equation for the ionization
of 0.10 mol/L HCN(aq). The percent ionization at SATP is
7.8 103 %.
(b) Calculate the hydronium ion concentration and the pH
of a 0.10 mol/L solution of HCN(aq).
5. At 25 °C, the hydronium ion concentration in vinegar is
1.3 mmol/L. Calculate the hydroxide ion concentration.
6. At 25 °C, the hydroxide ion concentration in normal human
blood is 2.5 107 mol/L. Calculate the hydronium ion
concentration and the pH of blood.
7. Acid rain has a pH less than that of normal rain. The
presence of dissolved carbon dioxide, which forms carbonic
acid, gives normal rain a pH of 5.6. What is the hydronium
ion concentration in normal rain?
8. If the pH of a solution changes by 3 pH units as a result of
adding a weak acid, by how much does the hydronium ion
concentration change?
NEL
9. If 8.50 g of sodium hydroxide is dissolved to make 500 mL
of cleaning solution, determine the pOH of the solution.
10. What mass of hydrogen chloride gas is required to produce
250 mL of a hydrochloric acid solution with a pH of 1.57?
11. Determine the pH of a 0.10 mol/L hypochlorous acid
solution, which has 0.054% ionization at 25 °C.
12. Calculate the pH and pOH of a hydrochloric acid solution
prepared by dissolving 30.5 kg of hydrogen chloride gas to
make 806 L of solution. What assumption is made when
doing this calculation?
13. Acetic (ethanoic) acid is the most common weak acid used
in industry. Determine the pH and pOH of an acetic acid
solution prepared by dissolving 60.0 kg of pure, liquid
acetic acid to make 1.25 kL of solution. The percent reaction
with water at this concentration is 0.48%.
14. Determine the mass of sodium hydroxide that must be
dissolved to make 2.00 L of a solution with a pH of 10.35.
15. Write an experimental design for the identification of four
colourless solutions: a strong acid solution, a weak acid
solution, a neutral molecular solution, and a neutral ionic
solution. Write sentences, create a flow chart, or design a
table to describe the required diagnostic tests.
16. Sketch a flow chart or concept map that summarizes the
conversion of [H3O(aq)] to and from [OH(aq)], pH, and
percent reaction (ionization) of acid solute. Make your flow
chart large enough that you can write the procedure
between the quantity symbols in the diagram.
Equilibrium in Acid–Base Systems 721
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
16.2
Page 722
The Brønsted–Lowry Acid–Base
Concept
By now, our much-revised acid and base theory includes concepts of hydronium ions,
reaction with water, reaction equilibrium, and the ionization equilibrium of water. Our
modified theory is, thus, much more comprehensive, and much better at describing,
explaining, and predicting acid–base reactions, than the original theory proposed by
Arrhenius. We find, however, that there are still problems, and our theory must still be
considered too restrictive. There is no provision in it for reactions that do not occur in
aqueous solution. In addition, we find that there are some substances that seem to have
both acid and base properties.
As a common example, sodium hydrogen carbonate (sodium bicarbonate, baking
soda) forms a basic solution (raises the pH) in water, but the same compound will partly
neutralize (lower the pH of) a sodium hydroxide (lye) solution. When we observe that
NaHCO3(s) forms a basic aqueous solution, we conclude that this happens because
some hydrogen carbonate ions react with water molecules to produce hydroxide ions. We
know from many other observations that sodium ions have no acidic or basic properties. The following reaction equation seems to explain our observation easily:
(a)
HCO3(aq) H2O(l)
0 H2CO3(aq) OH(aq)
But, if adding sodium hydrogen carbonate makes a (strongly basic) sodium hydroxide
solution less basic, the concepts we are using lead us to conclude that something must
be reacting to decrease the concentration of the hydroxide ions. It must be the hydrogen
carbonate ions because there are no other entities in this system except sodium ions
and water molecules. We can easily write an equation to explain this observation, as
well:
HCO3(aq) OH(aq)
(b)
0 CO32(aq) H2O(l)
This equation seems to explain how the hydroxide ions can be partially consumed. But,
if it is correct, it begs the question, “Is sodium bicarbonate a base, or is it an acid?”
Clearly, our theory still needs some work! We need a broader, more comprehensive concept to successfully explain these kinds of observations.
The Proton Transfer Concept
Figure 1
Johannes Brønsted (1879–1947) (a)
and Thomas Lowry (1874–1936) (b)
independently created new
theoretical definitions for acids and
bases, based upon proton transfer
during a reaction.
New theories in science usually result from looking at the evidence in a way that has
not occurred to other observers. In 1923, two European scientists independently developed a new approach to acids and bases (Figure 1). These scientists focused on the role
of an acid and a base in a reaction rather than on the acidic or basic properties of their
aqueous solutions. An acid, such as hydrogen chloride, functions in a way opposite to a
base, such as ammonia. According to the Brønsted–Lowry concept, hydrogen chloride,
upon dissolving, donates a proton to a water molecule:
H
HCl(aq) H2O(l) → H3O(aq) Cl(aq)
acid
and ammonia, upon dissolving, accepts a proton from a water molecule.
H
NH3(aq) H2O(l)
base
722
Chapter 16
0 OH(aq) NH4(aq)
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Page 723
Section 16.2
Water does not have to be one of the reactants. For example, hydronium ions present in
a hydrochloric acid solution can react directly with dissolved ammonia molecules.
H
H3O(aq) NH3(aq) → H2O(l) NH4(aq)
acid
base
We can describe this reaction as NH3 molecules removing protons from H3O ions.
Hydronium ions act as the acid, and ammonia molecules act as the base. Water is present
as the solvent but not as a primary reactant. In fact, water does not even have to be
present, as evidenced by the reaction of hydrogen chloride and ammonia gases
(Figure 2).
H
HCl(g) NH3(g) → NH4Cl(s)
acid
base
According to the Brønsted–Lowry concept, a Brønsted–Lowry acid is a proton
donor and a Brønsted–Lowry base is a proton acceptor. A Brønsted–Lowry neutralization
is a competition for protons that results in a proton transfer from the strongest acid
present to the strongest base present. A Brønsted–Lowry reaction equation is an equation written to show an acid–base reaction involving the transfer of a proton from one
entity (an acid) to another (a base).
The Brønsted–Lowry concept does away with defining a substance as being an acid
or base. Only an entity that is involved in a proton transfer in a reaction can be defined
as an acid or base—and only for that particular reaction. This last point is extremely
important: protons may be gained in a reaction with one entity, but lost in a reaction with
another entity. For example, in the reaction of HCl with water shown on the previous
page, water acts as the base; whereas, in its reaction with NH3, water acts as the acid. A
substance that appears to act as a Brønsted–Lowry acid in some reactions and as a
Brønsted–Lowry base in other reactions is called amphoteric, or sometimes (incorrectly)
amphiprotic. The empirical term, amphoteric, properly refers to a chemical substance
with the ability to react as either an acid or a base. The theoretical term, amphiprotic,
describes an entity (ion or molecule) having the ability to either accept or donate a
proton. The hydrogen carbonate ion in baking soda (Figure 3), like every other hydrogen
polyatomic ion, is amphiprotic, as shown by the following reactions:
8
HCO3 (aq) H2O(l)
0 OH (aq) H2CO3(aq)
Kc 2.2 10
HCO3(aq) H2O(l)
0 H3O(aq) CO32(aq)
Kc 4.7 1011 mol/L
base
acid
acid
base
mol/L
Figure 2
Invisible fumes of ammonia gas and
hydrogen chloride gas mix and react
above these beakers, producing tiny
visible particles of solid ammonium
chloride that are suspended in the
air.
Figure 3
Baking soda is a common
household chemical, but it requires
an uncommonly sophisticated
theory to describe, explain, or
predict all of its properties.
When bicarbonate ions are in aqueous solution, some react with the water molecules by
acting as an acid, and some react by acting as a base. Kc values given for these two equilibriums show that one of them predominates—so much so that the other reaction is
simply inconsequential. The number of ions reacting as a base is over 2000 times more
than the number reacting as an acid. As you might expect, the resulting solution is basic.
We can get a clearer idea of the amphoteric nature of baking soda (caused by the
amphiprotic nature of the hydrogen carbonate ion) by noting the evidence expressed in
the next two equations. Note that bicarbonate ions partly neutralize a strong acid, but
they can also partly neutralize a strong base.
HCO3(aq) H3O(aq)
base
acid
HCO3(aq) OH(aq)
acid
NEL
base
0 H2CO3(aq) H2O(l)
0 CO32(aq) H2O(l)
(raises the pH of a strong
acid solution)
(lowers the pH of a
strong base solution)
Equilibrium in Acid–Base Systems 723
Unit 8 - Ch 16 Chem30
11/2/06
DID YOU KNOW
11:09 AM
?
Superacids
In aqueous solution, all strong acids
are equal in strength, since they all
react instantly and completely with
water to form the strongest possible
acid entity that can exist in water—
the hydronium ion. Scientists have
long suspected, however, that the
well-known strong acids, HCl and
H2SO4, may not be equal in strength
when no water is present and that
much stronger acids could exist.
Dr. Ronald Gillespie of McMaster
University has done extensive
research on non-aqueous strong
acids. His definition of a superacid—
one that is stronger than pure
sulfuric acid—is now generally
accepted by scientists. Perchloric
acid, the only common superacid,
easily loses protons to H2SO4(l)
molecules. Fluorosulfonic acid,
HSO3F(l), is more than one
thousand times stronger than
H2SO4(l). It is the strongest
Brønsted–Lowry acid known.
Page 724
Scientists consider the Brønsted–Lowry concept to be a theoretical definition. It falls
short of being a comprehensive theory because it does not explain why a proton is
donated or accepted, and cannot predict theoretically which reaction will occur for a
given entity in any given new situation. The advantage of the Brønsted–Lowry definitions
is that they enable us to define acids and bases in terms of chemical reactions rather than
simply as substances that form acidic and basic aqueous solutions. A definition of acids
and bases in terms of chemical reactions allows us to describe, explain, and predict a
great many more reactions, whether they take place in aqueous solution, in solution in
some other solvent, or between two undissolved entities in pure chemical states.
Practice
1. Theories in science develop over a period of time. Illustrate this development by
writing a theoretical definition of an acid, using the following concepts. Begin your
answer with, “According to [authority], acids are substances that….”
(a) Arrhenius’ original theory
(b) the modified Arrhenius theory
(c) the Brønsted–Lowry concept
2. How does the definition of a base according to the modified Arrhenius theory
compare with the Brønsted–Lowry definition?
3. Classify each reactant in the following equations as a Brønsted–Lowry acid or base.
(a)
(b)
(c)
(d)
HF(aq) SO32(aq) 0 F(aq) HSO3(aq)
CO32(aq) CH3COOH(aq) 0 CH3COO(aq) HCO3(aq)
H3PO4(aq) OCl(aq) 0 H2PO4(aq) HOCl(aq)
HCO3(aq) HSO4(aq) 0 SO42(aq) H2CO3(aq)
4. Evidence indicates that the hydrogen sulfite ion is amphiprotic. A sodium hydrogen
sulfite solution can partly neutralize either a sodium hydroxide spill or a hydrochloric
acid spill.
(a) Write a net ionic equation for the reaction of aqueous hydrogen sulfite ions with
the hydroxide ions in solution. Label the reactants as acids or bases.
(b) Write a net ionic equation for the reaction of hydrogen sulfite ions with the
hydronium ions from a hydrochloric acid solution. Label the reactants as acids or
bases.
5. What restrictions to acid–base reactions do the Brønsted–Lowry definitions remove?
6. Why is the Brønsted–Lowry concept labelled a theoretical definition rather than a
theory?
Conjugate Acids and Bases
Figure 4
White vinegar is a 5% acetic
(ethanoic) acid solution. The amount
concentration is 0.83 mol/L. Such a
solution is only about 0.43% ionized
at 25 °C, making acetic acid a
typical weak acid, by definition.
724
Chapter 16
According to the Brønsted–Lowry concept, acid–base reactions involve the transfer of a
proton. These reactions are universally reversible and always result in an acid–base equilibrium.
In a proton transfer reaction at equilibrium, both forward and reverse reactions involve
Brønsted-Lowry acids and bases. For example, in an acetic acid solution (Figure 4), the
forward reaction is explained as a proton transfer from acetic acid to water molecules,
and the reverse reaction is a proton transfer from hydronium to acetate ions.
H
CH3COOH(aq) H2O(l)
acid
base
base
acid
0 CH3COO(aq) H3O(aq)
H
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Page 725
Section 16.2
This equilibrium is typical of all acid–base reactions. There will always be two acids (in
the example, CH3COOH and H3O) and two bases (in the example, H2O and CH3COO)
in any acid–base reaction equilibrium. Furthermore, the base on the right (CH3COO)
is formed by removal of a proton from the acid on the left (CH3COOH). The acid on the
right (H3O) is formed by the addition of a proton to the base on the left (H2O). A pair
of substances with formulas that differ only by a proton is called a conjugate acid–base
pair. An acetic acid molecule and an acetate ion are a conjugate acid–base pair. Acetic acid
is the conjugate acid of the acetate ion, and the acetate ion is the conjugate base of acetic
acid. The equation shows that there must always be two conjugate acid–base pairs in
any proton transfer reaction. The hydronium ion and water are the second conjugate
acid–base pair in this equilibrium. Conjugate acid–base pairs appear opposite each other
in a table of acids and bases, such as that in Appendix I.
conjugate pair
CH3COOH(aq) H2O(l)
1.3%
0 CH3COO(aq) H3O(aq) (0.10 mol/L at 25 °C)
conjugate pair
H+
CH3COO–
H
O
H
Figure 5
This pictorial analogy is used to explain why acetic acid is a weak acid in aqueous solution. We
believe that the proton (H) of the carboxyl group is attracted more strongly by the rest of the
acetic acid molecule than it is by the water molecule. We infer this conclusion because, at
equilibrium, pH evidence indicates that very few of the acetic acid molecules have lost protons
in their collisions with water molecules.
At equilibrium, only 1.3% of the CH3COOH molecules have reacted with water in a
0.10 mol/L solution at SATP. It appears that the ability of the CH3COO part of the
acetic acid molecule to keep its proton (H) is much greater than the ability of H2O to
attract the proton away (Figure 5). This means that CH3COO is a stronger base (it has
a greater attraction for protons) than H2O.
When HCl molecules react with water (Figure 6), the Cl of each HCl molecule has a
much weaker attraction for its proton (H) than any colliding water molecule has. The
water molecules “win” this “competition” for protons overwhelmingly. At equilibrium, essentially all of the HCl molecules have lost protons to water molecules. In this case, the
transfer of protons is quantitative, and, because its molecules have an extremely weak
attraction for their protons, HCl(aq) is called (somewhat confusingly) a strong acid.
Remember: The stronger the base, the more it attracts another proton. The stronger
an acid, the less it attracts its own proton.
+
H
NEL
O
+
Cl
H
O
H
H
H
H
–
+
Cl
Figure 6
When gaseous hydrogen chloride
dissolves in water, the HCl
molecules are thought to collide and
react quantitatively with water
molecules to form hydronium and
chloride ions.
Equilibrium in Acid–Base Systems 725
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Learning Tip
Acids were studied much
earlier than bases were, and
sometimes the old terminology
used to describe acid–base
situations can be misleading—
because it always focuses on
the nature of the acid. Strong
acids are very reactive, but we
believe it is because they are
very weak proton attractors. It is
common to speak of acids
“donating” protons, and of
bases “accepting” them, but
this terminology gives an
unrealistic view of what we
believe is really happening. It is
like saying that a bank robber
“accepts” money “donated” by a
bank teller. The tug-of-war
analogy in Figure 5 is a more
realistic way to think of proton
transfer reactions—as a
competition between two
bases, both attracting the same
proton.
Page 726
The terms strong acid and weak acid can be explained in terms of the Brønsted–Lowry
concept, and also by comparing the reactions of different acids with the same base—
for example, water. Using HA as the general symbol for any acid and A as its conjugate
base, the empirically derived Relative Strengths of Aqueous Acids and Bases table
(Appendix I) lists the position of equilibrium of aqueous solutions of many different acids.
They are ordered according to how much they ionize in (react with) the water solvent.
HA(aq) H2O(1)
0 H3O(aq) A(aq)
The extent of the proton transfer between HA and H2O determines the relative strength
of HA(aq). In Brønsted–Lowry terms, when a strong acid reacts with water, an almost
complete transfer of protons results for the forward reaction and almost no transfer of
protons occurs for the reverse reaction; a nearly 100% reaction with water. The equilibrium
constant value is found to be very large. Theoretically, a strong acid holds its proton
very weakly, and easily loses the proton to any base, even very weak bases such as water.
This result leads to the interpretation that the conjugate base, A, of a strong acid must
have a very weak (negligible) attraction for protons. A useful generalization regarding the
relative strengths of an acid–base conjugate pair is: the stronger an acid, the weaker its conjugate base; and conversely, the weaker an acid, the stronger its conjugate base. (See the
Relative Strengths of Aqueous Acids and Bases table in Appendix I.)
Chemists have no simple explanation, in terms of forces or bonds, for the differing abilities of acids to donate protons or of bases to accept them. The inability to predict acid
and base strengths for an entity not already included in an empirically determined table
of acids and bases is a major deficiency of all acid–base theories.
SUMMARY
•
•
•
•
•
•
•
Brønsted–Lowry Definitions
An acid is a proton donor and a base is a proton acceptor, in a specific reaction.
An acid–base reaction involves a single proton transfer from one entity (the acid) to
another (the base).
An amphiprotic entity (amphoteric substance) is one that acts as a Brønsted–Lowry
acid in some reactions and as a Brønsted–Lowry base in other reactions.
A conjugate acid–base pair consists of two entities with formulas that differ only by
a proton.
A strong acid has a very weak attraction for protons. A strong base has a very strong
attraction for protons.
The stronger an acid, the more weakly it holds its proton. The stronger a base, the
more it attracts another proton.
The stronger an acid, the weaker is its conjugate base. The stronger a base, the weaker
is its conjugate acid.
Practice
7. Use the Brønsted–Lowry definitions to identify the two conjugate acid–base pairs in
each of the following acid–base reactions.
(a) HCO3(aq) S2(aq) 0 HS(aq) CO32(aq)
(b) H2CO3(aq) OH(aq) 0 HCO3(aq) H2O(l)
(c) HSO4(aq) HPO42(aq) 0 H2PO4(aq) SO42(aq)
(d) H2O(l) H2O(l) 0 H3O(aq) OH(aq)
8. Some ions can form more than one conjugate acid–base pair. List the two conjugate
acid–base pairs involving a hydrogen carbonate ion in the reactions in question 7.
726
Chapter 16
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Unit 8 - Ch 16 Chem30
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11:09 AM
Page 727
Section 16.2
INVESTIGATION 16.1 Introduction
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Creating an Acid–Base Strength Table
An acid–base table organizes common acids (and their conjugate
bases) in order of decreasing acid strength. Acid strength can be
tested several ways, including by a carefully designed use of
indicators. Predict the order of strengths using the Relative
Strengths of Aqueous Acids and Bases table (Appendix I). Use
the indicators provided to create a valid and efficient Design, in
which you clearly identify the relevant variables. Evaluate the
Design (only), and suggest improvements if any problems are
identified.
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1)
Purpose
The purpose of this investigation is to test an experimental design
for using indicators to create a table of relative strengths of acids
and bases.
Problem
Can the indicators available be used to rank the acids and bases
provided in order of strength?
To perform this investigation, turn to page 768.
Predicting Acid–Base Reaction Equilibria
The Brønsted–Lowry concept unfortunately does not include any theoretical explanation about why any given entity attracts a proton more or less strongly. To predict the outcome of any acid–base combination, we must rely on empirical evidence, gained by
measuring and recording the relative strengths of acid and base entities. Predictions
must be restricted to only those acid–base combinations for which we already have data.
To help us, we can now look for a simple generalization that might allow us to predict
the approximate position of equilibrium in an acid–base proton transfer.
LAB EXERCISE 16.B
Predicting Acid–Base Equilibria
Is it possible to predict how far a reaction will proceed? Use the
table of Relative Strengths of Aqueous Acids and Bases in
Appendix I and the evidence of position of equilibrium to
complete the Analysis of the investigation report.
Purpose
The purpose of this investigation is to develop a generalization for
predicting the position of acid–base equilibria.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
2. HCl(aq) H2O(l)
99%
0
3. CH3COO(aq) H2O(l)
Problem
How do the positions of the reactant acid and base in the
acid–base table relate to the position of equilibrium?
50%
0
1. CH3COOH(aq) H2O(l)
4. H3PO4(aq) NH3(aq)
Analysis
Evaluation
H3O(aq) CH3COO(aq)
H3O(aq) Cl(aq)
50%
0
50%
0
H2PO4(aq) NH4(aq)
50%
5. HCO3(aq) SO32(aq)
CH3COOH(aq) OH(aq)
0
HSO3(aq) CO32(aq)
6. H3O(aq) OH(aq) → H2O(l) H2O(l)
WEB Activity
Web Quest—Pool Chemistry
A swimming pool can be an enjoyable place to spend a hot summer afternoon. Most of us are
familiar with the dangers of pools. This Web Quest introduces an unfamiliar risk: pool gas. What
is pool gas and how does it form? What are the dangers of pool gas, and how can they be
reduced?
www.science.nelson.com
NEL
GO
Equilibrium in Acid–Base Systems 727
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
CAREER CONNECTION
Chemistry Researcher
Farideh Jalilehvand (Figure 7) is
an associate professor of
chemistry at the University of
Calgary. Her research involves a
lot of X-ray absorption
spectroscopy. Find out how acidity
due to sulfur accumulation in
water-logged wood is connected
to ancient shipwrecks by visiting
her information site, and check out
her curriculum vitae (CV). Science
is international in scope!
Figure 7
Farideh Jalilehvand
www.science.nelson.com
GO
Page 728
Predicting Acid–Base Reactions
When making complex predictions, scientists often combine a variety of empirical and
theoretical concepts. We need to use a combination of concepts to predict both the
products and the extent of acid–base reactions. According to the collision–reaction
theory, a proton transfer may result from a collision between an acid and a base. In a system
that is a mixture of several different acid and/or base entities, there are countless random
collisions of all the entities present. So, in theory, in any such system, there are many
different possible acid–base reactions, and all of these reactions occur (to some extent)
all the time. Evidence indicates that one reaction predominates: it occurs to an extent so
much greater than the other reactions that it is the only observable reaction. We will be
able to explain which reaction will predominate if we use a combination of the collision–reaction theory and the Brønsted–Lowry concept.
According to collision–reaction theory, in an acid–base system, collisions of all entities present are constantly occurring. According to the Brønsted–Lowry concept, a proton
will only transfer if an acid entity collides with a base entity that is a better proton
attractor than itself. A proton could theoretically transfer several times (if there were
several different acids and bases in the system), transferring each time to a stronger
proton attractor. Once gained by an entity of the strongest base present, a proton will
remain there, because there is no other base present in the system that can attract that
proton strongly enough to remove it. By the same logic, once any entity of the strongest
acid present has lost its proton, its (remaining) conjugate base cannot gain one back
from any other entity present, because it is a weaker proton attractor than anything else
in the system. Overall, theory suggests that the predominant acid–base reaction should
be the one that involves proton transfer from the strongest acid to the strongest base
present in the system. Experimental evidence indicates that this explanation is correct.
Other possible proton transfer reactions occur to such a small extent that they have a negligible effect on the reactants and products, so they are normally ignored.
We theorize that the only significant reaction in any acid–base system involves proton
transfer from the strongest acid present to the strongest base present. For an aqueous solution system, we first must list all the entities present as they exist in aqueous solution.
Table 1 summarizes how to represent the entities that are present.
Table 1 Predominant Entities Present in Aqueous Solution
Substance dissolved
(example)
Kinds of entities
in solution
Predominant entities
present (example)
Ionic compounds
Ca(HCO3)2(aq)
cations, anions
Ca2(aq) and HCO3(aq)
Ionic oxides
Na2O(aq), CaO(aq)
cations, hydroxide ions
Na(aq), OH(aq) or
Ca2(aq), OH(aq)
Strong acids
HCl(aq) or HNO3(aq)
H3O(aq), conjugate base
H3O(aq), Cl(aq) or
H3O(aq), NO3(aq)
Weak acids
HF(aq) or HSO3(aq)
molecules or ions
HF(aq) or HSO3(aq)
Weak bases
NH3(aq) or HCO3(aq)
molecules or ions
NH3(aq) or HCO3(aq)
The process for determining the nature and extent of the predominant proton transfer
reaction in an aqueous acid–base system can be thought of as five distinct steps, for convenience, as shown in the following Sample Problem.
728
Chapter 16
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Page 729
Section 16.2
SAMPLE problem 16.1
What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution
is neutralized with vinegar (Figure 8)? Are reactants or products favoured in this reaction?
Step 1: List all entities present as they exist in aqueous solution. Refer to Table 1 if
necessary. The entity list for this situation is
Na(aq)
OH(aq)
CH3COOH(aq)
H2O(l)
Step 2: Use the entity lists of the Relative Strengths of Aqueous Acids and Bases table to
identify and label each entity present as a Brønsted–Lowry acid or base. Amphiprotic
entities are labelled for both possibilities. Conjugate bases of strong acids are not included
or labelled as bases because they cannot act as bases in aqueous solution. Metal ions are
treated as spectators.
Na(aq)
OH(aq)
B
A
CH3COOH(aq)
A
H2O(l)
B
Step 3: Use the order of the entities in the Relative Strengths of Aqueous Acids and
Bases table to identify and label the strongest Brønsted–Lowry acid (the highest one on
the table) and the strongest Brønsted–Lowry base (the lowest one on the table) that are
present in the solution.
Na(aq)
OH(aq)
SB
SA
CH3COOH(aq)
A
H2O(l)
B
Figure 8
The drain cleaner shown is a
concentrated solution of a very
strong base. A spill would be highly
corrosive and quite hazardous.
Excess weak acid, such as the
vinegar (5% acetic acid), is the best
choice to “neutralize” a spilled
strong base. The final solution will
not actually be neutral, but it will be
only mildly acidic, and much safer to
handle.
Step 4: Write a balanced equation to show a proton transfer from the strongest acid to
the strongest base, assuming that their respective conjugates are the reaction products.
H
CH3COOH(aq) OH(aq)
0
CH3COO(aq) H2O(l)
Step 5: Predict the position of equilibrium using the generalization developed in Lab
Exercise 16.B and illustrated in the margin Learning Tip. For this reaction, the strongest
acid is positioned higher on the table than the strongest base, so products are favoured.
Under certain assumed conditions (see the following discussion), the equilibrium percent
reaction may be labelled as greater than 50%.
50%
CH3COOH(aq) OH(aq) 0 CH3COO(aq) H2O(l)
The nature of water complicates the prediction of outcomes of acid–base reactions in
aqueous solution. We need to consider the specific restrictions that apply when predicting proton transfer reactions in aqueous solution, because they are much more
common than acid–base reactions in any other environment.
• Hydronium ion is the strongest acid entity that can exist in aqueous solution. If a
stronger acid than hydronium ion is dissolved in water, it reacts instantly and
completely with water molecules to form hydronium ions. For this reason, the six
strong acids are all written as H3O(aq) when in aqueous solution.
• Hydroxide ion is the strongest base entity that can exist in aqueous solution. If a
stronger base than hydroxide ion is dissolved in water, it reacts instantly and
completely to form hydroxide ions. The only common example is the dissolving of
soluble ionic oxide compounds such as Na2O(s). In such cases, the oxide ion is
written as OH(aq) when in aqueous solution.
NEL
Learning Tip
The relative positions of the
strongest acid and the strongest
base on an acid–base table can
be used to approximately
determine the position of an
acid–base equilibrium.
Products Favoured
SA
50%
0
SB
Reactants Favoured
SB
50%
0
SA
Equilibrium in Acid–Base Systems 729
Unit 8 - Ch 16 Chem30
11/2/06
11:09 AM
Learning Tip
Since hydroxide ion is the
strongest possible base in
aqueous solution, reaction of
any acid with this ion in
aqueous solution is
automatically one that favours
products. Similarly, the reaction
of any base with the strongest
possible acid, hydronium ion,
must favour products. Of
course, the reaction of
hydroxide ion with hydronium
ion is always quantitative.
Page 730
• No entity in aqueous solution can react as a base if it is a weaker base than water.
For this reason, the conjugate bases of the six strong acids are not considered as
bases, in aqueous solution.
Even though you find that products are favoured (as in Sample Problem 16.1), for
the predominant reaction, you cannot accurately predict the actual equilibrium position
of all cases of any specific acid–base reaction by this method. It simply means you know
that the forward reaction happens more readily than the reverse reaction, all things
being equal. But, recall (from Chapter 15) that the initial concentration of entities plays
a very large part in determining the percent reaction at equilibrium. To be able to
predict something meaningful about the position of an acid–base reaction equilibrium,
we must restrict the reaction conditions. Unless you are specifically informed otherwise
in a question, assume in this text that, for the predominant reaction in an acid–base
system,
• the equation represents a single proton transfer between two entities, neither of
which is water, and has a stoichiometric ratio of 1:1:1:1
• the strongest acid and the strongest base present in the reaction system are both
present in significant chemical amounts, in approximately equal concentrations
Under these circumstances (and only then), you may assume that a Brønsted–Lowry
acid–base reaction where products are favoured has a percent reaction greater than 50%,
or that a reaction where reactants are favoured may be labelled as “ 50%.” It is also
correct (given these restrictions) to state that the Kc value for any such reaction is 1
if products are favoured, and 1 if reactants are favoured. It is not possible to be more
precise about the equilibrium position, however, at our current level of theory.
Note that the above restrictions exclude any reaction equation written to describe the
ionization (reaction with water) of any single acid or base entity dissolved to make an
aqueous solution. In these cases, water is acting as one of the reactants, and the amount
concentration of the water is very much greater than that of the other entity.
Finally, recall that the reaction of hydronium ions with hydroxide ions is always quantitative. This equation should always be written with a single arrow. There are many
other acid–base reactions that are quantitative, particularly those where either hydronium
or hydroxide ions are involved, but it is not possible to easily predict whether any acid–base
reaction will be quantitative from the table of Relative Strengths of Aqueous Acids and
Bases. In this text, other quantitative reactions will always be identified for you, as in the
following Communication Example.
COMMUNICATION example
Ammonium nitrate fertilizer is produced by the quantitative reaction of aqueous ammonia
with nitric acid. Write a balanced acid–base equilibrium equation.
Solution
A
SA
NH3(aq), H2O(l), H3O(aq)
SB
B
NH3(aq) H3O(aq) → NH4(aq) H2O(l)
730
Chapter 16
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11:09 AM
Page 731
Section 16.2
SUMMARY
A Five-Step Method for Predicting the
Predominant Acid–Base Reaction
1. List all entities (ions, atoms, or molecules including H2O(l)) initially present as they
exist in aqueous solution. (Refer to Table 1, page 728.)
2. Identify and label all possible aqueous acids and bases, using the Brønsted–Lowry
definitions.
3. Identify the strongest acid and the strongest base present, using the table of Relative
Strengths of Aqueous Acids and Bases (Appendix I).
4. Write an equation showing a transfer of one proton from the strongest acid to the
strongest base, and predict the conjugate base and the conjugate acid to be the
products.
5. Predict the approximate position of the equilibrium, using the generalization developed in Lab Exercise 16.B on page 727 and the table of Relative Strengths of Aqueous
Acids and Bases (Appendix I).
Practice
Use the five-step method to predict the predominant reactions in the following chemical
systems:
9. Hydrofluoric acid and an aqueous solution of sodium sulfate are mixed to test the
five-step method of predicting acid–base reactions.
10. Strong acids, such as perchloric acid, have been shown to react quantitatively with
strong bases, such as sodium hydroxide.
11. Methanoic acid is added to an aqueous solution of sodium hydrogen sulfide.
12. A student mixes solutions of ammonium chloride and sodium nitrite in a chemistry
laboratory.
13. Empirical work has shown that nitric acid reacts quantitatively with a sodium acetate
solution.
14. A consumer attempts to neutralize an aqueous sodium hydrogen sulfate cleaner with
a solution of lye. (See Appendix J if you do not remember what lye is.)
15. Can ammonium nitrate fertilizer, added to water, be used to neutralize a muriatic acid
(hydrochloric acid) spill?
16. Predict the acid–base reaction of bleach with vinegar (Figure 9).
17. Commercial laundry bleach (Figure 9) is made by reacting chlorine gas with a sodium
hydroxide solution, according to the equation
Cl2(g) 2 OH(aq)
0
OCl(aq) Cl(aq) H2O(l)
As sold, the pH of laundry bleach solutions is always well above 8. You know that the
element chlorine has very low solubility in pure water. Explain, using Le Châtelier’s
principle, why bleach bottle labels warn so strongly against mixing bleach with other
cleaning agents, such as acidic toilet bowl cleaners.
Note: Household bleach also produces toxic chloramines if mixed with basic
ammonia cleaning solutions. The best rule is, NEVER mix bleach directly with any
other cleaning powder or solution. It is arguably the most dangerous common
household chemical.
NEL
Figure 9
Bottles of household bleach display
a warning against mixing the bleach
(aqueous sodium hypochlorite) with
acids. Does your prediction of the
reaction between vinegar and
hypochlorite ions provide any clues
about the reason for the warning?
Equilibrium in Acid–Base Systems 731
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11:09 AM
Page 732
LAB EXERCISE 16.C
Report Checklist
Aqueous Bicarbonate Ion Acid–Base
Reactions
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
An acid–base table organizes common acids (and their conjugate
bases) in a way that enables us to predict predominant acid–base
reactions. Evaluate the reliability of the five-step method (only)
using information from this investigation, by comparing your
analysis of the evidence with your predictions.
Purpose
The purpose of this investigation is to test the five-step method
for predicting reactions in acid–base systems.
Problem
What are the products and position of the equilibrium for sodium
hydrogen carbonate (Figure 10) with stomach acid, vinegar,
household ammonia, and lye, respectively?
Evidence
Table 2 The Addition of Baking Soda to Various Solutions
Reactant
Bubbles
Odour
pH
HCl(aq)
yes
none
increases
CH3COOH(aq)
yes
disappears
increases
NH3(aq)
no
remains
decreases
NaOH(aq)
no
none
decreases
INVESTIGATION 16.2 Introduction
Testing Brønsted–Lowry Reaction
Predictions
When predicting products for this investigation, list all entities
present as they normally exist in an aqueous environment. For
those reactants that are added in solid state, assume that they
will dissolve. Use the resulting entities for prediction. Evaluate the
predictions, the Brønsted–Lowry concept, and the five-step
method for acid–base reaction prediction.
Purpose
The purpose of this investigation is to test the Brønsted–Lowry
concept and the five-step method for reaction prediction from a
table of relative acid–base strength.
Figure 10
The versatility of baking soda is demonstrated by its use in
extinguishing fires, in baking biscuits, and in neutralizing
excess stomach acid. It is also used as a medium for local
anesthetics—baking soda reduces stinging sensations by
neutralizing the acidity of the anesthetic, with the result that
the speed and efficiency of the anesthetic are improved. The
broad range of uses for baking soda results, in part, from its
amphiprotic character.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Problem
What reactions occur when various pairs of substances are
mixed?
Design
A prediction is made for each of eleven pairs of substances. The
prediction is then tested using one or more diagnostic tests,
complete with controls. Additional diagnostic tests increase the
certainty of the evaluation.
To perform this investigation, turn to page 768.
732
Chapter 16
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Page 733
Section 16.2
LAB EXERCISE 16.D
Creating an Acid–Base Table
Complete the Analysis of the investigation report, including a
short table of the four acids and bases involved. Use entity
position generalizations in the acid–base table for reactions that
favour products or reactants. The evidence for reaction 1 is
interpreted as:
Acids
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation
Problem
What is the order of acid strength for the first four members of the
carboxylic acid family?
Bases
Evidence
CH3COOH(aq)
50%
CH3COOH(aq) C3H7COO(aq)
>
%
50
HCOOH(aq) CH3COO(aq)
C3H7COO–(aq)
Purpose
The purpose of this investigation is to test an experimental design
for using equilibrium position to create a table of relative
strengths of acids and bases.
0
CH3COO(aq) C3H7COOH(aq)
50%
0
HCOO(aq) CH3COOH(aq)
50%
C2H5COOH(aq) C3H7COO(aq) 0
C2H5COO(aq) C3H7COOH(aq)
C2H5COOH(aq) HCOO(aq)
50%
0
C2H5COO(aq) HCOOH(aq)
Case Study
Changing Ideas on Acids and
Bases—The Evolution of a Scientific
Theory
Historically, chemists have known the empirical properties of
substances long before any theory was developed to explain
and predict those properties. For example, chemists were
familiar with the distinguishing properties of several acids and
bases, and used them routinely, by the middle of the 17th
century. Early attempts at an acid–base theory tended to focus
on acids and to ignore bases. Over time, several theories
passed through cycles of formulation, testing, acceptance,
further testing, and eventual rejection. Following is a brief
historical summary of acid–base theories and the evidence
that led to their revision.
• Antoine Lavoisier (1743–1794) (Figure 11) believed that
the properties of acids could be traced back to a single
substance. Lavoisier studied the combustion of phosphorus
and sulfur and determined that these elements combine
with something in the atmosphere to produce compounds
that form acidic solutions when dissolved in water. When
Joseph Priestley identified the component of the
atmosphere that actively supports combustion, Lavoisier
(1777) named the gas oxygen, meaning “acid maker.”
Lavoisier assumed that oxygen was the substance
responsible for the generic properties of acids. There were
soon problems with this theory when it was found that
NEL
several acids, such as muriatic acid (HCl), do not contain
oxygen. Furthermore, many substances that form basic
solutions (such as lime, CaO) were found to contain oxygen.
This evidence led to the rejection of the oxygen theory, but it
is historically important because it is the first systematic
attempt to chemically characterize acids and bases. The
generalization that nonmetallic oxides form acidic solutions
is still useful in chemistry.
• Sir Humphry Davy (1778–1829) (Figure 12) conducted the
experiments that demonstrated the absence of oxygen in
muriatic acid (HCl), which led to the rejection of Lavoisier’s
theory. Davy (1810) advanced his own theory that the
Figure 11
Antoine Lavoisier
Figure 12
Humphry Davy
Equilibrium in Acid–Base Systems 733
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11:09 AM
Page 734
presence of hydrogen gives compounds acidic properties.
This theory, however, did not explain why many compounds
containing hydrogen have neutral properties (for example,
CH4) or basic properties (for example, NH3). Justus von
Liebig (1803–1873) revised Davy’s idea to define acids as
substances in which the hydrogen could be replaced by a
metal. This revision meant that acids could be thought of as
ionic compounds in which hydrogen had replaced the metal
ion. Liebig had no corresponding theoretical definition for
bases, which were still identified empirically as substances
that neutralized acids.
• Svante Arrhenius
(1859–1927) (Figure 13)
developed a theory in 1887
that provided the first useful
theoretical definitions of acids
and bases. He defined acids
as substances that ionize in
aqueous solution to form
hydrogen ions, H(aq).
Similarly, he defined bases as
substances that dissociate to
form hydroxide ions, OH(aq),
in solution. This theory
Figure 13
explained the process of
Svante Arrhenius
neutralization as the
combination of hydrogen ions
and hydroxide ions to form water
H(aq) OH(aq) → H2O(l)
Figure 15
Thomas Lowry
role of an acid and a base in a reaction rather than on the
properties of their aqueous solutions. According to the
Brønsted–Lowry concept, an acid is a proton (H ) donor and
a base is a proton acceptor. The solvent has a central role in
the Brønsted–Lowry concept. Water can be considered an
acid or a base since it can lose a proton to form a hydroxide
ion (OH) or accept a proton to form a hydronium ion
(H3O). In their view, neutralization is a competition for
protons that results in a proton transfer from the strongest
acid present to the strongest base present. For example, in
the reaction
HSO4(aq) NH3(aq) → SO42(aq) NH4(aq)
Arrhenius explained the strengths of acids in terms of
degrees of ionization. While these ideas were a major
development in chemists’ understanding of the properties of
acids and bases, there were some problems with Arrhenius’
theory.
the hydrogen sulfate ion acts as the acid (because it
donates a proton) and ammonia acts as the base (because
it accepts a proton). A weakness of the Brønsted–Lowry
concept is its limitation to solutions (gaseous or liquid).
• In Arrhenius’ theory, an acid is expected to be an acid in
any solvent, which was found not to be the case. For
example, when dissolved in water, HCl supposedly breaks
up into hydrogen ions and chloride ions, but when it is
dissolved in benzene, the HCl remains as intact
molecules. The nature of the solvent, therefore, had to
play a critical role in acid–base properties of substances.
• The need for hydroxide to always be the base led
Arrhenius to propose formulas such as NH4OH(aq) for the
formula for ammonia in water, which led to the
misconception that NH4OH(aq) was the base, not
NH3(aq).
• According to Arrhenius’ theory, all salts (ionic solids)
should produce neutral solutions, but many do not. For
example, solutions of ammonium chloride are acidic and
solutions of sodium acetate are basic.
• Bonding theory suggested that it was very unlikely that a
single proton could exist in aqueous solution without
being bonded to at least one water molecule.
• Gilbert Lewis (1875–1946)
(Figure 16) developed an
acid–base theory in 1923 that
includes all previous theories
and definitions of acids and
bases. He viewed acids and
bases in terms of the covalent
bond, a theory he had
developed in 1916. Lewis
defined acids as electron–pair
acceptors and bases as
electron–pair donors. It is
important to note that in
Figure 16
Lewis acid–base theory, no
Gilbert Lewis
hydrogen ion and no solvent
need be involved. The Lewis
definition is broader than all previous definitions and
explains more inorganic and organic reactions (Figure 17).
For example, in the reaction
• Johannes Brønsted (1879–1947) (Figure 14) and
Thomas Lowry (1874–1936) (Figure 15) independently
published (1923) essentially the same concept about how
acids and bases behave. These scientists focused on the
734
Figure 14
Johannes Brønsted
Chapter 16
BF3(g) NH3(g) → H3NBF3(g)
boron trifluoride acts as a Lewis acid because it accepts
(forms a bond with) a pair of electrons from the ammonia,
which acts as the Lewis base.
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Unit 8 - Ch 16 Chem30
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11:09 AM
Page 735
Section 16.2
Lewis
Arrhenius
BrønstedLowry
oxygen
hydrogen
Figure 17
This Venn diagram shows how comprehensive some different acid–base theories
are considered to be.
In science, it is unwise to assume that any scientific
concept is complete. Whenever scientists assume that they
understand a concept, two things usually happen. First,
conceptual knowledge tends to remain static for a while
because little conflicting evidence exists, or because
conflicting evidence (being somewhat discomfiting) is
ignored. Second, when enough conflicting evidence
accumulates, a change in thinking occurs within the
scientific community in which the current theory is
drastically revised or entirely replaced. Revolutionary
concepts most often tend to be formed in a moment of
insight, usually following a long period of incredibly hard
work done by a great many people.
Case Study Questions
1. Identify a word or phrase that describes the central idea in
each of the theories described above.
2. For each theory, describe the weakness that led to its
being replaced.
3. What was the first theoretical definition of a base, and
who developed it?
4. Write Lewis formulas for each substance in the reaction of
boron trifluoride with ammonia, to illustrate the “electron
pair transfer” concept.
Section 16.2 Questions
1. Aqueous solutions of nitric acid and nitrous acid of the
same concentration are prepared.
(a) Predict how their pH values compare.
(b) Explain your answer using the Brønsted–Lowry concept.
2. Briefly state the five steps involved in predicting the
predominant reaction and the approximate position of
equilibrium in an acid–base reaction system.
3. According to the Brønsted–Lowry concept, what
determines the position of equilibrium in an acid–base
reaction?
4. What generalization from the table of Relative Strengths of
Aqueous Acids and Bases (Appendix I) can be used to
predict the position of an acid–base equilibrium?
5. State two examples of conjugate acid–base pairs, each
6. Predict, with reasoning, including Brønsted–Lowry
equations, whether each of the following chemical systems
will be acidic, basic, or neutral.
(a) aqueous hydrogen bromide
(b) aqueous potassium nitrite
(c) aqueous ammonia
(d) aqueous sodium hydrogen sulfate
(e) carbonated beverages
(f) limewater
(g) vinegar
7. Write an experimental design to test each of the predictions
in question 6.
8. Write two experimental designs to rank a group of bases in
order of strength.
involving the hydrogen sulfite ion.
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Page 736
9. Ammonia molecules and hydronium ions have remarkably
similar shapes and structures (Figure 18). Both have three
hydrogen atoms bonded in a pyramidal structure to a small
central atom, which has one lone pair of electrons. Oxygen
and nitrogen atoms both have very high electronegativities,
and they are adjacent on the periodic table. Explain why an
ammonia molecule is a good proton attractor, whereas a
hydronium ion is an extremely poor proton attractor.
Figure 18
Evidence indicates that the ammonia
molecule, NH3, modelled here, has the same
pyramidal shape as the hydronium ion,
H3O.
10. Many household cleaning solutions claim to “dissolve
away” rust stains and hard water “lime” deposits
(Figure 19). These cleaners are simply acidic solutions (as
is vinegar) that react with slightly soluble substances such
as Fe2O3(s) and CaCO3(s). The word “react,” however,
sounds dangerous to consumers, and is hardly ever used in
advertising. These cleaning solutions must be used
carefully, and you should always read all of the label
instructions, or you may find the cleaner also reacting with
the item you are trying to clean, or even with you! These
cleaners have a variety of formulations, but most contain
hydroxyacetic acid (glycolic acid), CH2OHCOOH(aq). A
0.10 mol/L glycolic acid solution is 3.9% ionized at 25 °C,
compared to 0.10 mol/L acetic acid, which is 1.3% ionized
at the same temperature.
(a) Write a Brønsted–Lowry equation for the reaction of
aqueous glycolic acid with carbonate ions (from hard
water “scum”), and label both conjugate acid–base
pairs.
(b) Explain whether the equilibrium would favour products
more, or less, if acetic acid (vinegar) were used to do
the same cleaning job. Is it likely that any difference
would be significant? State your reasoning.
(c) Explain whether glycolic acid would react with a given
chemical amount of rust more quickly or more slowly
than an equal volume of equally concentrated acetic
acid solution.
(d) Explain whether glycolic acid would react with a
greater, or lesser, chemical amount of rust than an
equal volume of equally concentrated acetic acid
solution.
11. Since ancient times, people have known that strong
heating of natural limestone (calcium carbonate) will
produce quicklime (calcium oxide). This very useful
substance is both reactive and corrosive with many organic
substances because it has a high affinity for water. Adding
water to quicklime causes the formation of slaked lime
(calcium hydroxide)—the common name was derived from
the idea that the lime was “slaking” its thirst. This reaction
also produces a very considerable quantity of heat!
(a) Write a balanced chemical equation for the
decomposition of calcium carbonate to carbon dioxide
and calcium hydroxide.
(b) Write a balanced chemical equation for the addition of
water to calcium oxide to produce calcium hydroxide.
(c) Write a Brønsted–Lowry equation for the instantaneous
and overwhelmingly quantitative (and highly
exothermic) reaction of aqueous oxide ions with water.
Figure 19
Commercial rust removing solutions must be used
with care, but can be a very effective illustration of
the old TV slogan, “better living, through
chemistry.”
736
Chapter 16
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Page 737
Acid–Base Strength and the
Equilibrium Law
Modifying Arrhenius’ original concept of acids and bases allowed us to explain the
behaviour of many substances in solution that could not be explained by his original
theory. Then, problems with varying degrees of properties led to considering reactions
of acids and bases as examples of equilibrium, which allowed a much more comprehensive understanding of relative acid or base strengths. Apparent conflicts in the definition of acids and bases then led us to define them by what they do in a reaction (in terms
of proton transfer), rather than by what they are (in terms of some typical structure). This
process of continually testing (identifying problems) and then expanding and improving
concepts is typical of science—with the goal always being the formation of a more complete and comprehensive theory. Science assumes that no theory ever formed by the
human mind can be “perfect,” and that any theory, no matter how well supported by
evidence, must always be questioned and tested. We also realize that every step moves us
closer to a more complete understanding. Testing, and then restricting, revising, and
(perhaps) replacing concepts, and then testing again is the constant cycle of the scientific method. Next, we explore the possible value of using the equilibrium law to further increase our understanding of acid–base behaviours.
The Acid Ionization Constant, Ka
One important way that chemists communicate the strength of any weak acid is by using
the equilibrium constant expression for a Brønsted–Lowry reaction equation showing
the acid ionizing in (reacting with) water. This expression is another very important
special case of equilibrium. The constant is given its own unique name and symbol: the
acid ionization constant, Ka. See the Ka values listed in the Relative Strengths of Aqueous
Acids and Bases data table (Appendix I).
Note that in this table, the first six acids have Ka values given only as “very large.”
These acids are collectively called the strong acids because they all react quantitatively
(99.9%) with water to form hydronium ions. Because no acid stronger than H3O(aq)
can exist in aqueous solution, all of these acids are considered equivalent. For each of them,
the actual acid species present is H3O(aq).
All of the other acids listed on the table are considered to be weak acids, and they vary
greatly in extent of reaction with water at equilibrium. To empirically rank any weak
acid as stronger or weaker than any other, we normally compare the ionization of solutions of the same concentration, to see which is more strongly acidic (has a lower pH). This
is how Appendix I was created.
By considering the equilibrium established in solutions of weak acid to be a Brønsted–
Lowry proton transfer, we gain a more complete understanding about what is actually
going on among the entities reacting. By using Ka equilibrium constants, we can derive
and predict quantities that are very useful when working with acid–base reactions.
Consider a solution of hydrofluoric acid, HF(aq). The equation for the reaction in water
(ionization) of this weak acid is
HF(aq) H2O(l)
0 H3O(aq) F(aq)
The equilibrium law expression is
[H3O(aq)][F(aq)]
Ka [HF(aq)]
16.3
+ EXTENSION
Acid–Base Reactions
This quick simulation lets you
explore the percent reaction when
various acids and bases react
together. The acid ionization
constants are provided for two
strong and three weak acids.
www.science.nelson.com
GO
Learning Tip
Use of ionization terminology
(strong and weak) can create a
problem, caused by the
common use of the word
“strong.” For an acid, strength
refers only to the ease with
which a base can remove its
proton, and has nothing to do
with the quantity present in the
solution. Concentration refers to
the chemical quantity per unit
volume of solution, and has
nothing to do with strength.
Both the strength and the
concentration of an acid affect
how rapidly, and to what extent,
that acid will react.
Note that the concentration of liquid water is omitted from the Ka equilibrium law
expression. We make an assumption that this value will remain essentially constant for
NEL
Equilibrium in Acid–Base Systems 737
Unit 8 - Ch 16 Chem30
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Learning Tip
The great advantage of Ka
values over percent ionization
values is that, once determined,
Ka values are valid over a wide
range of acid solution
concentrations. A percent
ionization value is useful only
for one specific concentration
of one specific weak acid.
Page 738
aqueous solutions that are not highly concentrated. This assumption is not exactly true,
because the water is not a separate liquid phase in an aqueous solution. However, as
long as the amount of water solvent is much greater than the amount of acid solute,
making this assumption will cause negligible change in (and will not add uncertainty to)
the answers for any calculations.
All Ka values are calculated by making this assumption. For this reason (and for other
measurement reasons too involved to explain here), Ka values are necessarily somewhat
inaccurate, and become less accurate as the acid concentration increases. It is important to know that Ka values given in tables are normally given to only two significant
digits because they are usually only known to a certainty of about ± 5%.
Two calculations involving the Ka constant are common for weak acid solutions:
• calculating a Ka value from measured (empirical) amount concentration data
• using a Ka value to predict a concentration of hydronium ions for an aqueous
solution where the initial weak acid amount concentration is known
Calculating Ka from Amount Concentrations
SAMPLE problem 16.2
The pH of a 1.00 mol/L solution of acetic acid is carefully measured to be 2.38 at SATP.
What is the value of Ka for acetic acid?
Use the balanced equation to write the equilibrium law expression.
CH3COOH(aq) H2O(l)
0
H3O(aq) CH3COO(aq)
[H3O(aq)][CH3COO(aq)]
Ka [CH3COOH(aq)]
First, find the equilibrium concentration of aqueous hydronium ion from the pH.
[H3O(aq)] 10pH
102.38
0.0042 mol/L
Learning Tip
Because the value for the initial
acid concentration is precise
only to two decimal places, the
calculation of the equilibrium
concentration of the acid
(1.00 mol/L 0.0042 mol/L)
rounds to 1.00 mol/L. (See the
Precision Rule for Calculations,
Appendix F.3.) In other words,
the decrease in the initial acid
concentration is negligible, in
this particular case. A negligible
change in concentration is quite
often the case for Ka
calculations for weak acids, but
not always.
Recall that amount
concentrations must be used to
calculate any Kc value and, by
convention, units for the
equilibrium constant value are
simply ignored.
738
Chapter 16
Next, use the hydronium ion concentration and the balanced chemical equation to
calculate the concentration of the acetate ion. Since one acetate ion forms for each
hydronium ion, the equilibrium concentration of acetate ion must be identical to that of
the hydronium ion. The stoichiometric ratio is 1:1.
[CH3COO(aq)] [H3O(aq)] 0.0042 mol/L
An ICE table is useful to find the equilibrium concentration of acetic acid molecules.
Table 1 ICE Table for CH3COOH(aq) H2O(l)
Concentration
Initial
Change
Equilibrium
[CH3COOH(aq)]
(mol/L)
1.00
0.0042
1.00
0
H3O(aq) CH3COO(aq)
[H3O(aq)]
(mol/L)
0
[CH3COO(aq)]
(mol/L)
0
0.0042
0.0042
0.0042
0.0042
(0.0042 mol/L)(0.0042 mol/L)
Ka (1.00 mol/L)
0.000 017
Regardless of numerical size, Ka values are usually expressed in scientific notation.
The calculated Ka for acetic acid is 1.7 105.
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Page 739
Section 16.3
COMMUNICATION example 1
A student measures the pH of a 0.25 mol/L solution of carbonic acid to be 3.48.
Calculate the Ka for carbonic acid from this evidence.
Solution
H2CO3(aq) H2O(l)
0
H3O(aq) HCO3(aq)
[H3O(aq)][HCO3(aq)]
Ka [H2CO3(aq)]
At equilibrium,
[H3O(aq)] 10pH
103.48
3.3 104 mol/L
[HCO3(aq)] [H3O(aq)] 3.3 104 mol/L
[H2CO3(aq)] (0.25 0.000 33) mol/L
0.25 mol/L
Table 2 ICE Table for H2CO3(aq) H2O(l)
Amount
concentration
0
H3O(aq) HCO3(aq)
[H2CO3(aq)]
(mol/L)
Initial
[H3O(aq)]
(mol/L)
0.25
0.000 33
Change
Equilibrium
0
[HCO3(aq)]
(mol/L)
0
0.000 33
0.000 33
0.000 33
0.000 33
0.25
(0.000 33 mol/L)(0.000 33 mol/L)
Ka 0.000 000 44
(0.25 mol/L)
According to the equilibrium law, the Ka for carbonic acid is 4.4 107.
Learning Tip
Because a Brønsted-Lowry
equation is written to show a
single proton transfer, the
stoichiometric ratio will always
be 1:1:1:1. Thus, the units for K a
values will always be mol/L.
Because we can make this
assumption, units are often not
included with K a values in
reference tables or in final
answer statements.
COMMUNICATION example 2
The pH of a 0.400 mol/L solution of sulfurous acid is measured to be 1.17.
Calculate the Ka for sulfurous acid from this evidence.
Solution
H2SO3(aq) H2O(l)
0
H3O(aq) HSO3(aq)
[H3O(aq)][HSO3(aq)]
Ka [H2SO3(aq)]
At equilibrium,
[H3O(aq)] 10pH
101.17
0.068 mol/L
[HSO3 (aq)] [H3O(aq)] 0.068 mol/L
[H2SO3(aq)] (0.400 0.068) mol/L
0.332 mol/L
NEL
Equilibrium in Acid–Base Systems 739
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Page 740
Table 3 ICE Table for H2SO3(aq) H2O(l)
Amount
concentration
0
[H3O(aq)]
(mol/L)
[H2SO3(aq)]
(mol/L)
Initial
H3O(aq) HSO3(aq)
0.400
Change
Equilibrium
[HSO3(aq)]
(mol/L)
0
0
0.068
0.068
0.068
0.332
0.068
0.068
(0.068 mol/L)(0.068 mol/L)
Ka 0.014
(0.332 mol/L)
According to the equilibrium law, the Ka for sulfurous acid is 1.4 102.
Calculating [H3O(aq)] from Ka
The second type of calculation involving a Ka value allows us to predict the acidity of
any weak acid solution. We can calculate the concentration of hydronium ions from the
initial acid concentration and the Ka value as follows.
SAMPLE problem 16.3
Predict the [H3O(aq)] and pH for a 0.200 mol/L aqueous solution of methanoic acid.
Look up the value of Ka for methanoic (formic) acid in the Relative Strengths of Aqueous
Acids and Bases data table (Appendix I). The value is given as 1.8 104 mol/L.
Use the balanced equation to write the equilibrium law expression.
HCOOH(aq) H2O(l)
0
H3O(aq) HCOO(aq)
[H3O(aq)][HCOO(aq)]
Ka 1.8 104 [HCOOH(aq)]
Use x to represent the numerical value for the equilibrium amount concentration of
aqueous hydronium ions.
[H3O(aq)] x mol/L
An ICE table is useful to keep track of the concentration values.
Using the 1:1:1:1 stoichiometric ratio, calculate concentration increases and decreases for
the reaction to equilibrium, and note them in the ICE table.
Table 4 ICE Table for HCOOH(aq) H2O(l)
Concentration
Initial
Change
Equilibrium
[HCOOH(aq)]
(mol/L)
0.200
x
(0.200 x)*
0
H3O(aq) HCOO(aq)
[H3O(aq)]
(mol/L)
[HCOO(aq)]
(mol/L)
0
0
x
x
x
x
Substituting into the equilibrium law expression, ignoring units,
[H3O(aq)][HCOO(aq)]
1.8 104 [HCOOH(aq)]
(x) (x)
(0.200 x)
740
Chapter 16
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Page 741
Section 16.3
* Note:
Technically, solving for x (the amount concentration of hydronium ion) from this
b2 4
ac
b
expression requires the use of the quadratic formula, x ,
2a
which is quite tedious, unless you have a calculator that is preprogrammed to
solve such formulas.
Conveniently, for many (but not all) weak acid solutions, we can use an
approximation that greatly simplifies solving for x. If the percent reaction of the
weak acid is quite small, we can assume that the initial concentration of the acid
decreases by so little that the numerical value of the acid amount concentration
will remain effectively unchanged at equilibrium.
The problem, of course, is deciding when a percent reaction is small enough to
allow this assumption to be used. In this text, we will apply the mathematically
simplest (and most convenient) “rule of thumb” that is commonly used to make
this decision.
If the initial amount concentration of the acid is numerically at least 1000
times its Ka value, then you may assume that the initial and equilibrium acid
amount concentrations are numerically equal.
This limiting assumption is consistent with the 5% certainty of most Ka values,
meaning it ignores those amount concentration changes that would change the
Ka value by less than the uncertainty that already exists.
Note that the equation to be solved contains a squared value.
x2
0.000 18 (0.200 x)
DID YOU KNOW
?
Rule of Thumb
The very old English phrase “rule
of thumb,” meaning a rule for any
quick way to approximate a
quantity without measuring,
probably comes directly from use
of human thumbs. Since the width
of an adult’s thumb is
approximately one inch, and the
length of a thumb is approximately
four inches, you can make a fairly
good estimate of the length of
something (in Imperial units) by
simply placing your thumbs along
the object, one after the other.
What is the length of this textbook
in (your) thumb widths? What do
equestrians mean when they say a
horse stands 16 “hands” high?
A practical rule of thumb for
Americans driving their cars into
Mexico or Canada is to multiply
the posted speed by 6 and drop
the last digit, in order to quickly
convert the km/h speeds on road
signs into a fairly close miles-perhour equivalent.
Multiplying both sides by (0.200 x), collecting terms, and equating to zero gives
x2 0.000 18x 0.000 036 0 (expressed in the form of a quadratic equation).
Notice what happens when we test our present example for the simplifying assumption.
[HCOOH(aq)]initial
0.200
1111, which is greater than 1000
Ka
0.000 18
The test allows us to use the assumption (initial acid concentration equilibrium acid
concentration, or [HA(aq)]initial [HA(aq)]equilibrium). In this specific case, so little acid has
reacted with water at equilibrium that the initial acid concentration is decreased by a
negligible amount—less than the uncertainty of the initial value.
Numerically, for this example, we may assume that (0.200 x) 0.200, and we can then
simplify the equation to
x2
0.000 18 , which makes solving for x quick and easy.
0.200
Isolate x and take the square root of both sides of the equation:
0.000 1
8 0.200
x 0.0060
so,
and
[H3O(aq)] 0.0060 mol/L or 6.0 103 mol/L
pH log [H3O(aq)]
log (0.0060)
2.22
NEL
Equilibrium in Acid–Base Systems 741
Unit 8 - Ch 16 Chem30
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DID YOU KNOW
11:09 AM
?
pKa
For purposes of easy comparison,
the Ka values of different acids are
sometimes expressed as pK a values.
A pK a value is the negative
logarithm of the Ka. A reported pK a
value of 12 would represent an acid
with a Ka of 1 1012, a very weak
acid. In comparing any two weak
acids, the one with the lower pK a
value is the stronger of the two
acids.
Page 742
As shown in the next Communication Example, when the negligible ionization (percent reaction) assumption holds true, predictions of [H3O(aq)] using Ka values become
simple and straightforward. Note that you can perform the 1000:1 ratio test for the simplifying assumption before starting the problem, which is very convenient.
COMMUNICATION example 3
Predict the [H3O(aq)] and pH for a 0.500 mol/L aqueous solution of hydrocyanic acid.
Solution
Test the simplifying assumption:
0.500
[HCN(aq)]initial
8.1 107 (greater than 1000)
6.2 1010
Ka
The assumption holds, and may be used.
[H3O(aq)][CN(aq)]
Ka 6.2 1010 [HCN(aq)]
At equilibrium:
Let x [H3O(aq)] [CN(aq)]
Then [HCN(aq)] (0.500 x) 0.500 (using the assumption)
Table 5 ICE Table for HCN(aq) H2O(l)
Amount
concentration
Initial
Change
Equilibrium
0
[HCN(aq)]
(mol/L)
0.500
x
(0.500 x) 0.500
H3O(aq) CN(aq)
[H3O(aq)]
(mol/L)
[CN(aq)]
(mol/L)
0
0
x
x
x
x
x2
6.2 1010 0.500
10
6.2 10
0.500 1.8 105
x so,
[H3O(aq)] 1.8 105 mol/L
and
pH log [H3O(aq)] log (1.8 105) 4.75
According to the equilibrium law, the hydronium ion amount concentration of
0.500 mol/L hydrocyanic acid is 1.8 105 mol/L, and the pH of the solution is 4.75.
Note: No Practice, Section, or Review question in this text will require you to use the quadratic formula to solve acid ionization constant problems. However, you may be asked to
state whether a given problem would require use of this formula for solution (in other
words, whether the ionization assumption test fails). You should develop the habit of
always initially testing the assumption and making a qualifying statement before completing the solution to any Ka question where this assumption holds.
742
Chapter 16
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Page 743
Section 16.3
Practice
1. (a) Write a theoretical definition for the strength of an acid.
(b) State the empirical properties that provide evidence for differing acid strengths.
(c) Explain the difference in meaning between strength and concentration, as these
terms are used to refer to aqueous acids.
(d) Does the stronger of two acids, when dissolved, necessarily make a more acidic
aqueous solution? Explain.
2. Refer to Appendix I for required information to make the following predictions. For
each case, first decide and state whether a solution would require use of the
quadratic formula. For all cases where use of the quadratic formula is not required,
communicate the full solution.
(a) Predict [H3O(aq)] for 0.20 mol/L hydrobromic acid.
(b) Predict [H3O(aq)] for 0.20 mol/L hydrofluoric acid.
(c) Predict [H3O(aq)] for 0.20 mol/L ethanoic acid.
(d) Predict the pH of 2.3 mmol/L nitric acid.
(e) Predict the pH of 2.3 mmol/L nitrous acid.
(f) Predict the pH of 2.3 mmol/L hydrosulfuric acid.
3. For all of the cases (a–f) in question 2 where a prediction could be calculated, rank the
acid solutions in order of decreasing acidity.
4. For all of the cases (a–f) in question 2, rank the acids in order of decreasing acid
Figure 1
Hydrofluoric acid is a weak acid
but it etches glass. The etching is
not due to the presence of
hydronium ions, because, as you
know, even the strongest acids are
routinely stored in glass containers.
strength. Which two acids have essentially the same strength?
5. The hydronium ion concentration in 0.100 mol/L propanoic acid is determined (from a
pH measurement) to be 1.16 103 mol/L.
(a) Calculate the percent reaction (ionization) of this particular weak acid solution.
(b) Calculate Ka for aqueous propanoic acid.
(c) Is this Ka value constant for propanoic acid? Explain.
6. A 0.10 mol/L solution of lactic acid, a weak acid found in milk, has a measured pH of
2.43. The chemical name for this very common organic compound is
2-hydroxypropanoic acid.
(a) Find the percent reaction of this lactic acid solution.
(b) Calculate the Ka value for aqueous lactic acid.
(c) Compare the Ka values for 2-hydroxypropanoic acid and for propanoic acid. What
effect does adding an OH group to the central carbon atom of this molecule have
on the ability of the COOH group to attract a proton?
7. Unlike all other aqueous hydrogen halides, hydrofluoric acid is not a strong acid. It
does, however, have a special chemical property: it reacts with glass. This property is
used to etch frosted patterns on glassware (Figure 1).
(a) Write the Ka expression for hydrofluoric acid.
(b) Calculate the hydronium and fluoride ion amount concentrations, the pH, and the
percent reaction in a commercial 2.0 mol/L HF(aq) solution at 25 °C.
8. Phosphoric acid is used in rust-remover solutions. The aqueous acid is available for
purchase by high schools in concentrations of about 15 mol/L.
(a) Predict the hydronium ion amount concentration, the pH, and the percent
reaction of a 10 mol/L solution of phosphoric acid.
(b) Suggest a reason (having to do with Ka values and solution concentrations) why
you might well suspect that these values could be inaccurate.
9. Ascorbic acid is the chemical name for Vitamin C (Figure 2). A student prepares a
0.200 mol/L aqueous solution of ascorbic acid, tests its pH, and reads a value of 2.40
from the pH meter.
(a) From the student’s evidence, calculate the Ka for ascorbic acid.
(b) Compare your result to the value listed in the table of Relative Strengths of
Aqueous Acids and Bases (Appendix I). The Ka value for ascorbic acid increases
with increasing temperature. State whether the student’s aqueous solution was
likely warmer or colder than standard temperature (25 °C) when the pH was
tested.
NEL
Figure 2
Vitamin C (ascorbic acid) is present
in many fresh foods, particularly
citrus fruits. It is essential to the
body to promote healing of injuries
and to fight infection.
In the winter of 1535, the men of
Jacques Cartier’s expedition were
suffering from scurvy (vitamin C
deficiency) in their fort at Stadacona
(now Montreal). Cartier lost 25 men
before learning of a simple cure
from the Iroquois. The Aboriginal
people prepared a tea made from
white cedar needles—rich in
ascorbic acid—that cured sufferers
within days.
Equilibrium in Acid–Base Systems 743
Unit 8 - Ch 16 Chem30
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DID YOU KNOW
11:09 AM
?
Sailors and Scurvy
Vitamin C deficiency (from a dietary
lack of fresh fruits and vegetables)
causes scurvy, a disease where the
body cannot fight infections, and
cuts, sores, and bruises do not heal.
Until the last half of the 18th
century, this disease was a terrible
problem for sailors, and a major
cause of death on long voyages. For
example, on his famous voyage of
1498, Vasco da Gama lost 100 of
160 crewmen to scurvy. The diet of
European sailors was primarily
salted meat and hard biscuits—items
selected for their resistance to
spoilage.
In 1747, James Lind, a Scottish
physician, wrote that feeding citrus
fruit to scurvy victims effected an
amazingly rapid cure. Captain Cook
tried this remedy on his crew during
his famous voyages of exploration in
the 1770s, and reported that he was
astounded to have lost only one
man on a three-year voyage!
By 1795, the British Royal Navy
(after decades of deliberation)
formally adopted the practice of
providing lemon juice (incorrectly
referred to as lime juice) for all
hands. Scurvy in the fleet was wiped
out, and British sailors have been
called “limeys” ever since.
Note that this story illustrates
another case of technology leading
science: The cure for scurvy was
known long before the cause of the
disease was explained.
Interestingly, sailors on the long
sea voyages undertaken by the
Chinese during the Ming Dynasty
(1368–1644) had no problems with
scurvy because their traditional diet
included fresh germinated soya
beans, the shoots of which are
naturally rich in Vitamin C.
744
Chapter 16
Page 744
Base Strength and the Ionization Constant, Kb
The Brønsted–Lowry concept specifies that the strongest base possible in aqueous solution must be hydroxide ions. So, hydroxide ion is considered to be “the” strong base—
somewhat parallel to the strong acids. A difference is that many ionic substances contain
hydroxide ions initially, before being dissolved; whereas hydronium ions are always
formed by the reaction of some entity with water, after a substance dissolves. Recall that
for solutions of all ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), we assume
that the compound dissociates completely upon dissolving. Therefore, finding the
hydroxide ion concentration does not involve any reaction with water or any reaction equilibrium. Rather, in these cases, we find the hydroxide ion concentration directly from a
dissociation equation, as shown in the next Communication Example.
COMMUNICATION example 4
Find the hydroxide ion concentration of a 0.064 mol/L solution of barium hydroxide.
Solution
Ba(OH)2(aq) → Ba2(aq) 2 OH(aq) (complete dissociation)
2
[OH(aq)] [Ba(OH)2(aq)] 1
2
0.064 mol/L 1
0.13 mol/L
According to the stoichiometric ratio, the hydroxide ion concentration is 0.13 mol/L.
Other entities that act as bases are collectively called weak bases, because empirical
evidence indicates that they attract protons less than hydroxide ions do. Turn to the
table of Relative Strengths of Aqueous Acids and Bases (Appendix I), and observe that,
to the right of every acid entity formula listed, there is a formula for an entity that is
identical to the acid’s formula, except that one proton (represented as H) is missing. As
you have learned, such an entity is called the conjugate base of the acid. While the list of
these entities (reading upward) can be taken to be a list of bases and their (decreasing)
relative strengths, note that there is no corresponding value given to show the extent of
reaction of these entities with water. This convention is common in chemistry—information about bases in solution must be derived from a table of acid values.
Bases vary in strength much as acids do, and the system used to identify, classify, and
rank the strength of bases is quite similar to the one you have just learned for acids.
Note that the table in Appendix I shows that the weaker an acid is, the stronger its conjugate base (and vice versa). This observation is common sense. If it is very easy to
remove the proton from an acid, then the entity remaining (the conjugate base) must not
attract protons very well.
As you know, weak bases react only partially (usually much less than 50%) with water
in aqueous solution to produce hydroxide ions. We can communicate the strength of
any weak base using the equilibrium constant expression for its reaction with water in
dilute aqueous solution. This equilibrium constant for weak bases is another special
case, where the Kc constant is called the base ionization constant, Kb.
Consider a solution of sodium citrate, Na3C6H5O7(aq). We assume that in aqueous
solution, any ionic compound dissociates completely into its component ions. In this case,
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Section 16.3
the ions are aqueous sodium and citrate ions: Na(aq) and C6H5O73(aq). Sodium ions
are “spectators,” with no apparent acidic or basic properties. The citrate ions, however,
are found on the Relative Strengths of Aqueous Acids and Bases data table (Appendix I)
in the conjugate bases column. So the citrate ion is a weak Brønsted–Lowry base and
reacts with water to form a basic solution at equilibrium.
The reaction equilibrium equation for aqueous sodium citrate is
C6H5O73(aq) H2O(l)
0 HC6H5O72(aq) OH(aq)
The equilibrium law expression is
[HC6H5O72(aq)][OH(aq)]
Kb [C6H5O73(aq)]
Just as with Ka values, there are two kinds of weak base solution calculations that
involve the Kb constant:
• calculating a Kb value from empirical (measured) amount concentration data
• using a Kb value to predict an amount concentration of hydroxide ions for an
aqueous solution where the initial weak base concentration is known
Calculating Kb from Amount Concentrations
Calculating Kb for a weak base uses essentially the same method as calculating Ka for a
weak acid. Often, an extra (simple) calculation step must be included because measured pH values may need to be converted to find the equilibrium hydroxide ion concentration, as shown in the following Communication Example.
COMMUNICATION example 5
Learning Tip
The concentration of liquid
water is omitted from Kb
equilibrium law expressions for
the same reasons that it is
omitted from Ka expressions. As
with Ka values, Kb values are
also only certain to about ±5%.
For weak base Kb values, we
also assume that aqueous
solutions are not highly
concentrated, just as we do for
weak acids. Just as with Ka
values, Kb value units are
ignored, by convention.
A student measures the pH of a 0.250 mol/L solution of aqueous ammonia and finds it to
be 11.32.
Calculate the Kb for ammonia from this evidence. Show the establishment of the reaction
equilibrium using an ICE table.
Solution
NH3(aq) H2O(l)
0
NH4(aq) OH(aq)
[NH4(aq)][OH(aq)]
Kb = [NH3(aq)]
At equilibrium:
[H3O(aq)] 10pH
1011.32
4.8 1012 mol/L
Kw [H3O(aq)][OH(aq)]
Kw
[OH(aq)] [H3O(aq)]
1.0 1014
4.8 1012 mol/L
0.0021 mol/L
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[NH4(aq)] [OH(aq)]
0.0021 mol/L
[NH3(aq)] (0.250 0.0021) mol/L
0.248 mol/L
Table 6 ICE Table for NH3(aq) H2O(l)
Concentration
Initial
Change
Equilibrium
0
NH4(aq) OH(aq)
[NH3 (aq)]
(mol/L)
0.250
0.0021
0.248
[NH4(aq)]
(mol/L)
[OH(aq)]
(mol/L)
0
0
0.0021
0.0021
0.0021
0.0021
[0.0021 mol/L][0.0021 mol/L]
Kb = 0.000 018
[0.248 mol/L]
From this evidence, and according to the equilibrium law, the Kb for aqueous ammonia
is 1.8 105.
Figure 3
Dyes with molecular structures
based on aniline make possible a
variety of bright, stable colours for
fabrics and leathers.
Practice
10. List some empirical properties that would be useful when distinguishing strong bases
from weak bases.
11. For each of the following weak bases, write the chemical equilibrium equation and
the equilibrium law expression for Kb.
(a) CN(aq)
(b) SO42(aq)
Learning Tip
The use of chemical (IUPAC)
and common names can be
confusing if you have not
memorized a few examples. In
this textbook, the normal
practice is to give the chemical
name, followed by a common
name in parentheses. A few
substances are so widely used,
however, that the common
name is ubiquitous (used
everywhere, even by chemists).
Baking soda (sodium
bicarbonate, sodium hydrogen
carbonate) and vinegar (acetic
acid, ethanoic acid) are two
examples of such chemicals;
the common name is often
given, rather than the IUPAC
name. You are expected to be
familiar with such examples,
and others listed in Appendix J.
746
Chapter 16
12. The hydroxide ion concentration in a 0.157 mol/L solution of sodium propanoate,
NaC2H5COO(aq), is found to be 1.1 105 mol/L. Calculate the base ionization
constant for the propanoate ion.
13. Aniline, C6H5NH2, is used to make a wide variety of drugs and dyes (Figure 3). It has
the structure of an ammonia molecule, with a phenyl group substituted for one
hydrogen, and, like ammonia, acts as a weak base. If the pH of a 0.10 mol/L aniline
solution was found to be 8.81, what is its Kb?
Calculating [OH(aq)] from Kb
The second type of calculation involving a Kb value is the prediction of the basicity of any
weak base solution. We can calculate the concentration of hydroxide ions from the initial weak base concentration and the Kb value. There is an automatic problem, however,
because tables of relative acid–base strengths do not normally list Kb values. We use the
automatic relationship between conjugate acid–base pairs to deal with this problem.
The KaKb Relationship for Conjugate Acid–Base Pairs
To develop the next useful concept, we use the common weak acid, acetic acid,
CH3COOH(aq), and its conjugate base, the acetate ion, CH3COO(aq). Of course, it
would be just as correct to state that we will use the weak base, acetate ion, and its conjugate acid, acetic acid.
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Section 16.3
The equilibrium reaction and law expression for aqueous acetic acid are
CH3COOH(aq) H2O(l)
0 H3O(aq) CH3COO(aq)
[H3O(aq)][CH3COO(aq)]
Ka [CH3COOH(aq)]
For the weak base, aqueous acetate ion,
CH3COO(aq) H2O(l)
0 CH3COOH(aq) OH(aq)
[CH3COOH(aq)][OH(aq)]
Kb [CH3COO(aq)]
Notice what happens when we multiply these equilibrium constant expressions:
[H3O(aq)][CH3COO(aq)]
[CH3COOH(aq)][OH(aq)]
Ka Kb [CH3COOH(aq)]
[CH3COO(aq)]
[H3O(aq)][OH(aq)]
(Does this expression look familiar?)
Kw
Kw
So, for any conjugate acid–base pair, Kw KaKb and Kb .
Ka
This last equation is particularly useful because now, to find a Kb value for any base
listed on the Relative Strengths of Aqueous Acids and Bases table (Appendix I), you need
only identify its conjugate acid, and then divide Kw, 1.0 1014, by the Ka value given
for that conjugate acid.
SAMPLE problem 16.4
Solid sodium benzoate forms a basic solution. Determine Kb for the weak base present.
NaC6H5COO(s) → Na(aq) C6H5COO(aq) (complete dissociation)
First identify, from the Relative Strengths of Aqueous Acids and Bases table, and the
entities present in solution, which entity is reacting as a base.
The benzoate ion must be the weak base entity. Its equilibrium reaction with water is
C6H5COO(aq) H2O(l)
0
C6H5COOH(aq) OH(aq)
From the equilibrium equation, the conjugate acid of the benzoate ion is identified as
benzoic acid. From the acid–base table,
Ka for C6H5COOH(aq) 6.3 105
Using the Kw relationship for conjugate acid–base pairs, find Kb for C6H5COO(aq).
DID YOU KNOW
?
Why Acidity?
It should be obvious by now that
pH values are used much more
than pOH values, and Ka constants
are listed in tables rather than Kb
constants. You might wonder why
we choose to emphasize the
hydronium ion properties of
aqueous solutions—after all,
hydroxide ions play exactly the
opposite (and an equally
important) role. But historically,
chemists have always been much
more concerned with acidic
properties than with basic
properties. Acids dissolve (react
with) many metals to form
common ionic compounds and
hydrogen. Hydrogen gas was a
fascinating product to early
chemists, both because it is lighter
than air and because it is violently
explosive. The stronger acids tend
to have painfully sharp odours and
flavours, and react destructively
with human tissue—all properties
that make acids quite noticeable.
Even when it became clearly
understood that acids and bases
were opposite aspects of the same
concept, it seemed easier to
choose acid properties as the
initial basis for studying acid–base
relationships. We talk about the
“acidity” of something routinely; it
is a common term, found in any
pocket dictionary. What is the
equivalent term for how basic
something is? There is an
accepted word for it, but you’ll
need a really good (very big)
dictionary to find it. Or a good
Chemistry text, of course...
Kw
Kb Ka
1.0 1014
6.3 105
1.6 1010
According to the K w relationship, the benzoate ion has a Kb value of 1.6 1010.
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COMMUNICATION example 6
Calculate Kb for the weak base aqueous ammonia, commonly used in commercial window
cleaning solutions. Write the equation for the equilibrium.
Solution
NH3(aq) H2O(l)
0
NH4(aq) OH(aq)
For NH3(aq), NH4 (aq) is the conjugate acid:
Ka 5.6 1010
Kw
Kb Ka
1.0 1014
5.6 1010
1.8 105
According to the table of Relative Strengths of Aqueous Acids and Bases, and the Ka–Kb
relationship, the Kb value for aqueous ammonia is 1.8 105.
Now we can easily predict how basic a solution of known concentration will be from
its Kb value. Predicting basicity always involves calculating the hydroxide ion concentration,
and may also involve calculating a value for pOH, pH, and/or percent reaction. The first
step may be determining the Kb value by using a table of Ka values. The form of the calculation then follows the same format as the equivalent type of question for weak acids,
with a similar assumption made, and the same restrictions.
• Assume that the initial weak base aqueous concentration decreases so little that it is
numerically unchanged at equilibrium, but only if the initial amount concentration
of the weak base is at least 1000 times greater than its Kb value.
• For questions in this text, you are not required to do calculations using the quadratic formula.
The Sample Problem that follows shows the format for calculations of this type, and
also an example of each conversion that may be required.
Recall, from Section 16.2, that, in aqueous solution, the amphiprotic ion HCO3(aq)
reacts with water as a base to a very much greater extent than it does as an acid. You
may, therefore, assume that the solution is basic because the ion reacts as a base, and
ignore its negligible reaction extent as an acid.
SAMPLE problem 16.5
Find the hydroxide ion amount concentration, pOH, pH, and percent reaction (ionization)
of a 1.20 mol/L solution of baking soda.
The compound is sodium hydrogen carbonate, NaHCO3(s). The weak base is the
HCO3(aq) ion, produced by dissolving NaHCO3(s) in water.
For HCO3(aq), H2CO3(aq) is the conjugate acid:
Ka 4.5 107
Find the Kb value using the Ka–Kb–Kw relationship.
Kw
Kb Ka
1.0 1014
4.5 107
2.2 108
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Chapter 16
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Section 16.3
Write the equation for the reaction and the equilibrium expression for Kb for hydrogen
carbonate ion.
HCO3(aq) H2O(l)
0
H2CO3 (aq) OH(aq)
[H2CO3(aq)][OH (aq)]
Kb [HCO3(aq)]
At equilibrium, let x be the numerical value of [OH(aq)] and also [H2CO3(aq)].
Table 7 ICE Table for HCO3(aq) H2O(l)
0
[HCO3(aq)]
(mol/L)
Amount
concentration
Initial
H2CO3(aq) OH(aq)
[H2CO3 (aq)]
(mol/L)
1.20
Change
x
Equilibrium
(1.20 x)
[OH(aq)]
(mol/L)
0
0
x
x
x
x
Then [HCO3(aq)] (1.20 x)
Test the assumption that (1.20 x) is numerically equal to 1.20.
1.20
[HCO3(aq)]initial
5.5 107 (much greater than 1000)
2.2
108
Kb
The assumption holds, so substitute into the Kb expression and solve for x.
x2
2.2 108 1.20
2.2 108 1.20
x 1.6 104
[OH(aq)] 1.6 104 mol/L
Use [OH(aq)] to find the value of pOH.
DID YOU KNOW
?
Visualizing Reaction Extents
From any percent reaction value,
you can get a better idea of the
reaction extent at equilibrium by
converting mentally to a wholenumber ratio. If bicarbonate ions
react with water 0.013% at
equilibrium, the ratio is 0.013:100,
or 13:100 000. This means that at
any given instant, about 13 of
every 100 000 initial bicarbonate
ions have changed to carbonic
acid molecules, but 99 987 of them
have not. The few ions that have
changed are responsible for all the
basicity of the solution!
If you really want to test your
powers of imagination, try to
visualize that for every 100 000
bicarbonate ions in this aqueous
solution, there are approximately
5 500 000 water molecules. Now
think about all of these entities
constantly moving at speeds
averaging about 1500 km/h, and
then imagine every one of them
colliding with another molecule or
ion at a rate of approximately
7 000 000 000 times every
second…
pOH log [OH(aq)]
log (1.6 104)
3.80
Use the pOH value to find pH.
pH 14.00 pOH
14.00 3.80
10.20
Percent reaction is the percent ratio of the equilibrium concentration of produced
hydroxide ions to the initial concentration of hydrogen carbonate ions.
[OH(aq)]equilibrium
percent reaction 100%
[HCO3(aq)]initial
1.6 104 mol/L
100%
1.20 mol/L
0.013%
A 1.20 mol/L sodium hydrogen carbonate solution has a hydroxide ion concentration of
1.6 104 mol/L, a pOH of 3.80, a pH of 10.20, and a percent reaction of 0.013%.
The Effect of Amphoteric Entities
Finally, we deal with the problem of amphoteric entities. If an entity can react as either
a Brønsted–Lowry acid or base, how do you know which will be the predominant reaction in its aqueous solution? You can determine the answer with one simple calculation.
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Read the Ka value for the entity from the table of Relative Strengths of Aqueous Acids
and Bases. Calculate the Kb value for the entity from Kw and the Ka value (from the table)
of its conjugate acid, as shown in Sample Problem 16.5. The higher of the “K” values
tells you which reaction predominates, and, therefore, whether the entity reacts as an
acid or as a base in aqueous solution.
COMMUNICATION example 7
Which reaction predominates when NaHSO3(s) is dissolved in water to produce
HSO3(aq) solution?
Solution
Ka 6.3 108
The conjugate acid is H2SO3(aq), with Ka 1.4 102.
Figure 4
Codeine (methylmorphine,
C18H21NO3) is available by
prescription as an ingredient in
some analgesic (pain relief)
medicines. This narcotic alkaloid
was initially extracted from opium.
Other alkaloids include nicotine,
quinine, cocaine, strychnine, and
caffeine. All are common plantderived organic compounds based
on nitrogen-containing ring
structures.
So, for HSO3(aq),
Kw
Kb Ka
1.0 1014
1.4 102
7.1 1013
The Ka value for HSO3(aq) far exceeds its calculated Kb value.
According to the Kw relation and the equilibrium law, an aqueous solution of this
substance will be acidic because a hydrogen sulfite ion will react predominately as a
Brønsted–Lowry acid.
Section 16.3 Questions
1. Codeine, an ingredient in these migraine pills (Figure 4),
has a Kb of 1.73 106. Calculate the pH of a 0.020 mol/L
codeine solution (use Cod as a chemical shorthand symbol
for this complex weak base entity.).
2. What is the pH of a 0.18 mol/L cyanide ion solution?
3. Acetylsalicylic acid (ASA) is a painkiller used in many
headache tablets. This drug forms an acidic solution that
attacks the digestive system lining. The Merck Index lists its
Ka at 25 °C to be 3.27 104. Explain the difficulty in
calculating the pH of a saturated 0.018 mol/L solution of
acetylsalicylic acid, C6H4COOCH3COOH(aq).
4. Boric acid is used for weatherproofing wood and
fireproofing fabrics. Very dilute aqueous boric acid is used
in eyewash solution as a preservative. Predict the pH of a
0.50 mol/L solution of boric acid.
5. Salicylic acid (2-hydroxybenzoic acid, C6H5OHCOOH(s)) is
an active ingredient of medications, such as Clearasil®,
that are used to treat acne. Since the Ka for this acid was
not listed in any convenient references, a student tried to
determine the value experimentally. She kept adding water
slowly to a 1.00 g sample, with constant stirring, until the
crystals all dissolved. The volume of solution was 460 mL.
She found that the pH of this solution of salicylic acid was
2.40 at 25 °C. Calculate the ionization constant for this acid.
750
Chapter 16
6. Sodium ascorbate is the sodium salt of ascorbic acid and is
used as an antioxidant in food products. The pH of a
0.15 mol/L solution of the ascorbate ion, HC6H6O6(aq), is
8.65. Calculate the Kb of the ascorbate ion.
7. Sodium hypochlorite is a strong oxidizing agent that is a
fire hazard when in contact with organic materials.
Solutions of sodium hypochlorite are used as bleach and
disinfectant. Determine the hydroxide ion amount
concentration of a sodium hypochlorite solution sold as
household bleach. The bleach bottle label reads “5.25% (by
mass) when packed.”
8. Write the equilibrium law expression for acetic acid reacting
(ionizing) in water. Use Le Châtelier’s principle to explain, in
terms of equilibrium shift, why dilute acetic acid solutions
have a higher percent reaction than more concentrated
solutions. Consider that diluting a solution containing
reacting aqueous entities is very similar to increasing the
volume of a container of reacting gaseous entities.
9. Predict and write the predominant Brønsted–Lowry
reaction in aqueous solutions of the following substances:
(a) K2HPO4(s)
(b) NaH2PO4(s)
(c) Na2HC6H5O7(s)
10. Calculate the pH of 5.0 102 mol/L solutions of each of
the reagents in question 9.
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Interpreting pH Curves
For many acid–base reactions, the appearance of the products resembles that of the
reactants, so we cannot directly observe the progress of a reaction. Also, acids cannot
easily be distinguished from bases except by measuring pH. As you first learned in
Chapter 8, a graph showing the continuous change of pH during an acid–base titration, continued until the titrant is in great excess, is called a pH curve (titration curve)
for the reaction. The pH values and changes provide important information about the
nature of acids and bases, the properties of conjugate acid–base pairs and indicators,
and the stoichiometric relationships in acid–base reactions.
Buffering Regions, Endpoints, and Indicators
The pH curves for acid–base reactions have characteristic shapes. You have learned that
a common reason for plotting a pH curve is to determine the value of the solution’s pH
when an acid–base reaction reaches its equivalence point. We can then use this information
to select an appropriate indicator—one that will produce an easily seen endpoint when
a solution reaches this pH value. Then we can use the same acid–base reaction, done as
a titration to that indicator’s endpoint, for titration analysis.
In all titration analyses, the critical measured quantity is the titrant volume required
to reach the titration endpoint. For accurate analysis, this value needs to be as close as
possible to the theoretically exact value needed to reach the reaction equivalence point.
Analysis titrations, like all empirical work, necessarily involve some uncertainty. We call
any variation between endpoint values and equivalence point values the titration error
of an experiment. Good technique and careful application of knowledge can ensure that
this variation is as small as possible—preferably negligible.
Remember:
16.4
Learning Tip
When you began to study acids
and bases, the term “neutralize”
was understood to refer to an
acid–base reaction that had the
same final pH as neutral pure
water: a pH of 7. Later, you
learned that the pH at the
equivalence point for an acid–
base reaction is almost never 7,
except for the strong acid–
strong base reaction of hydronium ions with hydroxide ions.
As commonly used, the word
“neutralize” is taken to have a
more general meaning: that one
substance can completely or
partially lessen the acidic or
basic properties of another
substance. When we say
sodium carbonate can be used
to neutralize spilled nitric acid
(Figure 1), we mean that
adding the base brings the pH
value up much closer to neutral.
• Endpoint refers to that point in a titration analysis where the addition of titrant is
stopped. The endpoint is defined (empirically) by the observed colour change of an
indicator.
• Equivalence point refers to that point in any chemical reaction where chemically
equivalent amounts of the reactants have been combined. The equivalence point is
defined (theoretically) by the stoichiometric ratio from the reaction equation.
Endpoints are easily detectable because pH changes a great deal, and very abruptly, as the
reaction solution changes (at the equivalence point) from a tiny excess of acid to a tiny
excess of base (or vice versa). But this effect begs the question: Why, for most of a titration, does the pH hardly change at all?
To better understand the information that a pH curve provides, we will use a familiar
reaction about which we already know a great deal. The pH curve for the strong
acid–strong base reaction (Figure 2) has three regions of interest. In the course of this
excess titration, the pH first changes very slowly, then very rapidly, and finally very slowly
again, as titrant is steadily added. Notice that the addition of acid titrant to the base
sample initially has very little effect on the pH of the solution in the flask. In fact, the high
pH has not changed much even after enough acid has been added to react with 90% of
the original amount of the base sample. This nearly level region on a pH curve identifies a buffering region. Buffering is the property of some solutions (often called buffer
solutions) of resisting (counteracting) any significant change in pH when an acid or a
base is added. In this particular case, buffering occurs because any strong acid added
immediately reacts with excess hydroxide ions. Consider that before any acid is added,
the sample solution is primarily water molecules and hydroxide ions.
NEL
Figure 1
Soda ash (sodium carbonate)
is blown onto nitric acid that
has spilled from some railway
tank cars, to neutralize the
highly reactive and dangerous
acid.
Equilibrium in Acid–Base Systems 751
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25.0 mL of 0.48 mol/L NaOH(aq) Titrated
with 0.50 mol/L HCl(aq)
14
12
buffering region
10
8
the curve inflection point
represents the reaction
equivalence point
pH
6
4
Learning Tip
Recall from Unit 4 that the
inflection point of any plotted
curve is the point where the
curvature of the line changes
direction. For instance, on the
strong acid–strong base pH
curve in Figure 2, the line
curves downward until about
24 mL of acid has been added.
At some observable point, the
line curvature changes to
upward.
For this specific titration, the
change in pH is so large and so
rapid that the pH plot becomes
a nearly vertical line. In such a
case the equivalence point can
also be identified as the “midpoint” of the steep drop in the
plotted line.
Also recall that, for this
particular reaction (only), we
already know (from memory)
that the equivalence point pH
must be 7.0 (at SATP), without
actually needing to plot a
titration curve: The reaction of
hydroxide and hydronium ions
is highly quantitative, and the
only product is water.
buffering region
2
0
0
5
Chapter 16
15
20
25
30
35
40
45
50
Volume of HCl (mL)
Figure 2
This pH curve for the continuous and excess addition of 0.50 mol/L HCl(aq) to a 25 mL sample
of 0.48 mol/L NaOH(aq) helps chemists to understand the nature of acid–base reactions.
As long as a large amount of hydroxide ions is present, any added acid is immediately
converted to water, producing a solution that still consists primarily of water molecules
and hydroxide ions. The hydroxide ion concentration decreases a little, but that affects
the solution pH value only very slightly. This pH “levelling effect” finally fails near the
equivalence point—when the hydroxide ions in the solution become almost completely
consumed. But, until then, the solution maintains a (nearly) constant pH. This buffering
property turns out to be critically important for a great number of reactions in applied
chemistry and biology. Note that once excess acid has been added, the solution consists
primarily of water molecules and hydronium ions. Again, the pH remains stable because
adding more hydronium ions now does not change the nature of the solution; it only
increases the hydronium ion concentration slightly. Another buffering region is established—this time at a low pH. You will learn more about the causes of buffering regions
and the nature of buffer solutions later in this section.
Following the first buffering region there is a very rapid change in pH for a very small
additional volume of the titrant. The inflection point on the pH curve represents the
equivalence point of this reaction: At this point, we know from theory that the pH must
be 7. We also know from experience (Chapter 8) that this reaction can be used for titration analysis, if an indicator is selected that changes colour in solution at a pH very close
to 7. As shown in Figure 3, bromothymol blue indicator has a colour change pH range
with a middle value very close to (just slightly below) pH 7, which makes it an appropriate
indicator for this particular reaction. The values below show that the (theoretical) titrant
volume required for complete reaction, as predicted by calculation, should be 24 mL.
Figure 3 shows that if the colour change of bromothymol blue to a green intermediate
colour is the endpoint, then the titrant volume actually measured at the endpoint (24 mL)
is negligibly different from the theoretical predicted value. So, this titration analysis gives
very accurate results.
OH(aq)
752
10
H3O(aq)
25 mL 0.48 mol/L
24 mL 0.50 mol/L
12 mmol
12 mmol
→ 2 H2O(l)
NEL
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Section 16.4
Figure 3
Alizarin yellow is not a suitable
indicator because it will change
colour long before the equivalence
point of this strong acid–strong base
reaction, which theoretically has a
pH of exactly 7. Orange IV is also
unsuitable; its colour change would
occur too late. The pH at the middle
of the colour change range for
bromothymol blue is 6.8, which very
closely matches the equivalence
point pH; so, a titration analysis
endpoint for this reaction, as
indicated by bromothymol blue,
should give accurate results.
25.0 mL of 0.48 mol/L NaOH(aq) Titrated
with 0.50 mol/L HCl(aq)
14
12
alizarin yellow
10
8
bromothymol blue
pH
6
4
orange IV
2
0
0
5
10
15
20
25
30
35
40
45
50
Volume of HCl (mL)
Acid–Base Indicator Equilibrium
Acid–base indicators may be better understood and explained by using Brønsted–Lowry
definitions. Any acid–base indicator is really two entities for which we use the same
name: a Brønsted–Lowry conjugate acid–base pair. At least one (most often both) of
the entities is visibly coloured, so you can tell simply by looking when it forms or is consumed in a reaction. Phenolphthalein is a common example (Figure 4), where the conjugate base form is bright red and the conjugate acid form is colourless. Typically,
common indicators such as methyl red (Figure 5) have quite complex molecular formulas,
so we use a (very simplified) shorthand to identify them, for convenience. For example,
we symbolize the (invisible) acid form of phenolphthalein as HPh, and the (bright red)
base form as Ph. When an indicator equilibrium is shifted to the point where equal
concentrations of both forms exist, an intermediate colour may be observed. For example,
equal concentrations of the blue and yellow forms of bromothymol blue (at a pH of
6.8) appear green to the human eye.
The explanation of the behaviour of acid–base indicators depends, in part, on both the
Brønsted–Lowry concept and the equilibrium concept. An indicator is a conjugate weak
acid–weak base pair formed when an indicator dye dissolves in water. Using HIn(aq) to
Figure 4
Sodium hydroxide solution has been
added to hydrochloric acid
containing phenolphthalein
indicator, which is colourless in
acids and red in bases. The red
colour, indicating the temporary
presence of some unreacted sodium
hydroxide, is rapidly disappearing as
the flask contents swirl and mix.
Figure 5
A few common acid–base indicators
are shown here. Each indicator has
its own pH range over which it
changes colour from the acid form
(HIn) at the lower pH value to the
base form (In) at the higher pH
value. Material Safety Data Sheets
are available for these chemicals.
NEL
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Page 754
represent the acid form and In(aq) to represent the base form of any indicator, we can
write the following equilibrium equation. (Litmus colours are given below the equation as an example.)
conjugate pair
HIn(aq) H2O(l)
Learning Tip
Typically, acid–base indicators
such as methyl red (Figure 6)
are large molecules with quite
complex formulas. Smaller molecules and ions are less likely
to interact with light waves in
the visible spectrum.
The shorthand symbolism we
use to represent indicators is
partly for convenience, but also
partly to reinforce the concept
that indicators are conjugate
acid–base pairs. Consider the
actual formulas for the methyl
red entities, which we represent
as HMr for the acid form, and
Mr for the base form. This
simplified symbolism makes the
proton transfer, which causes
the colour change, more
obvious.
0 In(aq) H3O(aq)
acid
base
red
(litmus colour)
base
acid
blue
According to Le Châtelier’s principle, an increase in the hydronium ion concentration will shift the equilibrium to the left. Then more indicator will change to the colour
of the acid form (HIn(aq)). This change happens, for example, when litmus is added
to an acidic solution. Similarly, in basic solutions the hydroxide ions remove hydronium
ions by reacting with them to form water, with the result that the equilibrium shifts to
the right. Then the base colour of the indicator (In) predominates. Since different indicators have different acid strengths, the acidity or pH of the solution at which an indicator changes colour varies (Figures 5 and 6). These pH values have been measured and
are reported in the table of acid–base indicators on the inside back cover of this book.
H3C
HMr
N
N
N
COOH + H2O
pH < 4.8
H3C
H3C
Mr –
N
N
H3C
N
COO– + H3O+
pH > 6.0
Figure 6
The visible colour of methyl red indicator depends on the equilibrium proportions of its two
coloured forms at the pH of the solution in which it is placed. Methyl red exists predominantly
in its red (acid) form at pH values less than 4.8, and in its yellow (base) form at pH values
greater than 6.0. Between pH values of 4.8 and 6.0, intermediate orange colours occur, as both
forms of the indicator are present in detectable quantities.
Practice
1. The shape of a pH curve is interpreted to describe the change of properties
throughout the course of an acid–base reaction.
(a) In terms of curve shape, describe the characteristics of a buffering region.
(b) In terms of pH change and titrant volume, explain what a buffering region
represents.
(c) In terms of curve shape, describe the characteristics near an equivalence point.
(d) In terms of pH change and titrant volume, explain what an equivalence point
represents.
Use this information to answer questions 2 to 4.
According to the Brønsted–Lowry concept, acid–base indicators are simply coloured
conjugate acid–base pairs. As for all other weak acids, the conjugate acid forms of
these indicators must differ in strength for different entities.
2. For each indicator following, write a Brønsted–Lowry equilibrium equation for the
reaction of the acid form of the indicator with water, to form a hydronium ion and the
754
Chapter 16
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Section 16.4
base form of the indicator. Use the same indicator symbolism as used in the table of
Acid–Base Indicators on the inside back cover of this book. Identify both conjugate
acid–base pairs for each equation, and label all entities “acid” or “base.”
(a) methyl orange
(b) indigo carmine
(c) thymol blue (the entity with two “H’s” in the symbolized formula)
(d) thymol blue (the entity with one “H” in the symbolized formula)
(e) bromothymol blue
3. Which indicator entity, in your answers to question 2, can behave as both an acid and
a base? What chemical term is used to describe this property of an entity?
4. (a) Refer to the table of Acid–Base Indicators to determine which of the five
conjugate acid forms for the indicators in question 2 is the strongest acid. (Note
the colour change pH ranges.) Then rank the other four indicators beneath it in
order of decreasing strength, as in a relative strengths of acids table.
(b) Suppose that five identical samples of an aqueous hydroxide ion solution each
have four drops of a different indicator (from question 2) added. If each sample is
now titrated with the same HCl(aq) titrant, in which indicator/sample combination
will the least amount of titrant cause a colour change?
Polyprotic Entities and Sequential Reactions
Expanding on what you learned in Chapter 8, and adding the Brønsted–Lowry concept,
we now examine some acids and bases that can lose (or gain) more than one proton.
Polyprotic acids can lose more than one proton, and polyprotic bases can gain more
than one proton, in Brønsted–Lowry transfers. If more than one proton transfer actually occurs in the course of a titration, chemists believe the process occurs as a series of
single-proton transfer reactions. Typical pH curves for reactions of diprotic or triprotic
acids and bases differ from those of monoprotic acids and bases—and can provide useful
information about the reactions going on.
We observe (Figure 7) that the pH curve for the addition of HCl(aq) to Na2CO3(aq)
shows two equivalence points—two significant changes in pH. Chemists interpret such
curves as indicating how many quantitative reactions have occurred, sequentially, for
that particular acid–base titration. Here, for example, two successive quantitative reactions have occurred. The two equivalence points evident in Figure 7 can be explained by
two different proton transfer equations.
25.0 mL of 0.50 mol/L Na2CO3(aq) Titrated
with 0.50 mol/L HCl(aq)
12
first reaction
equivalence point
10
8
second reaction
equivalence point
pH 6
methyl orange
4
2
0
0
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
Volume of HCl (mL)
NEL
Figure 7
A pH curve for the addition of
0.50 mol/L HCl(aq) to a 25.0 mL
sample of 0.50 mol/L Na2CO3(aq)
can be used to select an
appropriate indicator for this
reaction done as a titration analysis.
The colour change of methyl orange
(from pH 4.4 to pH 3.2) means it will
show an endpoint that corresponds
closely to the second reaction
equivalence point.
Equilibrium in Acid–Base Systems 755
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Page 756
First, protons transfer from hydronium ions to carbonate ions, the strongest base
present in the initial sample. The first significant drop in pH is evidence that the first reaction is quantitative. We write the Brønsted–Lowry equation in the usual way, starting with
the acid and base entities present in the sample solution, and also listing the H3O(aq)
that is being added.
SA
A
H3O(aq), Cl(aq), Na(aq), CO32(aq), H2O(l)
SB
B
H3O(aq) CO32(aq) → H2O(l) HCO3(aq)
nitric acid,
HNO3(aq)
Because the first reaction is quantitative, essentially all the carbonate ions will have been
consumed at the equivalence point, and an equal quantity of (new) hydrogen carbonate
ions will have been formed. Then, in a second reaction, protons transfer from additional
(continuosly added) hydronium ions to the hydrogen carbonate ions that were formed
in the first reaction.
SA
A
A
B
SB
H3O(aq), Cl(aq), Na(aq), H2O(l), HCO3(aq)
H3O(aq) HCO3(aq) → H2O(l) H2CO3(aq)
sulfuric acid,
H2SO4(aq)
Because the second reaction is also quantitative, all of the hydrogen carbonate ions have
reacted at the equivalence point, and only newly formed carbonic acid molecules remain
in the solution. Continuing to add more hydronium ion after this point will cause no significant further change (no third reaction), because the strongest base in the system
now is water.
The carbonate ion is a diprotic base because it can accept a total of two protons. Other
polyprotic bases include sulfide ions and phosphate ions. The conjugate entities formed
when these bases gain successive protons are:
S2(aq) — HS(aq) — H2S(aq)
phosphoric acid,
H3PO4(aq)
Figure 8
For oxyacids, the protons that
transfer in Brønsted–Lowry
reactions are covalently bonded to
oxygen atoms (identified by a red
colour in the models above). Of
these three examples, the
monoprotic and diprotic acids are
strong acids, and the triprotic acid is
a weak acid.
Acid formulas were originally
derived from stoichiometric
evidence. Arguably, it would be
more informative to write the
molecular formulas differently, as
NO2OH, SO2(OH)2, and PO(OH)3, to
clarify the bonding concept. In both
science and society, however, once
conventions are established, they
are difficult to change.
756
Chapter 16
PO43(aq) — HPO42(aq) — H2PO4(aq) — H3PO4(aq)
Other polyprotic acids include oxalic acid and phosphoric acid (Figure 8).
HOOCCOOH(aq) — HOOCCOO(aq) — OOCCOO2(aq)
H3PO4(aq) — H2PO4(aq) — HPO42(aq) — PO43(aq)
Evidence from pH measurements clearly shows that, for every proton transferred by
a polyprotic entity, the strength of the new acid or base entity formed greatly decreases.
The new entity may be anywhere from 100 to 100 000 times weaker! Chemists believe
that electric charge and electrostatic attraction explain this effect. For example, it should
logically be much easier to pull a positively charged proton away from a neutral oxalic acid
molecule, than it is to pull one away from a negatively charged hydrogen oxalate ion.
Figure 9 shows the pH curve for phosphoric acid titrated with sodium hydroxide.
Phosphoric acid is triprotic, so three reactions should be possible. However, note that only
two equivalence points are evident. We describe the process as before. At the first equivalence point (pH 4), equal chemical amounts of H3PO4(aq) and OH(aq) have been
combined in solution. As the last of the H3PO4(aq) reacts, the pH rises abruptly, because
when the reaction moves past the equivalence point, only newly formed H2PO4(aq)
is now present, and it is a much weaker acid than H3PO4(aq).
NEL
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Section 16.4
25.0 mL of 0.50 mol/L H3PO4(aq) Titrated
with 0.48 mol/L NaOH(aq)
12
11
10
9
8
7
pH 6
5
4
3
2
1
0
0
Figure 9
A pH curve for the addition of
NaOH(aq) to a sample of H3PO4(aq)
displays only two rapid changes in
pH. This result is interpreted as
indicating that there are only two
quantitative proton transfer
reactions of phosphoric acid with
hydroxide ions. The third proton
transfer reaction never goes to
completion, but instead establishes
an equilibrium.
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80
Volume of NaOH (mL)
To write the reaction for the transfer of the first proton, start by listing the entities.
SA
A
Na(aq), OH(aq), H3PO4(aq), H2O(l)
SB
B
OH(aq) H3PO4(aq) → H2O(l) H2PO4(aq)
Since all the H3PO4(aq) has reacted when the reaction has passed the first equivalence
point, the second plateau must represent the reaction of OH(aq) with H2PO4(aq).
The second equivalence point (pH 9) corresponds to the completion of the reaction of
H2PO4(aq) with additional OH(aq) solution. As the last of the H2PO4(aq) reacts, the
pH begins to rise abruptly again because, when the reaction has passed this (second)
equivalence point, only newly formed HPO42(aq) is left, and it is an even weaker acid.
A
SA
B
B
Na(aq), OH(aq), H2O(l), H2PO4(aq)
SB
OH(aq) H2PO4(aq) → H2O(l) HPO42(aq)
No pH endpoint is apparent for the possible reaction of HPO42(aq) with additional
OH(aq). A clue to this lack of a third endpoint can be obtained from the table of acids
and bases. The hydrogen phosphate ion is an extremely weak acid and apparently does
not quantitatively lose protons to OH(aq). For the third reaction, an equilibrium is
established that gradually shifts right as more base is added. There is no pH increase at
a third equivalence point because the reaction never goes to completion.
A
SA
B
B
Learning Tip
When any weak polyprotic acid
or base is initially dissolved in
water, we always assume that
the ionization equilibrium only
involves the first proton
transfer—to or from water. This
is essentially true because the
tendency of any entity that is
formed by a first ionization to
lose or gain a (second) proton
is very much less than that of
the original polyprotic entity
dissolved; so much so that it
can safely be considered
negligible.
Na(aq), OH(aq), H2O(l), HPO42(aq)
SB
OH(aq) H2PO4(aq)
HPO42(aq) OH(aq)
0 H2O(l) HPO42(aq)
>50%
0 PO43(aq) H2O(l)
As a general rule, only quantitative reactions produce detectable equivalence points in an
acid–base titration.
NEL
Equilibrium in Acid–Base Systems 757
Unit 8 - Ch 16 Chem30
11/2/06
DID YOU KNOW
11:10 AM
?
Sulfuric Acid
Sulfuric acid (Figure 10) is probably
the world’s most important
industrial chemical. It is used for so
many things that the industrial
development and standard of living
of a country may be thought of as
being proportional to its sulfuric
acid production. Canada produces
approximately four million tonnes
every year. U.S. production (the
world’s largest) is about 10 times as
much. The main direct consumer
use of sulfuric acid is the 4.5 mol/L
solution inside every standard car
and truck battery.
Sulfuric acid plants (Figure 11)
can be quite compact, and are
assembled from “off-the-shelf”
technology. They involve only a few
simple reactions, starting with sulfur
(in Alberta, mostly extracted from
fossil fuels) and oxygen.
Figure 10
A model of sulfuric acid, showing
relative atom size, bonding, and
molecular shape
Page 758
Sulfuric acid, H2SO4(aq), is a unique polyprotic acid because it is the only common
one for which the first proton loss is already quantitative in aqueous solution; that is, it
is the only strong acid that is polyprotic. Sulfuric acid acts like any of the other five
common strong acids, except that its 100% ionization produces hydrogen sulfate ions,
HSO4(aq), along with hydronium ions, H3O(aq). Recall that of all the anions of this
type (like HCO3(aq), or HPO42(aq)), the hydrogen sulfate ion is the only one that is
a weaker base than water, so it cannot react as a base in aqueous solution. When reacting
as an acid, however, hydrogen sulfate ion is one of the stronger weak acids, and usually
reacts quantitatively with bases (except for very weak bases, of course). So sulfuric acid
will usually, but not always, react with a base (assuming the base is in excess) in two
complete proton transfer reactions, one after the other. When you first began a study
of reactions of acids and bases, you would have written a complete reaction of sulfuric
acid with sodium hydroxide as
H2SO4(aq) 2 NaOH(aq) → Na2SO4(aq) 2 H2O(l)
This reaction equation, if you think about it now, implies that you assumed that both of
this acid’s “hydrogens” (protons) were “replaced” (transferred). You had no information at that time, however, about how this transfer occurs. For simple stoichiometric
calculations based on this reaction, this limited understanding (and simplified equation) worked perfectly well. Explaining and predicting other reactions of sulfuric acid,
however, is a different story. To do so, understanding how the neutralization process
occurs (as two distinct and sequential reactions) is necessary, because both reactions
may not be quantitative when a different (weaker) base is involved, and stoichiometric
calculation can only be done for a reaction that is quantitative.
There are almost no overall reactions of common polyprotic acids or bases that have
more than two definite endpoints. For any such reactions that are actually done as
analysis titrations, an indicator can be selected so that the titration can be stopped at
either chosen equivalence point. For the titration of sodium carbonate solution with
hydrochloric acid, this would mean titrating through two sequential quantitative reactions to the second equivalence point, as shown in Figure 7. For this titration, the second
equivalence point involves a greater pH change, and thus is easier to detect (more accurate). As we found earlier, methyl orange indictor is suitable for detecting this second equivalence point.
For purposes of stoichiometric calculation, we can combine sequential quantitative
Brønsted–Lowry reaction equations into a single “overall” reaction equation. This example
is for the titration of sodium carbonate solution with hydrochloric acid.
1. H3O(aq) CO32(aq) → HCO3(aq) H2O(l) followed by
2. H3O(aq) HCO3(aq) → H2CO3(aq) H2O(l)“totals” to
2 H3O(aq) CO32(aq) → H2CO3 2 H2O(l) (overall reaction)
For purposes of stoichiometric calculation, this equation is equivalent to writing
2 HCl(aq) Na2CO3(aq) → H2CO3(aq) 2 NaCl(aq)
Figure 11
More sulfuric acid is manufactured
in North America than any other
chemical.
758
Chapter 16
because, either way, the reactant acid–base stoichiometric ratio is found to be 2:1.
Chemists normally use the simplest representation that will be useful, so Brønsted–Lowry
equations are more often used for correctly predicting products and equilibria, whereas
standard chemical formula equations are more often used for “doing” stoichiometry as,
for example, in titration analyses.
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Section 16.4
pH Curve Reaction Information
SUMMARY
Empirical pH curves provide a wealth of information:
• initial pH of sample solutions
• pH when excess titrant is added
• number of quantitative reactions
• non-quantitative (equilibrium) reactions
• equivalence point(s) for indicator selection
• buffering regions
Practice
5. How is buffering action displayed on a pH curve?
6. How are quantitative reactions displayed on a pH curve?
7. How is a pH curve used to choose an indicator for a titration?
8. An acetic acid sample is titrated with sodium hydroxide (Figure 12.)
CH3COOH(aq) Titrated with NaOH(aq)
14
12
10
pH
8
6
4
2
0
5
10
15
20
25
30
35
40
45
50
Volume of NaOH (mL)
Figure 12
The pH curve for the addition of 0.48 mol/L NaOH(aq) to 25.0 mL of 0.49 mol/L
CH3COOH(aq) illustrates pH changes during the reaction of a weak acid with a
strong base.
(a)
(b)
(c)
(d)
Based on Figure 12, estimate the pH at the equivalence point.
Choose an appropriate indicator for this titration.
Write a Brønsted–Lowry equation for this reaction.
At the very beginning of the titration, before the curve levels off, it rises. Explain
this rise in terms of entities present in the mixture before and after beginning the
titration.
9. A sodium phosphate solution is titrated with hydrochloric acid (Figure 13 on next
page).
(a) Why are only two equivalence points evident?
(b) Write three Brønsted–Lowry equations for the sequential reactions shown on the
pH curve in Figure 13. Communicate the position of equilibrium for each of the
three reactions.
10. Oxalic acid reacts quantitatively in a two-step overall reaction with a sodium
hydroxide solution. Assuming that an excess of sodium hydroxide is added, sketch a
pH curve (without any numbers) for all possible reactions.
NEL
Equilibrium in Acid–Base Systems 759
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 760
25.0 mL of 0.51 mol/L Na3PO4(aq) Titrated
with 0.50 mol/L HCl(aq)
14
12
10
pH
8
6
4
2
0
0
10
20
30
40
50
60
70
80
90
100
Volume of HCl (mL)
Figure 13
The pH curve for the addition of HCl(aq) to Na3PO4(aq) can be interpreted
using the Brønsted–Lowry acid–base concept.
Learning Tip
pH Curve Shape versus Acid and Base Strength
Recall that, for the examples
you have studied, bromothymol
blue has been an appropriate
indicator for the SA–SB
reaction, because it detects an
endpoint with a pH of 7 very
well. Methyl orange detects
endpoint pH values around 4
well, and is, therefore, useful for
many SA–WB reactions. Finally,
phenolphthalein detects
endpoints with pH values
around 10 well, and is,
therefore, useful for many
WA–SB titrations.
You have learned that the reaction of the strongest possible acid in aqueous solution,
H3O(aq), and the strongest possible base in aqueous solution, OH(aq), is overwhelmingly quantitative, and always results in an equivalence point pH of 7.00 at
25 °C. Not all titrations involve the hydronium and the hydroxide ions, however.
pH curves show that some weak acids (such as acetic acid) react quantitatively with
OH(aq) (Figure 12). The reactions of some weak bases (such as PO43(aq) and
HPO42(aq)) with H3O(aq), can also produce a quantitative reaction (Figure 13). The
strongest of the weak acids also react quantitatively with the strongest of the weak bases.
For example, the reaction of HSO4(aq) with CO32(aq) is a quantitative reaction. The
farther apart the reacting acid and base are on the Relative Strengths of Aqueous Acids
and Bases table (i.e., the larger the difference in Ka), the more likely it is that the reaction
will be quantitative. After plotting many pH curves, chemists have developed the general pH curve reference diagram shown in Figure 14.
Weak and Strong Acids and Bases
SB
Figure 14
Laboratory evidence is generalized
here to show approximations of the
pH curves for quantitative reactions
of weak and strong acids and bases.
Note the initial change in pH for
curves involving “weak” entities, due
to the immediate change of the
kinds of entities present in solution
when the titration begins.
760
Chapter 16
>7
WA — SB
7
SA — SB
WB
pH
<7
SA — WB
SA
WA
SB
WB
WA
SA
strong acid
weak acid
strong base
weak base
Volume of base
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 761
Section 16.4
The (stoichiometric) equivalence point pH for a strong acid–strong base (SA–SB)
reaction is seven (7). The equivalence point pH of any quantitative strong acid–weak
base (SA–WB) reaction is less than seven (7), whereas that of any quantitative weak
acid–strong base reaction (WA–SB) is greater than seven (7).
The composite curves in Figure 14 are not acceptable (do not predict well) for any reactions between a weak acid and a weak base. Weak acid–weak base reactions often do not
have a detectable equivalence point because they are usually not quantitative. For all
intents and purposes, analysis reactions and stoichiometric calculations are not done
for weak acid–weak base reactions, so a pH curve for such a reaction is simply irrelevant.
The pH of a solution at the equivalence point of an acid–base reaction may be explained
by considering the nature of those entities that are present in significant quantities at that
point. The entities present at an equivalence point can be predicted either by following
the five-step Brønsted–Lowry method, or simply by converting a chemical substance
formula equation into a net ionic equation.
Now let’s examine a case of each of the common acid–base reaction types, and explain
the equivalence point pH in each case. Each of the examples used is a quantitative
reaction.
Strong Acid–Strong Base Reaction
For example, nitric acid reacts with potassium hydroxide: SA–SB pH 7.
SA
A
H3O(aq) , NO3(aq), K(aq), OH(aq), H2O(l)
SB
B
H3O(aq) OH(aq) → 2 H2O(l)
or
HNO3(aq) KOH(aq) → H2O(l) KNO3(aq)
H (aq) NO3 (aq) K(aq) OH(aq) → H2O(l) K(aq) NO3(aq)
H(aq) OH(aq) → H2O(l)
At the equivalence point, water is the only acid or base entity present, which produces
a neutral solution (pH 7).
Strong Acid–Weak Base Reaction
For example, hydrochloric acid reacts with aqueous ammonia: SA–WB pH 7.
SA
A
H3O(aq), Cl(aq), NH3(aq), H2O(l)
SB
B
H3O(aq) NH3(aq) → H2O(l) NH4(aq)
or
HCl(aq) NH3(aq) → NH4Cl(aq)
H(aq) Cl(aq) NH3(aq) → NH4(aq) Cl(aq)
H(aq) NH3(aq) → NH4(aq)
At the equivalence point, the only acid or base entity present (other than water) is the
ammonium ion, which is a weak acid, resulting in an acidic solution (pH 7).
NEL
Learning Tip
Recall that entities common to
both sides of the equation are
cancelled out and the
coefficients of the remaining
entities are reduced to the
simplest ratio, in a net ionic
equation.
Equilibrium in Acid–Base Systems 761
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 762
Weak Acid–Strong Base Reaction
For example, acetic acid reacts with barium hydroxide: WA–SB pH 7.
SA
A
CH3COOH(aq), Ba2(aq), OH(aq), H2O(l)
SB
B
CH3COOH(aq) OH (aq) → H2O(l) CH3COO(aq)
or
2 CH3COOH(aq) Ba(OH)2(aq) → 2 H2O(l) Ba(CH3COO)2(aq)
2 CH3COOH(aq) Ba2(aq) 2 OH(aq) →
2 H2O(l) Ba2(aq) 2 CH3COO(aq)
CH3COOH(aq) OH(aq) → H2O(l) CH3COO(aq)
At the equivalence point, the only acid or base entity present (other than water) is the
acetate ion, which is a weak base, resulting in a basic solution (pH >7).
Learning Tip
In an equation for a titration, a
single arrow, →, is used. The
reaction represented by the
equation must be quantitative
in order for any stoichiometric
calculations based on the
equation to be valid. In such
cases, chemical formula
equations are often the most
convenient form.
SUMMARY
•
•
•
•
Titration Generalizations
Strong acid–strong base reactions are quantitative (100%) and have an equivalence point
pH 7.
Strong acid–weak base quantitative reaction equivalence points have a pH 7.
Weak acid–strong base quantitative reaction equivalence points have a pH 7.
Polyprotic entity samples produce sequential reactions in titrations, each of which
may or may not be quantitative.
WEB Activity
Simulation—Titration of Polyprotic Acids and Bases
These two simulations enable you to explore the pH curves that result from various titrations.
Of course, polyprotic acids and bases each have more than one K value, so the shapes of their
titration curves will reflect this. Predict the shapes of the curves before working through the
exercises.
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Practice
11. For the first quantitative reaction in each of the following acid–base titrations, predict
(where possible) whether the equivalence point pH will be greater than, less than, or
equal to 7.
(a) hydroiodic acid aqueous sodium hydrogen phosphate →
(b) boric acid aqueous sodium hydroxide →
(c) aqueous sodium hydrogen sulfate aqueous potassium hydroxide →
(d) hydrochloric acid solid magnesium hydroxide →
(e) hydrosulfuric acid aqueous sodium hydrogen carbonate →
(f) sulfuric acid aqueous ammonia →
762
Chapter 16
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 763
Section 16.4
12. Assume only three indicator solutions are available in your lab: methyl orange,
bromothymol blue, and phenolphthalein. Choose the most suitable indicator for
detecting the equivalence point of each of the following acid–base titrations (for the
first quantitative reaction only).
(a) HBr(aq) Ca(OH)2(s) →
(b) HNO3(aq) Na2CO3(aq) →
(c) KOH(aq) HNO2(aq) →
13. The following formulas each represent a solution at an acid–base reaction
equivalence point. From the entities present in each solution, predict the observed
pH as being greater than, less than, or equal to 7.
(a) NH4Cl(aq)
(b) Na2S(aq)
(c) KNO3(aq)
(d) NaHSO4(aq)
14. Use the five-step Brønsted–Lowry method to predict the overall reaction net ionic
equation when the following chemicals are mixed.
(a) solutions of perchloric acid and sodium carbonate (A pH curve shows two
protons are transferred quantitatively in successive reactions.)
(b) solutions of nitrous acid and potassium hydroxide
(c) solutions of phosphoric acid and sodium hydroxide (A pH curve shows two
protons are transferred quantitatively in successive reactions.)
15. Write a standard chemical formula equation, and also the net ionic equation, for the
predominant reaction that occurs when each of the following pairs of reagents mix.
(a) Stomach acid is neutralized by solid magnesium hydroxide.
(b) Aqueous ammonia is added to sulfurous acid.
(c) A solution of sulfuric acid is neutralized by sodium hydroxide. (A pH curve
indicates two quantitative reactions.)
pH Curve Buffering Regions and Buffer Solutions
A titration pH curve for any quantitative acid–base reaction shows at least one region
where buffering action occurs between the beginning of the titration and the equivalence
point. If the titration is continued much past the equivalence point, it enters another
region of buffering. Such a region represents a solution in which partial reaction has
occurred, and, therefore, contains significant amounts of both entities of a conjugate
acid–base pair. The term buffer refers specifically to the combination of any weak acid
with its conjugate base, in the same solution. Buffer solutions have a specific pH due to
the nature and concentration of the entities present, and change pH very little when
other acids or bases are added—provided that the quantity added is much less than the
quantities of the conjugate pair entities present in the buffer.
For example, consider the titration of acetic acid with sodium hydroxide (Figure 15).
The solution pH is approximately 4.8 when 10 mL of sodium hydroxide solution has
been added. This volume represents one-half of the (calculated) equivalence point
volume of titrant. At this point in the reaction, one-half of the acetic acid originally
present will have reacted (changed to acetate ions), according to the following equation:
OH(aq) CH3COOH(aq) → H2O(l) CH3COO(aq)
The mixture, therefore, now contains approximately equal amounts of the remaining
unreacted weak acid, CH3COOH, and of the conjugate base, CH3COO, produced in the
reaction. Chemists would describe this mixture as an acetic acid–acetate ion buffer.
NEL
Learning Tip
You have observed (Figure 2,
page 752) that the strong acid—
strong base reaction pH curve
shows that the conjugate pair
H3O(aq)—H2O(l) will maintain
a stable (very low) pH; and that
the H2O(l)—OH(aq) pair will
stabilize pH at a very high
value.
Even though these solutions
certainly do produce a
buffering effect, they are not
technically considered to be
buffers, because in each case
one of the two conjugate pair
entities (water) is also the
solution solvent, and is present
in great excess.
A buffer is normally thought
of as an aqueous solution
mixture of a weak acid and its
conjugate base, with both
entities of the conjugate pair
present in significant quantity,
but with both quantities much
less than the quantity of the
solution solvent (water) present.
Equilibrium in Acid–Base Systems 763
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Figure 15
The highlighted plateau shows an
effective buffering region during this
titration of aqueous acetic acid with
sodium hydroxide. At the point
halfway to the equivalence point,
the solution is a 1:1 ratio mix of an
acetic acid–acetate ion buffer.
Learning Tip
Buffers are normally prepared
as mixtures of a weak acid with
a solution of some salt of that
weak acid. Thus, dissolving
solid sodium acetate in a
solution of acetic acid would
produce a buffer solution with
properties the same as those of
the highlighted buffering region
of Figure 15.
Page 764
Titration Curve for Titrating 0.300 mol/L CH3COOH(aq)
with 0.300 mol/L NaOH(aq)
15
14
13
12
11
10
9
8
pH 7
6
5
4
3
2
1
equivalence point
0
5
20
35
15
25
30
Volume of NaOH(aq) (mL)
10
Acetic Acid Buffer
45
50
Buffering can be readily explained using Brønsted–Lowry equations. Suppose a small
amount of NaOH(aq) is added to the acetic acid–acetate ion buffer. Using the five-step
method for predicting the predominant acid–base reaction (page 731), the following
equation is obtained:
SA
(a)
40
A
Na(aq), OH(aq), CH3COOH(aq), CH3COO(aq), H2O(l)
SB
B
B
OH(aq) CH3COOH(aq) → H2O(l) CH3COO(aq)
pH
buffer
capacity
Volume of NaOH(aq)
Figure 15 shows that this reaction is quantitative. A small amount of OH would convert a small amount of acetic acid to acetate ions. The overall effect will be just a small
decrease in the ratio of acetic acid to acetate ions in the buffer and a slight increase in the
pH. The very small change in concentration of each entity of the acid–base conjugate pair
present, and the complete consumption of the added hydroxide ions in the process,
explains why the pH change is small. This buffer would work equally well if a small
amount of a strong acid, such as HCl(aq), were to be added. Evidence indicates that the
reaction that occurs in that case is also quantitative.
SA
(b)
Acetic Acid Buffer
A
A
H3O(aq), Cl(aq), CH3COOH(aq), CH3COO(aq), H2O(l)
SB
B
H3O (aq) CH3COO (aq) → H2O(l) CH3COOH(aq)
pH
buffer
capacity
Volume of HCI(aq)
Figure 16
When an initial chemical amount of
a buffer is eventually depleted, the
pH then changes very quickly.
764
Chapter 16
Added hydronium ion is consumed and the mixture then has a slightly higher ratio of
acetic acid to acetate ions and a slightly lower pH. Figure 16 illustrates the concept of
buffer capacity—the limit of the ability of a buffer to maintain a pH level. When the entity
of the conjugate acid–base pair that reacts with an added reagent is completely consumed, the buffering fails and the pH changes dramatically.
The ability of buffers to maintain a relatively constant pH is important in many biological processes where certain chemical reactions can only occur at a specific pH value.
Many aspects of cell functions and metabolism in living organisms are very sensitive to
pH changes. For example, each enzyme carries out its function optimally over a small pH
range. One essential buffer operates to maintain a stabilized pH in the internal fluid of
all living cells. This critically important buffer is a mixture of dihydrogen phosphate
ions and hydrogen phosphate ions. In mammals, cellular fluid has a pH in the range
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 765
Section 16.4
of 6.9 to 7.4. The H2PO4(aq)–HPO42(aq) buffer system has a pH of 7.2 when the
concentrations of these two conjugate entities are equal. With small variations in concentration, this buffer is effective in maintaining optimum pH levels for the innumerable reactions going on within any given cell. The major buffer system in the blood and
other body fluids (except the cytoplasm within cells) is the conjugate acid–base pair
H2CO3(aq)–HCO3(aq). Blood plasma has a remarkable buffering ability, as shown by
the empirical results in Table 1. Since production of the entities of this buffer in the
body is a continuous process, the buffer is not a limiting reagent, and you do not ordinarily have to worry about using up your buffering capacity.
BIOLOGY CONNECTION
Homeostasis
Initial pH of mixture
Final pH after adding
1 mL of 10 mol/L HCI
neutral saline
7.0
2.0
Many reactions in the human body
produce or consume carbonic acid
or hydrogen carbonate ions. When
one of these entities is depleted in
the blood, Le Châtelier’s principle
comes into effect, and more of that
entity is produced by an
equilibrium shift to keep
everything in balance.
The carbonic acid in solution in
blood is really in equilibrium with
dissolved carbon dioxide:
blood plasma
7.4
7.2
CO2(aq) H2O(l) 0 H2CO2(aq)
Table 1 Buffering Action of Neutral Saline Solution and of Blood Plasma
Solution (1.0 L)
Human blood plasma normally has a pH of about 7.4. Any change of more than 0.4 pH
units, induced by poisoning or disease, can be lethal. If the blood were not buffered, the
acid absorbed from a glass of orange juice would probably be fatal.
WEB Activity
Simulation—Preparation of Buffer Solutions
This simulation shows you how to select an appropriate acid–base conjugate pair to make a
buffer of the desired pH.
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Buffers are also important in many consumer, commercial, and industrial applications. Fermentation and the manufacture of antibiotics require buffering to optimize
yields and to avoid undesirable side reactions. The production of various cheeses, yogurt,
and sour cream are very dependent on controlling pH levels, since an optimum pH is
needed to control the growth of micro-organisms and to allow enzymes to catalyze fermentation processes. Sodium nitrite and vinegar are widely used to preserve food; part
of their function is to prevent the fermentation that takes place only at certain pH values.
The CRC Handbook of Chemistry and Physics provides recipes for preparing buffer
solutions. For example, a buffer with a pH of 10.00 can be prepared by mixing 50 mL of
0.050 mol/L NaHCO3(aq) with 10.7 mL of 0.10 mol/L NaOH(aq). The reaction to establish the buffer is described in the same way as all other acid–base reactions.
SA
SB
GO
+ EXTENSION
The Hydrogen Carbonate
Buffer System
B
The pH of your blood is maintained
at a constant level, courtesy of a
series of equilibrium reactions,
including a buffer system. This
extension outlines the reactions,
and gives an example of how
climbing at high altitude can upset
this delicate balance.
HCO3(aq) OH(aq) → H2O(l) CO32(aq)
3
The buffer preparation recipe reaction converts about 5 of the initial hydrogen carbonate ions into carbonate ions. An aqueous mixture of a 3:2 ratio of these two acid–base
conjugates at the concentrations specified has been found empirically to have an equilibrium pH of precisely 10.00 at 25 °C. Buffer mixtures like this one—with highly
precise pH values—are routinely used to calibrate pH meters for accuracy.
NEL
www.science.nelson.com
A
Na(aq), HCO3(aq), OH(aq), H2O(l)
B
The enzyme carbonic anhydrase
acts as a catalyst to allow this
equilibrium to establish rapidly, or
reestablish quickly, once shifted. It
is possible to make your blood
temporarily too basic by
deliberately breathing heavily for a
long time, thereby causing your
body to lose too much carbon
dioxide. Doctors call this condition
respiratory alkalosis.
There are countless equilibrium
reactions interconnected in this
way in and around living cells.
Biologists refer to this
interdependent network of
reaction equilibria as an example
of homeostasis—the condition of
automatic continual readjustment
of cells, systems, and whole
organisms to very specific
conditions. Biology courses will tell
you a lot more about homeostasis
and the factors that affect it.
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Equilibrium in Acid–Base Systems 765
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 766
WEB Activity
Canadian Achievers—Maud Menten
Maud Menten (Figure 17) was a world-renowned pioneer in explaining the chemical action of
enzymes. What key concept was her chief contribution to understanding enzyme activity?
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Practice
Figure 17
Maud Menten (1879–1960)
16. Give both an empirical and a theoretical definition of a buffer.
17. List two buffers that help maintain a normal pH level in your body.
DID YOU KNOW
?
Commercial Buffers
18. Use the five-step method to predict the quantitative reaction of a carbonic
Buffers are fairly common,
commercially. You can buy buffer
mixtures to adjust the pH of
aquarium water to suit the type of
fish you have. Buffer mixes are sold
to add to hot tubs to control the pH
of the water. Many food recipes mix
ingredients that form natural buffers
within the dough, batter, or sauce.
acid–hydrogen carbonate ion buffer
(a) when a small amount of HCl(aq) is added
(b) when a small amount of NaOH(aq) is added
19. What happens if a large (excess) amount of a strong acid or base is added to a buffer?
20. Use Le Châtelier’s principle to predict what will happen to a benzoic acid–benzoate
ion buffer when a small amount of each of the following substances is added:
(a) HCl(aq)
(b) NaOH(aq)
21. Which of the following solution pairs, when mixed in equal quantities, will not form an
effective buffer?
(a) HNO3(aq) and NaNO3(aq)
(b) C6H5COOH(aq) and NaC6H5COO(aq)
INVESTIGATION 16.3 Introduction
Testing a Buffer Effect
References provide “recipes” for preparing standard buffer
solutions of any desired pH from 1.0 to 13.0. The one used in this
investigation has a pH of precisely 7.0, and might be used to
calibrate a pH meter, for example. If our theory of buffers is
correct, this solution should resist significant change in pH upon
gradual addition of outside acid or base entities.
Purpose
The purpose of this investigation is to test our concept of buffers.
Write the Design, Materials, and table of evidence to match the
Procedure that is provided. The Materials list should include the
size of the equipment used. The buffer is prepared by a reaction
communicated by the following chemical equation:
(c) NH3(aq) and NH4Cl(aq)
(d) HCl(aq) and NaOH(aq)
Report Checklist
Purpose
Problem
Hypothesis
Prediction
H2PO4(aq)
excess
(acid part
of buffer)
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
OH(aq) → HPO42(aq) H2O(l)
limiting
reagent
(base part
of buffer)
Problem
How does the pH change when a strong acid and a strong base
are slowly added separately to an H2PO4(aq)—HPO42(aq)
buffer?
To perform this investigation, turn to page 769.
CAREER CONNECTION
Microbiologist
Microbiologists study, test for, and
isolate microorganisms such as
bacteria, fungi, and viruses. As
part of a medical or research
team, their work is essential for
public health and safety.
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766
Chapter 16
GO
SUMMARY
Brønsted–Lowry Is a Unifying Concept
The five-step Brønsted–Lowry method to explain and predict acid–base reactions is a preferred, acceptable method because it works for all quantitative and non-quantitative
reactions studied so far:
• neutralization reactions
• buffer reactions
• excess reactions
indicator
reactions
polyprotic
reactions
•
•
• titration reactions
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 767
Section 16.4
Section 16.4 Questions
1. Draw a pH curve to illustrate the following information
about acid–base reaction systems.
(a) What is buffering?
(b) Where does buffering appear on a pH curve?
(c) How are quantitative reactions represented on a pH
curve?
(d) Define endpoint and equivalence point.
(e) How is a suitable indicator chosen for a titration?
(f) Do non-quantitative reactions have an endpoint?
Explain your answer briefly.
2. Sketch and label generalized pH curves on a single set of
axes to illustrate the addition of a strong and a weak base
to a strong and a weak acid.
3. Sketch a generalized pH curve for the addition of a strong
base to a weak acid. Label the approximate equivalence
point pH, and suggest an indicator that could be suitable
for titration analysis using this reaction.
4. If the pH of a solution is 6.8, what is the colour of each of
the following indicators in this solution?
(a) methyl red
(d) phenolphthalein
(b) chlorophenol red
(e) methyl orange
(c) bromothymol blue
5. Use Figure 18 to answer the following questions.
(a) How many quantitative reactions have occurred?
(b) Write the chemical equation for each quantitative
reaction.
(c) State the equivalence point pH for each quantitative
reaction.
(d) Choose a suitable indicator to correspond to the
equivalence point pH value(s).
(e) Identify the buffering region(s) and state the chemical
formulas for the entities present in each region.
25.0 mL of 0.46 mol/L Na2SO3(aq) Titrated
with 0.50 mol/L HCl(aq)
10
acid–base reactions. (Use any method that you find
convenient to derive the net ionic equation.)
(a) Oxalic acid is titrated with aqueous sodium hydroxide.
(b) Sodium phosphate is titrated with hydrochloric acid.
(The titration is stopped at the first equivalence point.)
(c) Sodium hydrogen phosphate is titrated with
hydrochloric acid. (An indicator is chosen to detect the
first equivalence point.)
(d) Nitric acid is titrated with aqueous barium hydroxide.
(e) A sulfuric acid spill is neutralized by adding excess lye.
7. Write a formula (non-ionic) equation, and also the
corresponding Brønsted–Lowry equation, to represent the
following acid–base reactions.
(a) Acetic acid is titrated with aqueous sodium hydroxide.
(b) Nitrous acid is titrated with aqueous barium hydroxide.
(c) Sodium carbonate is titrated with hydrobromic acid.
(The titration is stopped after the first quantitative
reaction.)
(d) Carbonic acid is titrated with aqueous sodium
hydroxide. (An indicator endpoint is used to stop the
reaction with a stoichiometric ratio of 1:1.)
(e) A sulfuric acid spill is neutralized by adding excess lye
(caustic soda).
8. State two different applications of buffers.
9. Suggest a compound that could be dissolved in a sulfurous
acid solution to make an effective buffer.
10. Assume that you have a hydrogen carbonate ion solution.
Suggest substances to add to make
(a) a buffer with a lower pH than the original solution
(b) a buffer with a higher pH than the original solution
11. Complete the Design of the following investigation report.
Purpose
The purpose of this investigation is to test buffer concepts
by quantitatively measuring equilibrium shifts for a buffer
system.
Problem
Does the Brønsted–Lowry concept, as applied to buffers,
predict changes in equilibrium concentration of H3O(aq)
in a standard buffer solution, when strong acids or bases
are added?
8
6
pH
Materials
pH 7.00 H2PO4/HPO42
buffer solution
pH meter
1.00 mol/L HCl(aq)
1.00 mol/L NaOH(aq)
4
2
0
0
10
20
30
40
50
60
Volume of HCl (mL)
Figure 18
pH curve for the titration of sodium sulfite solution with
hydrochloric acid
NEL
6. Write a net ionic equation to represent the following
70
80 Extension
12. From a CRC Handbook or any other reference, locate and
copy at least one “recipe” for preparing a buffer solution
that will have a pH of precisely 5.00 at 25 °C.
Equilibrium in Acid–Base Systems 767
Unit 8 - Ch 16 Chem30
11/2/06
Chapter 16
11:10 AM
Page 768
INVESTIGATIONS
INVESTIGATION 16.1
Creating an Acid–Base Strength
Table
An acid–base table organizes common acids (and their conjugate bases) in order of decreasing acid strength (Figure 1).
Acid strength can be tested several ways, including by a carefully designed use of indicators. Predict the order of strengths
using the Relative Strengths of Aqueous Acids and Bases Table
(Appendix I). Use the indicators provided to create a valid
and efficient Design, in which you clearly identify the relevant variables. Evaluate the Design (only), and suggest
improvements if any problems are identified.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (1)
(6) 13 100 mm test tubes
for each indicator or spot
plate (microchem)
test-tube rack
0.10 mol/L Na2CO3(aq)
0.10 mol/L NaOH(aq)
indicators, for example:
methyl orange, methyl
violet, bromothymol
blue, phenolphthalein
Chemicals used include toxic, corrosive, and irritant
materials. Avoid eye and skin contact. If you spill any
of the chemical solutions on your skin, immediately
rinse the area with lots of cool water. In the unlikely
situation of getting some of the chemicals in your
eye, immediately rinse your eye for at least 15 min
and inform your teacher.
Purpose
The purpose of this investigation is to test an experimental
design for using indicators to create a table of relative strengths
of acids and bases.
Problem
Acid–Base Table
Can the indicators available be used to rank the acids and
bases provided in order of strength?
SA
WB
H3O+(aq)
H2O(l)
Materials
lab apron
eye protection
0.10 mol/L HCl(aq)
0.10 mol/L NaHSO4(aq)
0.10 mol/L CH3COOH(aq)
0.10 mol/L NaHSO3(aq)
INVESTIGATION 16.2
Testing Brønsted–Lowry Reaction
Predictions
When predicting products for this investigation, since at least
one reactant is always in solution, list all entities present as they
normally exist in an aqueous environment. For those reactants that are added in solid state, assume that they will dissolve. Use the resulting entities for prediction. Evaluate the
predictions, the Brønsted–Lowry concept, and the five-step
method for acid–base reaction prediction.
Purpose
The purpose of this investigation is to test the Brønsted–Lowry
concept and the five-step method for reaction prediction
from a table of relative acid–base strength.
768
Chapter 16
H2O(l)
OH—(aq)
WA
SB
Figure 1
Simplified acid–base table
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Problem
What reactions occur when the following substances are
mixed? (Hints for diagnostic tests are in parentheses.)
1. ammonium chloride and sodium hydroxide solutions
(odour)
2. hydrochloric acid and sodium acetate solutions
(odour)
3. sodium benzoate and sodium hydrogen sulfate solutions (benzoic acid has low solubility)
4. hydrochloric acid and aqueous ammonium chloride
(odour)
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 769
Chapter 16
INVESTIGATION 16.2 continued
5.
6.
7.
8.
solid sodium chloride added to water (litmus)
solid aluminium sulfate added to water (litmus)
solid sodium phosphate added to water (litmus)
solid sodium hydrogen sulfate added to water
(litmus)
9. solid sodium hydrogen carbonate added to
hydrochloric acid (pH)
10. solid sodium hydrogen carbonate added to sodium
hydroxide solution (pH)
11. solid sodium hydrogen carbonate added to sodium
hydrogen sulfate solution (pH)
INVESTIGATION 16.3
Testing a Buffer Effect
References provide “recipes” for preparing standard buffer
solutions of any desired pH from 1.0 to 13.0. The one used in
this investigation has a pH of precisely 7.0, and might be used
to calibrate a pH meter, for example. If our theory of buffers
is correct, this solution should resist significant change in pH
upon gradual addition of outside acid or base entities.
Design
A prediction is made for each of eleven pairs of substances.
The prediction is then tested using one or more diagnostic
tests, complete with controls. Additional diagnostic tests
increase the certainty of the evaluation.
Chemicals used include toxic, corrosive, and irritant
materials. Avoid eye and skin contact. If you spill any
of the chemical solutions on your skin, immediately
rinse the area with lots of cool water. In the unlikely
situation of getting some of the chemicals in your
eye, immediately rinse your eye for at least 15 min
and inform your teacher. Remember to detect odours
cautiously by wafting air toward your nose from the
container.
Report Checklist
Purpose
Problem
Hypothesis
Prediction
Design
Materials
Procedure
Evidence
Analysis
Evaluation (2, 3)
Procedure
1. Obtain 50 mL of 0.10 mol/L KH2PO4(aq) and 29 mL of
0.10 mol/L NaOH(aq) in separate graduated cylinders.
2. Pour the KH2PO4(aq) and then the NaOH(aq) into a
beaker to prepare a buffer with a pH of 7.
Purpose
The purpose of this investigation is to test our concept of
buffers. Write the Design, Materials, and table of evidence to
match the Procedure that is provided. The Materials list should
include the size of the equipment used. The buffer is prepared by a reaction communicated by the following chemical equation:
4. Add 0.10 mol/L NaCl(aq) as a control into a third
and a fourth test tube.
H2PO4(aq) OH(aq) → HPO42(aq) H2O(l)
6. Add and count drops of 0.10 mol/L HCl(aq) until the
colour changes.
excess
(acid part
of buffer)
limiting
reagent
(base part
of buffer)
Problem
How does the pH change when a strong acid and a strong base
are slowly added separately to an H2PO4(aq)–HPO42(aq)
buffer?
3. Pour an equal volume of the buffer into two test tubes.
5. Add two drops of bromocresol green to one buffer
test tube and one control test tube.
7. Repeat steps 5 and 6 with phenolphthalein and
0.10 mol/L NaOH(aq) using the other two test tubes.
8. Dispose of all solutions down the drain with running
water.
Acids and bases are corrosive and toxic. Avoid skin
and eye contact. If you spill any of the chemical
solutions on your skin, immediately rinse the area
with lots of cool water. In the unlikely situation of
getting some of the chemicals in your eye,
immediately rinse your eye for at least 15 min and
inform your teacher.
NEL
Equilibrium in Acid–Base Systems 769
Unit 8 - Ch 16 Chem30
11/2/06
Chapter 16
11:10 AM
Page 770
SUMMARY
Outcomes
Key Terms
Knowledge
16.1
•
describe Brønsted–Lowry acids as proton donors and bases
as proton acceptors (16.2)
•
write Brønsted–Lowry equations and predict whether
reactants or products are favoured for acid–base equilibrium
reactions (including indicators and polyprotic acids and
bases) (16.2, 16.4)
•
identify polyprotic acids, polyprotic bases, conjugate pairs,
and amphiprotic entities (16.2, 16.4)
•
define a buffer as relatively large amounts of a conjugate
acid–base pair in equilibrium that maintain a relatively
constant pH when small amounts of acid or base are added
(16.4)
•
sketch and qualitatively interpret titration curves of
monoprotic and polyprotic acids and bases, identifying
equivalence points and regions of buffering for weak
acid–strong base, strong acid–weak base, and strong
acid–strong base reactions (16.4)
ionization constant for water, Kw
16.2
Brønsted–Lowry concept
Brønsted–Lowry acid
Brønsted–Lowry base
Brønsted–Lowry reaction equation
amphiprotic
amphoteric
conjugate acid–base pair
16.3
acid ionization constant, Ka
base ionization constant, Kb
16.4
•
define Kw , Ka , and Kb and use them to determine pH, pOH,
[H3O], and [OH–] of acidic and basic solutions (16.1, 16.3)
pH curve
•
calculate equilibrium constants and concentrations for
homogeneous systems and Brønsted–Lowry acids and bases
(excluding buffers) when concentrations at equilibrium are
known, when initial concentrations and one equilibrium
concentration are known, and when the equilibrium constant
and one equilibrium concentration are known (16.1, 16.3)
buffer
•
state that the goal of science is knowledge about the natural
world (16.1, 16.2, 16.3, 16.4)
state that a goal of technology is to solve practical problems
(16.2, 16.3, 16.4)
Skills
•
initiating and planning: design an experiment to show
quantitative equilibrium shifts in concentration under a given
set of conditions (16.4); describe procedures for safe
handling, storage, and disposal of materials used in the
laboratory, with reference to WHMIS and consumer product
labelling information (16.2, 16.4)
•
performing and recording: prepare a buffer to investigate the
relative abilities of a buffer and a control to resist a pH
change when a small amount of strong acid or strong base is
added (16.4)
•
analyzing and interpreting: use experimental data to
calculate equilibrium constants (16.3)
•
communication and teamwork: work cooperatively in
addressing problems and apply the skills and conventions of
science in communicating information and ideas and in
assessing results (16.2, 16.4)
770
buffer capacity
Key Equations
(All at SATP)
Kw [H3O(aq)][OH(aq)] 1.0 1014
STS
•
buffering
Chapter 16
(16.1)
pH pOH 14.00
(16.1)
[H3O (aq)][A (aq)]
(16.3)
For any aqueous acid, HA(aq), Ka = [HA(aq)]
[HB(aq)][OH(aq)]
For any aqueous base, B(aq), Kb = (16.3)
[B(aq)]
For any conjugate acid–base pair,
KaKb Kw
(16.3)
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 771
Chapter 16
MAKE a summary
1. Use a blank page to create a “star” (radial) chart for
quantities and calculations related to Brønsted–Lowry
acid–base reaction systems. Begin by writing H3O(aq)
and OH(aq) symbols about 10 cm apart horizontally,
centred on the page and connected by a line. Draw
lines from each of these symbols to other quantity
symbols that are calculated from them (and from each
other, in some cases). Along each connecting line,
indicate what information the calculation requires and
how it is performed. Include Ka, Kb, pH, and pOH
quantities, but not percent ionization (because it is not
specific to acid–base reactions).
2. Revisit your answers to the Starting Points questions at
the beginning of this chapter. How would you answer
the questions differently now? Why?
Go To
www.science.nelson.com
GO
The following components are available on the Nelson
Web site. Follow the links for Nelson Chemistry Alberta 20–30.
• an interactive Self Quiz for Chapter 16
• additional Diploma Exam-style Review questions
• Illustrated Glossary
• additional IB-related material
There is more information on the Web site wherever you see
the Go icon in this chapter.
+ EXTENSION
Lost Treasures of Tibet
What does acid–base chemistry have to do with ancient works
of art? Watch this video to discover how archeologists and
conservators made use of chemical reactions to restore the
brilliance to Tibetan temple paintings.
www.science.nelson.com
NEL
GO
Equilibrium in Acid–Base Systems 771
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 772
REVIEW
Chapter 16
Many of these questions are in the style of the Diploma
Exam. You will find guidance for writing Diploma Exams in
Appendix H. Exam study tips and test-taking suggestions
are on the Nelson Web site. Science Directing Words used
in Diploma Exams are in bold type.
www.science.nelson.com
GO
DO NOT WRITE IN THIS TEXTBOOK.
Part 1
6. Listed from strongest to weakest, the acids are
1. A hydrated proton is referred to as a
A.
B.
C.
D.
Use the names given and Appendix G to answer questions
6 and 7. Note that some of the ions listed are amphiprotic.
1. phenol
2. sulfide ion
3. cyanide ion
4. ammonium ion
5. hydrogen sulfate ion
6. hydrogen carbonate ion
hydronium ion
hydroxide ion
hydroxyl group
Brønsted–Lowry base
2. The hydroxide ion concentration in a solution of window
cleaner is 2.1 mmol/L. The hydronium ion concentration in
this solution is calculated to be
A. 2.1 1015 mol/L
B. 4.8 1015 mol/L
C. 2.1 1012 mol/L
D. 4.8 1012 mol/L
3. A pH meter indicates that the pH of a soft drink is 3.46. The
pOH of the soft drink is calculated to be
A. 10.46
B. 10.54
C. 14.46
D. 14.54
4. The recipe for preparing a cleaning solution calls for
dissolving 5.0 g of sodium hydroxide in 4.0 L of water. The
pH of the resulting solution is calculated to be
A. 1.51
B. 12.49
C. 13.10
D. 13.90
5. Acid–base theories have developed over the last two
centuries. Which of the following chemists did not make a
significant contribution to acid–base theory?
A. Gilbert Lewis
B. Svante Arrhenius
C. Ernest Rutherford
D. Johannes Brønsted
NR
___ ___ ___ ___.
7. Listed from strongest to weakest, the bases are
NR
___ ___ ___ ___.
8. The nearly level region on a pH curve represents
A.
B.
C.
D.
the endpoint
a region of buffering
an indicator point
the equivalence point
9. The titration of a weak acid with a strong base has an
equivalence point pH of 9.0. Which of the following
indicators would be most suitable for this titration, done as
an analysis?
A. litmus
B. methyl red
C. phenolphthalein
D. alizarin yellow R
10. Which of the following combinations of ions does not
represent a conjugate acid–base pair?
A. H3PO4(aq) and H2PO4(aq)
B. H2PO4(aq) and HPO42(aq)
C. HPO42(aq) and PO43(aq)
D. H3PO4(aq) and HPO42(aq)
11. Which of the following statements about acid–base
titrations is not true?
A. Strong acid–strong base reactions have an
equivalence point pH of 7.
B. Strong acid–weak base reaction equivalence points
have a pH 7.
C. Weak acid–strong base reaction equivalence points
have a pH 7.
D. Strong acid–strong base reactions are quantitative.
12. Which of the following statements about buffers is not
true?
A. Buffers maintain a solution pH at approximately 7.
B. Buffers can be formed by partially neutralizing a weak
acid with a strong base.
C. Buffers maintain a relatively constant pH when small
amounts of acid or base are added.
D. Buffers contain relatively large amounts of a conjugate
acid–base pair in equilibrium.
772
Chapter 16
NEL
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 773
Chapter 16
Part 2
13. Define each of the following substances, according to the
Brønsted–Lowry theory.
(a) an acid
(b) a base
(c) an acid–base reaction
(d) an amphiprotic substance
(e) a strong acid
(f) a strong base
14. Write net ionic equations for each of the following
reactions in aqueous solution. Label the reactants as
Brønsted–Lowry acids or bases. Identify any amphiprotic
ions.
(a) Solutions of sodium hydrogen sulfate and sodium
carbonate are mixed.
(b) Aqueous ammonia is added to a solution of potassium
hydrogen sulfite.
(c) Solutions of sodium hydrogen phosphate and acetic
acid are mixed.
(d) Aqueous sodium hydroxide is added to a solution of
sodium hydrogen phosphate.
15. Identify all acids, bases, and conjugate pairs, and predict
the position of equilibrium (reactants or products favoured)
in each of the following reactions.
(a) CH3COOH(aq) CN(aq) 0 CH3COO(aq) HCN(aq)
(b) HSO3(aq) HPO42(aq) 0 SO32(aq) H2PO4(aq)
(c) NH4+(aq) CO32(aq) 0 NH3(aq) HCO3(aq)
16. Write equilibrium law expressions for each of the following
equilibrium situations.
(a) HF(aq) H2O(l) 0 H3O(aq) F(aq)
(b) NH3(aq) H2O(l) 0 OH(aq) NH4(aq)
(c) H2SO3(aq) H2O(l) 0 H3O(aq) HSO3(aq)
(d) Ca(OH)2(aq) H2O(l) 0 2OH(aq) Ca2(aq)
19. If a sample of acid rain has a pH of 5.0, predict the colour
of each of the following indicators in this solution.
(a) litmus
(b) methyl red
(c) methyl orange
(d) phenolphthalein
20. Aqueous 0.10 mol/L solutions of potassium sulfate and
potassium benzoate are prepared. Predict how the pH
values of the solutions would compare, using the
Brønsted–Lowry proton transfer concept to justify your
answer.
21. Separate samples of a household cleaning solution were
tested with two indicators. Indigo carmine was blue and
phenolphthalein was red in the solution. Estimate the
approximate pH and approximate hydroxide ion
concentration in the solution.
22. Determine the percent reaction, pH, pOH, and hydroxide
concentration in a 0.015 mol/L solution of sodium acetate.
23. Sketch a pH (titration) curve for each of the following
titrations, assuming all the acids and bases have a
concentration of 0.10 mol/L, and all reaction proton
transfers are quantitative.
(a) A diprotic acid is titrated with sodium hydroxide.
(b) A diprotic base is titrated with hydrochloric acid.
24. Design an experiment to determine the relative strengths
of four weak acids.
25. Use the five-step Brønsted–Lowry method to predict the
net ionic equation for the overall reaction when the
following chemicals are mixed.
(a) solutions of sodium sulfate and potassium benzoate
(b) aqueous ammonium nitrate fertilizer and aqueous
sodium phosphate
(c) solutions of citric acid and sodium hydrogen
carbonate
17. Refer to Appendix G for required information to predict
[H3O(aq)] and pH for each of these 0.10 mol/L solutions.
(a) hydroiodic acid
(b) methanoic acid
(c) hydrosulfuric acid
18. A student measures the pH of a 0.25 mol/L solution of
DE
NEL
potassium hydrogen citrate to be 3.42.
(a) Determine the Ka of the hydrogen citrate ion from
this evidence.
(b) How could a student find the pH of a solution without
access to a pH meter? Design a procedure, involving
three indicators, that would establish the approximate
pH of this solution. Criticize your design.
Equilibrium in Acid–Base Systems 773
Unit 8 - Ch 16 Chem30
Unit 8
11/2/06
11:10 AM
Page 774
REVIEW
Many of these questions are in the style of the Diploma
Exam. You will find guidance for writing Diploma Exams in
Appendix H. Exam study tips and test-taking suggestions
are on the Nelson Web site. Science Directing Words used
in Diploma Exams are in bold type.
www.science.nelson.com
GO
DO NOT WRITE IN THIS TEXTBOOK.
Part 1
Use this information to answer questions 6 to 9.
Pure solid arsenic acid, also called orthoarsenic acid, is a
hydrated substance at room temperature with a formula that
1
may be written as H3AsO4•2H2O(s), or alternatively as
(H3AsO4)2•H2O(s). This substance is used in commercial
glassmaking and in making wood-preservative solutions.
Arsenic acid is soluble in water and ionizes according to the
following equation:
H3AsO4(aq) H2O(l) 0 H3O(aq) H2AsO4(aq)
Ka 5.6 103 (18 °C)
1. The following aqueous solutions are each at equilibrium in
sealed containers that are half-full of liquid. For which one
of these may the volume of the liquid solution be
reasonably considered to be the boundary of the closed
system for the equilibrium?
A. HCl(aq)
B. NaOH(aq)
C. H2CO3(aq)
D. NH3(aq)
6. The conjugate base of arsenic acid in the ionization
reaction shown is
A. H3AsO4(aq)
B. H2AsO4(aq)
C. HAsO42(aq)
D. AsO43(aq)
7. Which is the correct form of the equilibrium law for this
acid ionization reaction?
2. Basicity of solutions has been explained by a variety of
theories, each one more comprehensive than its
predecessors. The Brønsted–Lowry concept defines a base
as an entity that is a
A. proton attractor in a specific acid–base reaction
B. proton donor in a specific acid–base reaction
C. hydronium attractor in any acid–base reaction
D. hydroxide producer in any acid–base reaction
A.
[H3O(aq)][H2AsO4(aq)]
Ka [H3AsO4(aq)]
B.
[H3O(aq)][H2AsO4(aq)]
Ka [H3AsO4(aq)][H2O(aq)]
C.
[H3O(aq)]
Ka [H3AsO4(aq)][H2AsO4(aq)]
D.
[H3O(aq)][H2AsO4(aq)]
Ka [H3AsO4(aq)][HAsO42(aq)]
3. For which of the following aqueous solution titrations
would you expect to measure a pH of 7 at the equivalence
point, at SATP?
A. acetic acid titrated with sodium hydroxide
B. nitric acid titrated with potassium carbonate
C. hydrochloric acid titrated with lithium hydroxide
D. ammonia titrated with hydrochloric acid
8. An empirical pH curve for a titration of a sample of
aqueous arsenic acid with a standardized sodium
hydroxide solution shows two distinct endpoints. From this
evidence, it can be inferred that
A. hydrogen arsenate ion is a very weak acid, with a
relatively strong conjugate base
B. arsenate ion is a very weak base, with a strong
conjugate acid
C. an arsenic acid molecule can lose two protons
simultaneously to one hydroxide ion
D. the third sequential proton transfer reaction is
quantitative
4. An industrial reaction process designer has a choice of
several different equilibrium reactions, all of which produce
the same desired chemical substance. These reactions
each have different reaction characteristics. The designer
wishes to produce product as fast as is safely possible. The
best choice from an economic perspective is the reaction
that
A. has a high rate (speed) of reaction and a low percent
yield
B. has a low rate (speed) of reaction and a high percent
yield
C. requires extreme high pressure to improve the percent
yield
D. requires an expensive catalyst to react noticeably
5. What energy condition must be met to maintain a dynamic
equilibrium system?
A. The temperature must be constant.
B. The forward reaction must be exothermic.
C. The container must not conduct heat.
D. The activation energy must be high.
774
Unit 8
9. A student discovers that a pure liquid substance that is
NR
sold as a common agricultural insecticide forms an acidic
solution in water. The measured pH of a 1.00 mol/L
aqueous solution of this weak acid is 5.22. The calculated
Ka value for this acid at this temperature may be expressed
numerically as a.b 10cd. The values (in order) of a, b, c,
and d are _____, _____, _____, and _____.
NEL
Unit 8 - Ch 16 Chem30
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11:10 AM
Page 775
Unit 8
10. In writing an equilibrium law expression, condensed (pure
solid and liquid) phases are ignored because
A. forward and reverse reaction rates must be equal if
they occur at the surface of a solid or liquid
B. solids and liquids are completely separate physically
from the rest of the reaction components, which are
dissolved in each other
C. solid or liquid pure substances have essentially
constant amount concentration values
D. other entities cannot move between molecules of
solids or liquids to collide with their entities
11. Consider the reaction N2(g) 3 H2(g)
0
2 NH3(g), at
equilibrium. If the system container volume were to be
suddenly decreased, the concentration of nitrogen gas in
the system would immediately
A. increase, then decrease to a new constant value
B. remain unchanged, because the reaction is at
equilibrium
C. increase, then remain steady at a new constant value
D. decrease, then remain steady at a new constant value
Use this information to answer questions 12 to 14.
Titration of acid samples with hydroxide ion solution produces
very different pH curves depending on the strength of the
acid sample (Figure 1). These curves represent data from two
separate titrations plotted on the same axes. The two acid
samples were standardized hydrochloric acid and acetic acid
solutions, both with the same initial concentration and
volume. The same sodium hydroxide titrant was used in both
titrations.
12. The strong acid titration curve does not increase noticeably
when the very first addition of NaOH(aq) is made, because
A. OH(aq) was already present in the original sample
solution
B. OH(aq) does not initially react with entities present in
the sample solution
C. reaction of the added OH(aq) creates no new entities
in the sample solution
D. hydronium ions resist any change to their structure
13. During most of these titrations, the pH curves remain
nearly level. The conjugate acid–base pair that is
responsible for creating this buffering region during the
strong acid–strong base titration is
A. H3O(aq)H2O(l)
B. H3O(aq)CH3COOH(aq)
C. CH3COOH(aq)CH3COO(aq)
D. CH3COO(aq)OH(aq)
14. The incorrect statement about the weak acid–strong base
titration is:
A. Initial addition of OH(aq) produces a new entity in
the sample solution.
B. The titrant volume at the reaction equivalence point is
the same as for the strong acid–strong base titration.
C. The solution pH at the reaction equivalence point will
be higher than for the strong acid–strong base
titration.
D. At the equivalence point, the solution pH will be
controlled by the concentration of hydroxide ion,
which is the strongest base present.
Part 2
Titration of Two Acids with NaOH(aq)
15. Formal concepts of acids have existed since the 18th
century. Explain the main idea and the limitations of each
of the following: the oxygen concept; the hydrogen
concept; Arrhenius’ concept; and the Brønsted–Lowry
concept of acids.
16. What happens when scientists find a theory, such as
pH
Arrhenius’ original theory of acids, to be unacceptable?
17. Describe two main ways in which a theory or a theoretical
definition may be tested, in terms of what an acceptable
theory is required to do.
18. According to modern evidence, what is the nature of a
Volume of NaOH (mL)
Figure 1
Titration curves for HCl(aq) and CH3COOH(aq) with
NaOH(aq)
“hydrogen ion” in aqueous solution?
19. Briefly outline how the theoretical definition of a base has
changed from Arrhenius’ original concept to the modified
Arrhenius concept, and subsequently to the
Brønsted–Lowry concept.
20. Aqueous solutions of sodium sulfite and of sodium
carbonate of equal concentrations are prepared. Explain
how the pH values would compare, using the
Brønsted–Lowry proton transfer concept to justify your
answer.
NEL
Chemical Equilibrium Focusing on Acid–Base Systems
775
Unit 8 - Ch 16 Chem30
11/2/06
11:10 AM
Page 776
21. Use the If [procedure], and [evidence], then [analysis]
format (Appendix C.4) to describe a diagnostic test that
could be used to determine which of two solutions (of
equal concentration) is sodium benzoate, and which is
sodium hydroxide.
22. Identify two different examples of conjugate acid–base
pairs, each involving the dihydrogen phosphate ion.
(b) 0.50 mol of PCl3 and 0.30 mol of Cl2 were placed into a
1.0 L container. Once equilibrium had been reached, it
was found that the equilibrium concentration of PCl3
was 0.36 mol/L. Figure 3 describes the change in
[PCl3] over time. Copy the graph. Sketch how the
concentrations of PCl5 and Cl2 change over the same
time period.
23. Which of the following statements are necessarily true
24. From the equation information provided, predict all of the
system changes you might introduce that, according to Le
Châtelier’s principle, will act to shift the equilibria to
maximize the percent yield of the specified product.
(a) production of propene (propylene)
C3H8(g) energy 0 C3H6(g) H2(g)
(b) production of iodine
5 Sn2(aq) 2 IO3(aq) 12 H3O(aq) 0
5 Sn4(aq) I2(s) 18 H2O(l)
25. Consider this system at equilibrium.
PCl5(g) 0 PCl3(g) Cl2(g); Kc 0.40 at 170 °C
(a) One mole of phosphorus pentachloride was initially
placed into a 1.0 L container. Once equilibrium had
been reached, it was found that the equilibrium
concentration of PCl5 was 0.54 mol/L. Figure 2
describes the change in [PCl5] over time. Copy the
graph. Sketch lines to indicate the changing
concentrations of PCl3 and Cl2 over the same time
period.
1.0
Concentration
of PCl5(g)
0.9
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Time
Figure 3
Reaction progress: PCl3(g) Cl2(g)
0
PCl5(g)
26. Consider the equilibrium reaction,
CO(g) H2O(g) 0 CO2(g) H2(g)
Kc 5.0 at 650 °C
In a rigid 1.00 L laboratory reaction vessel, a technician
places 1.00 mol of each of the four substances involved in
this equilibrium. The vessel is heated to 650 °C.
Determine the equilibrium amount concentrations of each
substance, organizing your values in an ICE table.
27. Write chemical formulas and net ionic equations for the
following overall reactions. (If necessary, refer to
Appendix J.)
(a) Vinegar is used to neutralize a drain cleaner spill.
(b) Baking soda solution is used to neutralize spilled rust
remover containing oxalic acid.
(c) An antacid tablet (flavoured calcium carbonate) is
used to neutralize excess stomach acid.
the position of equilibrium (reactants or products favoured)
in each of the following reactions:
(a) HCOOH(aq) CN(aq) 0 HCOO(aq) HCN(aq)
(b) HPO42(aq) HCO3 (aq) 0
H2PO4(aq) CO32(aq)
3
(c) Al(H2O)6 (aq) H2O(l) 0
H3O(aq) Al(H2O)5OH2(aq)
5
Ka 1 10
(d) C6H5NH2(aq) H2O(l) 0
C6H5NH3(aq) OH(aq)
Kb 4 1010
0.7
0.6
0.5
0.4
0.3
0.2
Time
Figure 2
Reaction progress: PCl5(g)
Unit 8
0.9
0.8
28. Identify all acids, bases, and conjugate pairs, and predict
0.8
0.1
776
1.0
Concentration
of PCl3(g)
when products are strongly favoured in an acid–base
equilibrium, assuming equal amount concentrations and
chemical amounts of the initial reactants?
(a) The stronger of the two Brønsted–Lowry bases is a
product.
(b) The equilibrium constant is greater than one.
(c) The forward reaction is exothermic.
(d) The stronger Brønsted–Lowry acid is a reactant.
(e) The percent reaction is greater than 50%.
(f) The pH of the final solution is greater than 7.
(g) The reactant acid is above the reactant base in an
acid–base table.
0
PCl3(g) Cl2(g)
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29. Separate samples of an unknown solution were tested with
hydrochloric acid changed from 1.0 to 4.5 after the addition
of baking soda. Explain these results.
Your response should include
chemical reaction equations to describe the reactions
identification of the amphiprotic entity
two indicators. Congo red was red and chlorophenol red
was yellow in the solution. Predict the approximate pH
and approximate hydronium ion concentration of the
solution.
30. The test for acceptability of any theory is its ability to
explain and predict a wide range of phenomena: to be a
unifying theory. Test the Brønsted–Lowry concept by using
the five-step procedure to write chemical equations
describing and predicting each of the following acid–base
reactions. Design one diagnostic test that could be used
to verify each prediction.
(a) the addition of hydrofluoric acid to a solution of
potassium sulfate
(b) the addition of a solution of sodium hydrogen sulfate
to a solution of sodium hydrogen sulfide
(c) the titration of methanoic acid with sodium hydroxide
solution
(d) the addition of a small amount of a strong acid to a
hydrogen phosphate ion–phosphate ion buffer solution
(e) the addition of colourless phenolphthalein indicator,
HPh(aq), to a strong base
(f) sodium sulfate is dissolved in water
(g) the addition of blue bromothymol blue, Bb(aq), to
vinegar
(h) the addition of washing soda to water
(i) the addition of baking soda to water
•
•
34. Each of seven unlabelled beakers was known to contain
one of the following 0.10 mol/L solutions: CH3COOH(aq),
Ba(OH)2(aq), NH3(aq), C2H4(OH)2(aq), H2SO4(aq), HCl(aq),
and NaOH(aq). Describe diagnostic test(s) required to
distinguish the solutions and label the beakers. Use the
“If ____, and ____, then ____” format (Appendix C.4), a flow
chart, or a table to communicate your answer.
35. Use Figure 4 to answer the following questions.
25.0 mL of 0.50 mol/L Na3PO4(aq)
Titrated with 0.50 mol/L HCl(aq)
14
31. Many acid–base phenomena can be described and/or
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predicted by using Le Châtelier’s principle. Evaluate this
generalization's ability to describe or predict a variety of
reactions.
Your response should include
reference to the following three examples:
NaOH(aq) is added to the following buffer equilibrium:
NH3(aq) H2O(l) 0 NH4(aq) OH(aq)
NaOH(aq) is added to a bromothymol blue indicator
solution:
HBb(aq) H2O(l) 0 H3O(aq) Bb(aq)
NaOH(aq) titrant is added to a vinegar sample:
CH3COOH(aq) H2O(l) 0
H3O(aq) CH3COO(aq)
descriptions of diagnostic tests that would provide
evidence for your predictions
•
•
32. One way to evaluate a theory is to test predictions with
new substances. Sodium methoxide, NaCH3O(s), is
dissolved in water. Predict whether the final solution will
be acidic, basic, or neutral. Explain your answer using a
net ionic equation. (Hint: Think of the methoxide ion as a
hydroxide ion, with a methyl group substituted for the
hydrogen.)
33. In an experimental investigation of amphoteric substances,
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samples of baking soda were added to a solution of
sodium hydroxide and to a solution of hydrochloric acid.
The pH of the sodium hydroxide changed from 13.0 to 9.5
after the addition of the baking soda. The pH of the
(a) Infer how many quantitative reactions have occurred.
(b) Write the Brønsted–Lowry equation for each
successive proton transfer reaction, with appropriate
“arrows” to indicate the extent of each reaction.
(c) Determine the titrant volume at each quantitative
reaction equivalence point.
(d) Identify a suitable indicator to correspond to the
equivalence point pH values.
(e) Identify the buffering region(s), and state the
chemical formulas for the entities present in solution
in each buffering region.
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12
10
pH
8
6
4
2
0
10
20 30 40 50 60 70 80
Volume of HCl(aq) added (mL)
90
100
Figure 4
Graph for question 35
36. Because chlorine–oxygen compounds are toxic to micro-
organisms and react readily with organic materials in food
stains, they find wide application in disinfectants and
bleaches. The reaction of hypochlorous acid with molecules
of coloured substances in stains often produces colourless
products, making the stain “disappear.” Hypochlorous acid
may be produced by the following reaction:
H2O(g) Cl2O(g) 0 2 HOCl(g)
Kc 0.090 (25 °C)
Determine the concentrations of each reagent at
equilibrium at 25 °C if the initial concentrations of both
water vapour and chlorine monoxide were 4.0 mol/L.
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37. Ethyl acetate (ethyl ethanoate) is an ester with a great
many different uses as an organic solvent—for everything
from paint to perfume. It is a product of the following
equilibrium reaction equation:
CH3COOH(os) C2H5OH(os) 0
CH3COOC2H5(os) H2O(os)
The equation shows the reaction taking place at
25 °C, with all components dissolved in a complex nonreacting liquid organic solvent, which we have symbolized
as “os” for convenience. Determine Kc for this reaction if, at
equilibrium, the reagent concentrations are
[CH3COOH(os)]
2.5 mol/L
[C2H5OH(os)]
1.7 mol/L
[CH3COOC2H5(os)]
3.1 mol/L
[H2O(os)]
3.1 mol/L
Hydrochloric acid has also been in common use for a very
long time. It is still sold under its historical name, muriatic acid,
usually as a solution of about 30–35% HCl. It is very useful as a
powerful rust remover, or to adjust pH levels in swimming
pools. It will also “etch” the surface of concrete by reacting
with carbonate ions in the solid mixture, with the result that
paint can adhere to the clean, rough concrete surface. A spill
of this very corrosive acid is always a serious problem
(Figure 5).
38. Phenolphthalein indicator was used for a titration of several
10.00 mL samples of hypochlorous acid with 0.350 mol/L
barium hydroxide solution. An average titrant volume of
12.6 mL was required to reach the observed endpoint of
these trials. According to this evidence, predict the
amount concentration of the hypochlorous acid solution.
39. In a titration, 0.20 mol/L HCl(aq) titrant is to be gradually
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added to a 20.0 mL sample of 0.10 mol/L NaOH(aq).
(a) Sketch a theoretical (calculated) pH curve.
(b) Include the following information as labels on your
sketch. Show all relevant calculations.
(i) the equivalence point pH and titrant volume for
the reaction
(ii) the initial pH of the sodium hydroxide sample
solution
(iii) the pH after adding 5.0 mL of HCl(aq) (treat this
part as a limiting reagent calculation)
(iv) the entities present at the equivalence point
(v) the pH after adding 9.0 mL of HCl(aq)
(vi) the pH after adding 11.0 mL of HCl(aq)
(c) Suggest a suitable indicator for an endpoint
determination for this titration. Indicate the pH values
for the indicator colour change range on your pH
curve.
Use this information to answer questions 40 to 43.
Lime is a very simple substance, and very easily prepared by
strongly heating natural chalk. It has been used throughout
recorded history, and by now the word “lime” appears in
common names for many substances. Lime is calcium oxide,
CaO(s), which is often sold as quicklime, or unslaked lime. This
oxide is hazardous to handle because it reacts very readily and
rapidly with water, releasing a lot of heat. When “slaked” with
water, it forms calcium hydroxide, Ca(OH)2(s), which is much
less dangerous to handle. This compound is widely sold in
garden stores as agricultural, or horticultural, lime—meaning a
rather impure form. It is commonly added to soils to raise the
soil pH value, because the absorption by plants of some
nutrients and trace elements depends heavily on soil pH levels.
Figure 5
Horticultural lime (calcium hydroxide) is quickly sprinkled on a
spill of concentrated muriatic (hydrochloric) acid.
40. The solubility of calcium hydroxide is expressed in the
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following equation, representing a saturated solution
equilibrium at 25 °C:
Ca(OH)2(s) 0 Ca2(aq) 2 OH(aq)
Ksp 1.3 106
(a) Predict whether calcium hydroxide is highly soluble or
only slightly soluble in water.
(b) What is the common name for a saturated calcium
hydroxide solution, and what diagnostic test can be
performed with it?
(c) Explain how the solubility makes this compound safe
to handle, even though hydroxide ion is the strongest
base possible in aqueous solution.
(d) Explain why “liming” of a (moist) garden soil will only
raise the pH by a small amount, but then will continue
to keep it elevated for months.
41. In a concentrated commercial hydrochloric acid solution,
about one of every four molecules is HCl, as calculated
from the mass percent printed on the label. Predict what
fraction of these HCl molecules is actually in the “ready to
react” form of H3O(aq).
42. When neutralizing this acid spill with calcium hydroxide,
identify which compound you would want to be in excess,
and explain why.
43. Write a chemical equation and a Brønsted–Lowry equation
for this neutralization.
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Unit 8
44. A series of experiments with a non-aqueous solvent
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determined that the products are highly favoured in each
of the following acid–base equilibria, as written.
(C6H5)3C C4H4NH 0 (C6H5)3CH C4H4N
CH3COOH HS 0 CH3COO H2S
O2 (C6H5)3CH 0 OH (C6H5)3C
C4H4N H2S 0 C4H4NH HS
51. Chloro-substituted acetic acids are used in organic
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(a) Identify the Brønsted–Lowry acids, bases, and
conjugate acid–base pairs in each of these chemical
reactions.
(b) Arrange the acids in these four chemical reactions in
order of decreasing acid strength (in the solvent
used), as a standard table of acids and conjugate
bases.
Table 1 Comparison of 0.100 mol/L Solutions of Acetic
Acid and the Chloroacetic Acids
45. The hydronium ion concentration of a 0.100 mol/L
n-butanoic (butyric) acid solution was measured to be
1.24 103 mol/L.
HO
C
C
C
C
Determine the percent reaction (ionization) of this
particular weak acid solution, and a Ka value for aqueous
n-butanoic acid at this ambient temperature.
46. A 0.10 mol/L solution of the essential amino acid
tryptophan (1-a-amino-3-indolepropanoic acid) has a
measured pH of 5.19 at 25 °C. Predict the percent reaction
of this tryptophan solution, and the Ka value for aqueous
tryptophan. The molecular formula for tryptophan may be
written as C10H11N2COOH.
47. Glycine, H2NCH2COOH(aq), is a nonessential amino acid,
48. Thioacetic acid, CH3COSH(aq), has a Ka 4.7 104
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at 25 °C.
(a) Write the Ka expression for thioacetic acid.
(b) Calculate the hydronium and thioacetate ion
concentrations, the pH, and the percent reaction in a
2.00 mol/L CH3COSH(aq) solution at 25 °C.
(c) Calculate Kb for the conjugate base, the thioacetate
ion.
49. Predict the percent reaction, pH, pOH, and hydroxide ion
concentration of a 0.012 mol/L solution of sodium
benzoate.
50. A 0.100 mol/L laboratory solution of sodium propanoate,
NaC2H5COO(aq), has a measured pH of 8.95 at 25 °C.
Calculate Kb for the propanoate ion.
Substance
Formula
pH
acetic acid
CH3COOH(aq)
2.89
chloroacetic acid
CH2ClCOOH(aq)
1.94
dichloroacetic acid
CHCl2COOH(aq)
1.30
trichloroacetic acid
CCl3COOH(aq)
1.14
(a) Calculate acid ionization constants for each
chloroacetic acid. Write chemical equations to
describe the reactions of these acids with water.
(b) Suggest a theoretical explanation for the relative
strengths of this series of acids. (Hint: Would adding
chlorines to the other end of the molecule make the
—COOH end of the molecule more, or less, negative?)
(c) A chemical technician is assigned the design of an
acid–base titration to determine the amounts of each
acid present in the mixture. She knows from
experience that the reaction of acetic acid with a
strong base is quantitative. Sketch a simplified pH
curve for titration of a mixture of the four acids
produced by chlorinating acetic acid. How could the
technician determine relative chemical amounts of the
acids present in the mixture?
O
with the simplest structure of all the amino acids, and has
a Ka 4.5 107 at 25 °C. Calculate the hydronium ion
concentration, the pH, and the percent reaction in a
0.050 mol/L aqueous solution of glycine at 25 °C.
synthesis, cleaners, and herbicides. These acids are
prepared by the chlorination of acetic acid in the presence
of small amounts of phosphorus. This process is known as
the Hell–Volhard–Zelinsky reaction. As is typical of organic
synthesis reactions, at equilibrium, the reaction vessel
contains a mixture of all the different possible reaction
products as well as some unreacted acetic acid and
chlorine. When these acids were studied separately, the
data in Table 1 were obtained.
52. Liquid ammonia can be used as a solvent for acid–base
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reactions.
(a) Predict the strongest acid species that could be
present in this solvent. (Consider the parallel with
liquid water, H2O(l). Also consider the likely reaction of
a very strong proton donor, such as hydrogen
chloride, when it dissolves and reacts quantitatively in
liquid ammonia.)
(b) The ionization equilibrium of pure liquid ammonia is
similar to that of pure liquid water. Write the
equilibrium equation for the ionization of liquid
ammonia.
(c) Predict the strongest base that could be present in
liquid ammonia solution.
(d) Sketch a titration curve for the addition of the
strongest acid in ammonia solution to the strongest
base. Suggest what value, instead of pH, might be
used on the vertical axis of your graph.
53. Review the focusing questions on page 670. Using the
knowledge you have gained from this unit, briefly outline a
response to each of these questions.
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779