Download 15.0 EquilibriumIHS2014

+
Unit D: Equilibrium
Textbook Reference: Chapters 15 and 16
+ Equilibrium DAY 1
1.1K: Define equilibrium and state the criteria that apply to a chemical system in
equilibrium; i.e., closed system, constancy of properties, equal rates of forward and reverse
reactions
1.2K: Identify, write and interpret chemical equations for systems at equilibrium
1.4K: Define Kc to predict the extent of the reaction and write equilibrium-law expressions for
given chemical equations, using lowest whole-number coefficients
+
Chemical Equilibrium
100%

Not all reactions are quantitative (reactants  products)


Evidence: For many reactions reactants are present even after the
reaction appears to have stopped
Recall the conditions necessary for a chemical reaction:

Particles must collide with the correct orientation and have
sufficient energy

If product particles can collide effectively also, a reaction is said
to be reversible

Rate of reaction depends on temperature, surface area and
concentration
+
Chemical Equilibrium

Consider the following reversible reaction:

The final state of this chemical system can be explained as a
competition between:
The collisions of
reactants to form
products

The collisions of
products to re-form
reactants
We assume this system is closed (so the reactants and products
cannot escape) and will eventually reach a:
DYNAMIC EQUILBRIUM
- Opposing changes occur simultaneously at the same rate
+
Chemical Equilibrium

Consider the following hypothetical system:
AB + CD  AD + BC forward reaction, therefore
AD + BC  AB + CD reverse reaction
•
Initially, only AB and CD are present. The forward reaction is occurring exclusively
at its highest rate.
•
As AB and CD react, their concentration decreases. This causes the reaction rate to
decrease as well.
•
As AD and BC form, the reverse reaction begins to occur slowly.
•
As AD and BC’s concentration increases, the reverse reaction speeds up.
•
Eventually, both the forward and reverse reaction occur at the same rate =
DYNAMIC EQUILIBRIUM
+
4 Conditions of Dynamic Equilibrium*
1.
Can be achieved in all reversible reactions when the rates
of the forward and reverse reaction become equal
Represented by
rather than by 
2.
All observable properties appear constant (colour, pH, etc)
3.
Can only be achieved in a closed system (no exchange of
matter and must have a constant temperature)
4.
Equilibrium can be approached from either direction. This
means that the equilibrium concentrations will be the same
regardless if you started with all reactants, all products, or a
mixture of the two
Types of Equilibrium
1.
2.
Phase Equilibrium: a single substance existing in
more than 1 phase

Example: Liquid water in a sealed container with
water vapour in the space above it

Water evaporates until the concentration of water
vapour rises to a maximum and then remains
constant
Solubility Equilibrium: a saturated solution

Rate of dissolving = rate of recrystallization
Types of Equilibrium
3.
Chemical Equilibrium – reactants and products in a closed system

Example: The Hydrogen-Iodine Equilibrium System
The rate of reaction of the reactants decreases as the number of reactant molecules
decrease. The rate at which the product turns back to reactants increases as the number
of product molecules increases. These two rates become equal at some point, after which
the quantity of each will not change.
+
Describing the Position of Equilibrium
1.
Percent Yield- the yield of product measured at equilibrium
compared with the maximum possible yield of product.

% yield = product eq’m
x 100 %
product max

The equilibrium concentration is determined experimentally,
the maximum concentration is determined with stoichiometry
+
Describing the Position of Equilibrium
1.
Percent Yield – Example

If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas
in a 1.00L vessel, what is the percent yield if 3.90 mol of
hydrogen iodide is present at equilibrium

% yield = product eq’m
x 100 %
product max

H2(g) + I2(g)
2HI(g)
2.50 mol x (2 mol HI) = 5.0 mol HI
1 mol H2(g)
% yield = 3.90 mol x 100 %
5.00 mol
= 78%
+
Describing the Position of Equilibrium
2.
Using an Equilibrium Constant, (Kc)

Think “products over reactants”

If the Kc > 1, the equilibrium favours products

If the Kc < 1, the equilibrium favours reactants

If the Kc > 100, the reaction is Quantitative

If the Kc > 1/100, the reaction is considered non-spontaneous

WHAT if Kc = 1 ??
+
Describing the Position of Equilibrium

Kc is a constant value at a given temperature. It only varies with
changes in temperature which is why temp is always stated for Kc.

The following do NOT affect Kc:
concentration, catalyst, or time.

Limitation- Kc tells you the position of a reaction it does not tell you
the rate of a reaction. Eg. a large Kc does not mean the reaction is
fast only that it favors products.
+
Describing the Position of Equilibrium
2.
Using an Equilibrium Constant, (Kc)

Example #1: Write the equilibrium law expression for the
reaction of nitrogen monoxide gas with oxygen gas to form
nitrogen dioxide gas.
+
Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
2.

Note: The Kc value describes the extent of the forward reaction.
Kc reverse =
1 .
= The reciprocal value
Kc forward

Example #2: The value of Kc for the formation of HI(g) from H2(g)
and I2(g) is 40, at a given temperature. What is the value of Kc for
the decomposition of HI(g) at the same temperature.
Kc reverse =
1
.
Kc forward
=
1
40
= 0.025
+
Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
2.

Note: For heterogeneous equilibrium systems, DO NOT
include liquids and solids in the expression. (They are
assumed to have fixed concentrations)

Example #3: Write the equilibrium law expression for the
decomposition of solid ammonium chloride to gaseous
ammonia and gaseous hydrogen chloride
The solid is
omitted from
the expression

Example #4: Write the equilibrium law expression for the
reaction of zinc in copper(II) chloride solution.
The solids, as well as
spectator ions (Cl-)
are omitted from the
expression
+
Describing the Position of Equilibrium
+
Describing the Position of Equilibrium
GUIDED PRACTICE
Determine the equilibrium law and constant for the
following reactions.
2NO(g) + O2(g) <--> 2 NO2(g)
0.3 mol/L
0.5 mol/L
0.8 mol/L
Mg(s) + Cu2+(aq) <--> Cu(s) + Mg2+ (aq)
2.0 M
4.0 M
Hg(l) + Zn2+(aq) <--> Zn(s) + Hg2+(aq)
3.5 M
7.0 M
C6H6(l) + Br2(l) <--> C6H5Br(l) + HBr(l)
+
PRACTICE

Read Section 15.1
Equilibrium Concentrations
+ DAY 2
2.3K: Calculate equilibrium constants and concentrations for homogeneous systems when
• concentrations at equilibrium are known
• initial concentrations and one equilibrium concentration are known
• the equilibrium constant and one equilibrium concentration are known.
+
Calculations in Equilibrium Systems

Using the equilibrium law expression to determine whether
a system is at equilibrium:

Substitute in the given concentrations to the equilibrium
expression. If the value is the equilibrium constant, the system is
at equilibrium

If the value is larger, this means there are more products that
reactants. To reach equilibrium, the reaction must proceed to the
left (towards the reactants)

If the value is smaller, this means there are more reactants than
products. To reach equilibrium, the reaction must proceed to the
right (towards the products)
+
Calculations in Equilibrium Systems

Example #1: In the following system:
N2(g) + 3H2(g) ↔ 2NH3(g)

0.249 mol N2(g), 3.21 X 10-2 mol H2(g) and 6.42 X 10-4 mol NH3(g)
are combined in a 1.00 L vessel at 375oC, Kc = 1.2

Is the system at equilibrium?
Kc = NH3 2
N2(g) H2(g) 3

= (6.42 x 10 -4)2 =0.0500
(0.249)(3.21 x 10-2)3
If not, predict the direction in which the reaction must
proceed.

Value is below Keq = therefore more reactants than products, so
reaction must proceed to the right
+
Calculations in Equilibrium Systems

Example #2: Find the equilibrium concentration of the ions
that are formed when solid silver chloride is dissolved in
water. The equilibrium constant for this reaction is
Kc = 5.4 X 10-4.
AgCl(s)  Ag+(aq) + Cl-(aq)
Kc = Ag+(aq)
Cl-(aq)
5.4 x 10 -4 = x2
0.023 mol/L = x = Ag+(aq) eq = Cl-(aq)
eq
+ ICE Tables and Equilibrium Calculations
STEPS:

Always write out the equilibrium reaction and equilibrium
law expression if not given.

Draw an ICE Table (Initial, Change in and Equilibrium
concentrations) (I + C = E)

Substitute values where appropriate

Solve for x

Solve for equilibrium concentrations
+ ICE Charts and Equilibrium Calculations

Example #3: Consider the following equilibrium at 100 oC:
N2O4(g) ↔ 2 NO2(g)

2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb.
After equilibrium was established, only 1.6 mol of N2O4(g)
remained. What is the value of Kc?
2.0 mol = 1.0 mol/L (I)
2.0L
1.6 mol = 0.8 mol/L (E)
2.0L
N2O4(g)
I:
1.0 mol/L
C:
E:
-x
Solve for Kc = (0.40)2
(0.80)
= 0.20
0
+ 2x
1.0 – x = 0.80
E: 1.0 – x = 0.80 solve for x
2NO2(g)
x = 0.20 2x = 0.40
2x
+ ICE Charts and Equilibrium Calculations
Example #4
 A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of
SO3(g). The gases then reach equilibrium according to the following
equation:
2SO2(g) + O2(g) ↔ 2SO3(g)
 At equilibrium, the bulb was found to contain 2.6 mol of SO2(g).
Calculate Kc for this reaction.
4.0 mol = 0.40 mol/L (I)
10.0L
2.6 mol = 0.26 mol/L (E)
10.0L
0.40 – 2x = 0.26
X = 0.07
2.2 mol = 0.22 mol/L (I)
10.0L
2SO2(g)
O2(g)
2SO3(g)
I:
0.40
0.22
0.56
C:
- 2x
- x
+ 2x
0.22 - x
0.56 + 2x
0.15
0.70
E: 0.40 – 2x = 0.26
E:

Kc
=
5.6 mol = 0.56 mol/L (I)
10.0L
(0.70)2
(0.15)(0.26)2
0.26
= 48
+ ICE Charts and Equilibrium Calculations
Example #5
 In a certain experiment 1.000 mol of NH3(g) was placed in an empty
1.000L flask at 500oC. After equilibrium was reached, 0.399 mol of N2(g)
was found to be in the flask. Calculate the concentrations of NH3(g)
andH2(g) at equilibrium by setting up an ICE table.
1.0 mol = 1.0 mol/L (I)
1.0L
0.399 mol = 0.399 mol/L (E)
1.0L
0.0 + x = 0.399
X = 0.399
0 mol = 0.0 mol/L (I)
1.0L
2NH3(g)
0 mol = 0.0 mol/L (I)
1.0L
N2(g)
3H2(g)
I:
1.0
0.0
0.0
C:
- 2x
+ x
+ 3x
0.0 + x
0.0 + 3x
0.399
1.197
E: 1.0 – 2x =
E:
0.202
+ ICE Charts and Equilibrium Calculations
Example #6
 In a certain experiment, 3.00 mol of SO2(g), 2.00 mol of O2(g) and 4.00
mol of SO3(g) were placed in an empty 1.000L vessel at 500oC. After
equilibrium was reached, 5.00 mol of SO3(g) was found to be in the flask.
Calculate the concentrations each entity at equilibrium by setting up an
ICE table.
+
PRACTICE
Chemical Equilibrium WS
+ ICE Charts and Equilibrium Calculations
Example #6
 In a certain experiment, 3.00 mol of SO2(g), 2.00 mol of O2(g) and 4.00
mol of SO3(g) were placed in an empty 1.000L vessel at 500oC. After
equilibrium was reached, 5.00 mol of SO3(g) was found to be in the flask.
Calculate the concentrations each entity at equilibrium by setting up an
ICE table.
3.00 mol = 3.00 mol/L (I)
1.0L
5.00 mol = 5.00 mol/L (E)
1.0L
4.00 + 2x = 5.00
X = 0.500
2.00 mol = 2.00 mol/L (I)
1.0L
4.00 mol = 4.00 mol/L (I)
1.0L
2SO2(g)
O2(g)
2SO3(g)
I:
3.00
2.00
4.00
C:
- 2x
- x
+ 2x
2.00 – x
4.00 + 2x
1.50
5.00
E: 3.0 – 2x =
E:
2.00
+ ICE Charts and Equilibrium Calculations


Example #7: Using a perfect square
Given the following reaction:
N2(g) + O2(g) ↔ 2NO(g)

Determine the equilibrium concentrations for all species present given that
the initial concentration of each reactant is 0.200 mol/L.
N2(g)
O2(g)
2NO(g)
I:
0.200
0.200
0
C:
-x
- x
+ 2x
0.200 - x
2x
0.195mol/L
0.00976mol/L
E:
E:


Kc = 0.00250
0.00250 =
0.200 - x
0.195mol/L
(2x)2 square root both sides
(0.200-x)2
0.05 =
2x
0.200 – x
= 0.01 – 0.05x = 2x
= 0.01 = 2.05x
HOW DOES THE Kc value relate to the amount of change ??
= 0.00488
+ ICE Charts and Equilibrium Calculations


Example #8: Using the Approximation Rule
Calculate the concentration of gases produced when 0.100 mol/L COCl2(g)
decomposes into carbon monoxide and chlorine gas. The Kc for this
reaction is 2.2 X 10-10.
0.100
= >1000
-10
2.2x 10


2.2 x10-10 =
COCl2(g)
CO(g)
Cl2(g)
I:
0.100
0
0
C:
-x
+ x
+x
x
x
E:
0.100 - x
E:
0.100 mol/L
(x)2
(0.100-x)
2.2 x 10-10 = x2 =
0.100
x
4.7 x 10-6 mol/L 4.7 x 10-6 mol/L
Approximation Rule: When the reactants/Kc is larger than 1000, you can
disregard the change in concentration for the initial entities.
So 0.100 – x = 0.100
= 4.69 x 10-6 mol/L
+ ICE Charts and Equilibrium Calculations


Example #9: Using Graphing to solve
Gaseous NOCl decomposes to form NO(g) and Cl2(g). At 35oC the equilibrium
constant Keq = 1.60 X 10-5. If 1.00 mol of NOCl(g) is placed in a 1.0 L flask, what
are the equilibrium concentrations of each species
2NOCl(g)
1.00 = >1000
1.6x 10 -5
1.00
0
0
C:
- 2x
+ 2x
+x
0 + 2x
0+x
1.00 – 2x
E:
1.6 x10-5 =
Cl2(g)
I:
E:

2NO(g)
(2x)2 (x)
(1.00-2x)2
+
PRACTICE

Pg 682 Q 4, 6

Lab Excerise 15.B pg 686

Text Questions # 1-10 pg 688-689

Finish Equilibrium WS

Quiz on ICE tables Friday 

Coming Up Tomorrow

Le Chatelier’s Principle 

Chapter 15 Exam
+ Le Chatelier’s Principle
1.3K: Predict, qualitatively, using Le Chatelier’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
+
Le Chatelier’s Principle

Le Chatelier’s Principle is very useful in predicting how a
system at equilibrium will respond to change.

It states that when a system at equilibrium is disturbed, the
equilibrium shifts in the direction that opposes the change,
until a new equilibrium is reached.

There are three common ways an equilibrium may be
disturbed:

Change in the concentration of one of the reactants or products

Change in the temperature

Change in the volume of a container

Addition of a catalyst (less common)
Changes in the temperature, will
change the Kc value. No other
changes will affect this value.
+

Le Chatelier’s Principle
Effect of Changes in Concentration

If a system at equilibrium is disturbed by the addition of a reactant (or the
removal of a product), Le Chatelier’s principle predicts that the equilibrium
will shift right.
2N2O(g) + 3O2(g)

If the disturbance is the removal of a reactant (or the addition of a product),
Le Chatelier’s principle predicts that the equilibrium will shift left.
2N2O(g) + 3O2(g)

4 NO2(g)
4 NO2(g)
Since concentrations of solids are constants and do not appear in
expressions for K, removing or adding some solid does not cause shifts.
+
Concentration Change
If you see spike in either a reactant or
product, which causes a gradual change
in the other entities – a substance has
been added or removed.

Addition of reactant, HF(g) –
will shift equilibrium to the
right

Removal of product, HCl(g),
will shift the equilibrium to the
right

More products will be
produced, and a new
equilibrium is established

More products will be
produced, and a new
equilibrium is established
+ Le Chatelier’s Principle

Effect of Changes in Temperature

In endothermic reaction, heat acts like a reactant. Increasing the
temperature shifts the reaction right. Decreasing the temperature, shifts the
reaction left
Heat + 2N2O(g) + 3O2(g)

In exothermic reactions, heat acts like a product. Increasing the
temperature shifts the reaction left. Decreasing the temperature, shifts the
reaction right.
4 NO2(g)

4 NO2(g)
2N2O(g) + 3O2(g) + Heat
The equilibrium constant, Kc is temperature dependent
Remember K gets
bigger if there are more
products being created
(i.e. shifts to the right)
Reaction Type
Role of heat
Effect of T 
Effect of T 
Endothermic
Reactants + heat  products
K
K
Exothermic
Reactants  products + heat
K
K
+
Temperature Change

The temperature decreases, at the time indicated by the
dotted line.

This will cause the equilibrium to shift to the right, creating
more products, until a new equilibrium is established.
+ energy
If you see a gradual change in
the reactants with an opposite
change in the products, you have
a temperature change going on
+ Le Chatelier’s Principle

Effect of Changes in the Volume of the Container

If volume is decreased, pressure increases (Boyle’s Law – Chem 20)
So the reaction will shift in the direction which contains the fewest moles of gas
Pressure

2N2O(g) + 3O2(g)
4 NO2(g)
4 moles are
fewer than 5
If volume is increased, pressure decreases (Boyle’s Law – Chem 20)
So the reaction will shift in the direction which contains the most moles of gas
Pressure

2N2O(g) + 3O2(g)
4 NO2(g)
5 moles are
more than 4
If both sides of the equation have the same number of moles of gas, the
change in volume of the container has no effect on the equilibrium.
+
Volume Change

The volume of the container decreased, at the time indicated
by the dotted line. This will cause a pressure increase.

This will cause the equilibrium to shift to the right, the side
with fewer moles of gas, creating more products, until a new
equilibrium is established.
If you see a spike in all of the
entities = P increase, V decrease
If you see a drop in all of the
entities = P decrease, V increase
+ Le Chatelier’s Principle

Effect of the Addition of a Catalyst

Catalysts speed up the rate at which equilibrium is obtained, but have
no effect on the magnitude of K. They increase both the forward and
backward rate of reaction.
+
+

Identify the nature of the changes imposed on the following
equilibrium system at the four times indicated by
coordinates A, B, C and D
• At A, the concentration (or pressure) of every chemical in the system is decreased by
increasing the container volume. Then the equilibrium shifts to the left (the side with
more moles of gas)
• At B, the temperature is increased. Then the equilibrium shifts to left.
• At C, C2H6(g) is added to the system. Then the equilibrium shifts to the left.
• At D, no shift in equilibrium position is apparent; the change imposed must be addition
of a catalyst, or of a substance that is not involved in the equilibrium reaction.
+
Practice
Change
Direction of Shift (, ,
or no change)
Effect on quantity
a) Decrease in volume
Kc
b) Raise temperature
Amount of CO(g)
c) Addition of I2O5(s)
Amount of CO(g)
d) Addition of CO2(g)
Amount of I2O5(s)
e) Removal of I2(g)
Amount of CO2(g)
Effect (increase, decrease,
or no change)
+
Practice

Le Chat WS

Text questions

Pg 695 Q 1-3

Pg 699 Q 1-3, 5-7
Coming Up

Quiz Tomorrow – ICE tables

Review Le Chat

Friday – Study day

Monday Exam Chapter 15
+ Acids and Bases - Kw
2.1K: Recall the concepts of pH and hydronium ion concentration and pOH and hydroxide ion
concentration, in relation to acids and bases
2.2K: Define Kw to determine pH, pOH, [H3O+] and [OH–] of acidic and basic solutions
+ The Water Ionization Constant, Kw

Even highly purified water has a very slight conductivity. This
is due to the ionization of some water molecules into hydronium
ions and hydroxide ions.

The water ionization equilibrium relationship is so important, it gets its own special
symbol and name: Ionization Constant for Water, Kw
+
The Water Ionization Constant, Kw

Since the mathematical relationship is simple, we can easily
use Kw to calculate either the hydronium or hydroxide ion
concentration, if the other concentration is know.
The presence of
substances other than
water decreases the
certainty of the Kw value
to two significant digits;
1.0 x 10 -14
+
Kw Calculations

Example: 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.
+
Kw Calculations


Example #2: Calculate the amount concentration of the
hydronium ion in a 0.25 mol/L solution of barium hydroxide.
Ba2+ = 0.25 mol/L x 2 mol = 0.50 mol/L
1 mol
+
Kw Calculations

Example #3: Determine the hydronium ion and hydroxide
ion amount concentration in 500 mL of an aqueous solution
for home soap-making containing 2.6 g of dissolved sodium
hydroxide.
+
Do You Remember? pH and pOH
+ Acid Strength as an Equilibrium Position

Do you remember the difference between strong and weak acids?

Strong acids – ionize completely (quantitatively), even though this
could be written with a double arrow, it is simpler to use a single
arrow to show the the reaction is >99.9%


Strong Acids: HClO4(aq), HI(aq), HBr(aq) , HCl(aq) , HSO4(aq), HNO3(aq)
Weak acids – ionize (react with water) only partially (<50%)
+
% Ionization

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.
+
% Ionization

The pH of 0.10 mol/L methanoic acid solution is 2.38.
Calculate the percent reaction for ionization of methanoic
acid.
+
Practice

pH and pOH Review WS

Kw, pH and pOH WS

Pg. 716 #1,-6

Pg. 718 #7-8
+ Bronsted-Lowry Acids and Bases
1.5K: Describe Brønsted–Lowry acids as proton donors and bases as proton acceptors
1.6K: Write Brønsted–Lowry equations, including indicators, and predict whether reactants or
products are favoured for acid-base equilibrium reactions for monoprotic and polyprotic
acids and bases
1.7K: Identify conjugate pairs and amphiprotic substances
+
Bronsted-Lowry Acid-Base Concept

Focuses on the role of the chemical species in a reaction
rather than on the acidic or basic properties of their aqueous
solutions.

Bronsted Lowry Definition for an Acid: proton donor

Bronsted Lowry Definition for an Base: proton acceptor
+
Bronsted-Lowry Acid-Base Concept

Bronsted-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)

Bronsted-Lowry Neutralization: is a competition for protons that
results in a proton transfer from the strongest acid present to the
strongest base present.

The Bronsted-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 a particular reaction
Remember this is just a theoretical definition, not a theory, because it
does not explain why a proton is donated or accepted, and cannot
predict which reaction will occur for a given entity in any new situation.
+
Bronsted-Lowry Acid-Base Concept

Protons may be gained in a reaction with one entity, but lost in
a reaction with another entity.

The empirical term, amphoteric, refers to a chemical substance with the
ability to react as either an acid or base.

The theoretical term, amphiprotic, describes an entity (ion or molecule)
having the ability to either accept or donate a proton.

Example: 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 reactions, show that one
predominates. The number of ions acting as a base is over 2000x more
than the number reacting as an acid. = BASIC Solution
+
Bronsted-Lowry Acid Base Concept

The Amphoteric Nature of Baking Soda

It can partially neutralize a strong acid, but also a strong base
Practice: pg. 724 #2-4
+ Conjugate Acids and Bases

In an acid-base reaction, there will always be two acids and
two bases.

The original acid on the left and the acid on the right created by
adding a proton.

The original base on the left and the base on the right created by
removing a proton

A pair of substances with formulas that differ only by a proton is
called a conjugate acid-base pair
This analogy shows why
acetic acid is a weak acid.
The proton is more
strongly attracted to the
acetic acid molecule than
it is to the water.
+
Acids/Bases

What about strong acids:

When HCl reacts with water, the water wins the competition
against the Cl- for the proton. This is why at equilibrium,
essentially all of the HCl molecules have lost protons to
water.
>99%

NOTE: The stronger the base, the more it attracts a proton
(proton acceptor). The stronger the acid, the less it attracts
its own proton (proton donor)
+
Conjugate Acids and Bases

RULE: The stronger the base, the more it attracts a proton
(proton acceptor). The stronger the acid, the less it attracts
its own proton (proton donor)

What does this mean about their conjugate pair??

The stronger an acid, the weaker is its conjugate base.


If you are good at donating a proton, this means the conjugate
base is not good at competing for it (weak attraction for protons)
The stronger a base, the weaker is its conjugate acid.

If you are good at accepting a proton, this means the conjugate
acid is not good at giving it up (strong attraction for protons).
+
+
Practice

Bronsted-Lowry WS

Pg 724 Q 2-4

Pg 726 Q 7
Predicting Acid and Base
+ Reactions
1.6K: Write Brønsted–Lowry equations, including indicators, and predict whether reactants or
products are favoured for acid-base equilibrium reactions for monoprotic and polyprotic
acids and bases
1.7K: Identify conjugate pairs and amphiprotic substances
+
Acid-Base Reactions

What is happening?

Collisions are constantly occurring. Each time, a proton is
transferred to a stronger proton attractor.

Theoretically, it could transfer several times (each time to a
stronger proton attractor.) But once, it is transferred to the
strongest base present, the proton will remain there as nothing
outcompetes it.

Likewise, once the strongest acid has given up its proton, its
conjugate base cannot gain one back (as it is the weakest proton
attractor in the whole system)

So what does this mean?

Proton transfer occurs between the strongest acid and strongest
base. All other transfers are negligible so are ignored.
+
Predicting Acid-Base Reactions
+ Predicting Acid-Base Reactions

1) List all entities present initially (ions, atoms, molecules, H2O(l))
as they exist in aqueous solution.
No entity can react as
a base if it is weaker
than water.
For this reason, the
conjugate bases of
the strong acids are
not considered bases
in aqueous solutions

H3O+(aq) is the SA that can exist. If a stronger acid is dissolved, it reacts instantly and
completely with water to form H3O+(aq). So all strong acids are written as H3O+(aq)

OH-(aq) is the SB that can exist. If a stronger base is dissolved, it reacts instantly and
completely with water to form OH-(aq). The only example of this: soluble ionic oxides,
write the cation and the oxide ion is written as OH-(aq)
+ Predicting Acid-Base Reactions

1) List all entities present initially (ions, atoms, molecules, H2O(l))
as they exist in aqueous solution.

Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?

List entities present:

Na+(aq)
OH-(aq)
CH3COOH(aq)
H2O(l)
+ Predicting Acid-Base Reactions

2) Identify and list all possible aqueous acids and bases,
using the Bronsted-Lowry definitions.

Use the entity lists of the Relative Strengths of Acids and Bases (in your
data booklet). Amphiprotic entities are labeled fro both possibilities.
Conjugate bases on SA’s are not included. Metal ions are treated as
spectators.

Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?


List entities present:
Na+(aq)
OH-(aq)
B
A
CH3COOH(aq)
A
H2O(l)
B
+ Predicting Acid-Base Reactions

3) Identify the strongest acid and the strongest base present,
using the table of Relative Strengths of Acids and Bases.

Use the order of the entities in the Relative Strengths of Acids and Bases
table to identify the SA (the highest one on the table) and the SB (the
lowest one on the table)

Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?

List entities present:
SA

Na+(aq)
OH-(aq)
SB
CH3COOH(aq)
H2O(l)
+ Predicting Acid-Base Reactions

4) Write an equation showing a transfer of one proton from
the SA to the SB, and predict the conjugate base and the
conjugate acid to be the products.

Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?


List entities present:
Na+(aq)
OH-(aq)
SB
SA
CH3COOH(aq)
H2O(l)
+ Predicting Acid-Base Reactions

5) Predict the approximate position of equilibrium,
using the following Learning Tip

Example: What will be the predominant reaction if spilled
drain cleaner (sodium hydroxide) solution is neutralized
by vinegar?

Na+(aq)
OH-(aq)
SA
CH3COOH(aq)
SB
The reaction of H3O+(aq) and OH-(aq) is always
quantitative (100%) so a single arrow is used
H2O(l)
+
Practice

Pg. 731 #9-17

Tomorrow:

Table Building and Thought Lab

Five-step method practice
+

Any SA will react
quantitatively with any
base from water down

Any SB will react
quantitatively with any
acid from water up

If WA above WB > 50%

If WA below WB <50%
+

For which of these
reactions would

Kc > 1?

Kc < 1 ?
+
Predict % reaction for these rxs
+ Table Building

Lab Exercise 16.D

If a reaction is >50%, it favours products and the SA is above the SB.

If a reaction is <50%, it favours reactants and the SA is below the SB.
+ Table Building

Lab Exercise 16.D
+
Practice
-
Work on 5 step method questions
-
Quiz Tuesday
+ Acids and Bases – Ka
2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O+] and [OH–] of acidic and
basic solutions
2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base
systems when:
• concentrations at equilibrium are known
• initial concentrations and one equilibrium concentration are known
• the equilibrium constant and one equilibrium concentration are known.
+ The Acid Ionization Constant, K
a

One important way that chemists communicate the strength
of any weak acid is by using the equilibrium constant
expression for the ionization of weak acids. (Ka)

Look at your Relative Strengths of Acids and Bases Table:

All of the strong acids have a Ka value listed as very large;
remember they ionize quantitatively, so the actual acid species
present is H3O+(aq)

All other acids are weak and vary greatly in the extent of reaction
at equilibrium.

We use the equilibrium law to write the formula for Ka
Ka Values have only two sig
digs because they are
somewhat inaccurate because
we calculate them based on
assumptions.
We omit liquid water because we
assume its value is essentially
constant for dilute solutions
+
Ka Calculations

Two common Ka calculations:

Calculating Ka from empirical data (Examples #1-3)

Using Ka to predict hydronium ion concentration from an initial
concentration of weak acid. (Example #4 and 5)
+ Ka Calculations

Example #1: 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?
1.00mol/L – 0.0042 mol/L = 0.9958
(rounds to 1.00 – precision rule)
Change in concentration is negligible
in this case – but not always
Regardless of size, Ka values are usually expressed in scientific notation = 1.7 x 10-5
+ Ka Calculations

Example #2: 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.
+ Ka Calculations

Example #3: 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.
According to the equilibrium law,
the Ka for sulfurous acid is1.4 x 10-2
+ Ka Calculations

Example #4: Predict the hydronium ion concentration and pH
for a 0.200 mol/L aqueous solution of methanoic acid.
Approximation Rule:
0.200
= >1000
-4
1.8 x 10
So (0.200-x) = 0.200
1.8 x 10-4 =
x2
(0.200)
x = 0.006 = H3O+(aq) concentration
+ Ka Calculations

Example #5: Predict the hydronium ion concentration and pH
for a 0.500 mol/L aqueous solution of hydrocyanic acid.
Approximation Rule:
0.500
= >1000
-10
6.2 x 10
So (0.500-x) = 0.500
6.2 x 10-10 =
x2
(0.500)
x = 1.8 x 10-5 = H3O+(aq) concentration
+
Practice Ka

Text pg 743 Q1-9 (any not finished from yesterday)

Read pg 744-750
+ Acids and Bases – Kb
2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O+] and [OH–] of acidic and
basic solutions
2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base
systems when:
• concentrations at equilibrium are known
• initial concentrations and one equilibrium concentration are known
• the equilibrium constant and one equilibrium concentration are known.
+ Base Strength & Ionization Constant, K

Ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), are
assumed to dissociate completely upon dissolving.


b
Finding the hydroxide ion concentration does not involve any
reaction with water
Example: Find the hydroxide ion concentration of a 0.064
mol/L solution of barium hydroxide
+ Base Strength & Ionization Constant, K
b

We can communicate the strength of weak bases, with the
equilibrium constant for its reaction with water.

The equilibrium constant is called the base ionization
constant, Kb

Two common Kb calculations:

Calculating Kb from empirical data (Example #1)

Using Kb to predict the concentration of hydroxide ions when the initial
concentration is known (Example #2)
+
Kb Calculations

We will use the same method as Ka calculations, but there
is usually one extra step because pH values need to be
converted to find hydroxide ion concentrations
Example #1: A student measures the pH of a 0.250 mol/L
solution of aqueous ammonia and finds it to 11.32. Calculate
the Kb for ammonia
14 = pH + pOH
pOH = 2.68
10-2.68 = 0.0021 = OH-(aq)
Remember Kb has
only 2 sig digs
Kb for ammonia is 1.8 x 10-5
+ Calculating OH- from Kb

First problem, the data booklet has Ka values not Kb values.
What do we do?

Kw = KaKb
So
Kb = Kw/Ka

Example #1: Solid sodium benzoate forms a basic solution.
Determine the Kb for the weak base present.
remember Kw = 1.00 x 10-14
+ Calculating OH- from Kb

Example #2: Find the hydroxide ion amount concentration, pOH, pH and the
percent reaction (ionization) of a 1.20 mol/L solution of baking soda.

Baking soda = NaHCO3(s)  Na+(aq) + HCO3-(aq)

For HCO3-(aq), the conjugate acid is H2CO3(aq) whose Ka is = 4.5 x 10-7
Approximation Rule:
1.20
= >1000
-8
2.2 x 10
So (1.20-x) = 01.20
2.2 x 10-8 =
x2
1.20
.
x = 1.6 x 10-4 = OH-(aq)
+ Calculating OH- from Kb

Example #2: Find the hydroxide ion amount concentration, pOH, pH and the
percent reaction (ionization) of a 1.20 mol/L solution of baking soda.
2.2 x 10-8 =
x2
1.20
.
x = 1.6 x 10-4 = OH-(aq)
+
Amphoteric Species

If an entity can react as either a Bronsted-Lowry acid or base, how
do you know which will be the predominant reaction?

Find the Ka value, calculate the Kb value, which ever is larger wins!

Example: Which reaction predominates when NaHSO3(s) is
dissolved in water to produce HSO3-(aq) solution? Will the solution
be acidic or basic?

Ka = 6.3 x 10-8
Kb = Kw
Ka
= 1.0 x 10-14 = 7.1 x 10-13
1.4 x 10-2
The Ka value far exceeds the Kb value, so an aqueous solution of this
substance will be acidic because the hydrogen sulfite ion will react
predominately as a Bronsted-Lowry acid.
+
Practice

Pg. 746 #11-13

Pg. 750 #1-6
+ Acids and Bases - Buffers
1.8K: Define a buffer as relatively large amounts of a weak acid or base and its conjugate in
equilibrium that maintain a relatively constant pH when small amounts of acid or base are
added.
+ Chem 20 Review:

A graph showing the continuous change of pH during an acid-base titration,
which continues until the titrant is in great excess, is called a pH curve

Endpoint refers to the 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 the point in any chemical reaction where
chemically equivalent amounts of the reactants have combined. This point
is determined by stoichiometry.
+
Interpreting pH Curves

Buffering: is the property of some solutions to resist any significant change
in pH when an acid or base is added.

Buffering region (nearly level portions of the graph)
Why is it buffering?
Initially, the solution is mainly all water and OH- ions.
Any additional acid added (H3O+) immediately reacts
with OH- to become water – which does not change the
pH significantly.
This “leveling effect” finally fails near the equivalence
point, when the OH- is almost completely consumed.
Once excess acid has been added, the solution
consists of water and H3O+ ions, so the pH has dropped
to the acid range.
Then any additional acid that is added, simple increase
the H3O+ concentration slightly, but does not change
the pH much.
+
Titration Analysis

Chem 20 Review: Selecting proper indicators
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
bromothymol blue should give accurate results.
+ Acid Base Indicators

Any acid base indicator is really two entities for which we use the same name:
a Bronsted-Lowry conjugate acid-base pair.

At lease one of the entities is visibly coloured, so you can tell simply by
looking when it forms or is consumed. Examples include:

Phenolphthalein: conjugate acid is colourless, conjugate base is bright pink

Bromothymol Blue: conjugate acid is yellow, conjugate base is blue, and when they are in equal
quantities – (appear green to the human eye)

Litmus Paper – red (HIn) to blue (In-)

We will use the designation HIn for the conjugate acid and In- for the conjugate base
as their actual formulas can be very complex.

Summary: An indicator is a conjugate weak acid-weak base pair formed when an
indicator dye dissolves in water.
+
Practice:
+ Polyprotic Entities

Chem 20 Review:

Polyprotic acids – can lose more than one proton

Polyprotic bases – can gain more than one proton

If more than one proton transfer occurs in a titration, chemists believe the
process occurs as a series of single-proton transfer reactions.

On a graph, this means there will be more than one equivalence point
First proton transfer = 100%
Second proton transfer = 100%
Carbonate ion is a diprotic base
+ Polyprotic Entities
A pH curve for the addition of NaOH to a sample of H3PO4(aq) displays only two rapid
changes in pH even though H3PO4(aq) is triprotic.
This is because only two of the transfers are quantitative. The third reaction never
goes to completion, but instead establishes an equilibrium.
General Rule: Only quantitative reactions produce detectable equivalence points in an
acid-base titration.
+
pH Curves
+
Practice
+ General Rule
Strong Acid to
Weak Base:
pH at
equivalence
point is always
lower than 7
Strong Base to
Weak Acid:
pH at
equivalence
point is always
higher than 7
+
+ pH Curve Shape

SA-SB: water is the only acid or base present = neutral solution

SA-WB: a weak acid (NH4+)is present along with water, at the equivalence
point, so the solution is acidic (pH < 7)

WA-SB: a weak base (CH3COO-) is present along with water, at the
equivalence point, so the solution is basic (pH > 7)

WA-WB: do not have detectable equivalence points, because the
reactions are usually not quantitative.
+
Practice:

Pg. 762 #11-14
+ Buffers

A buffer is a relatively large amount of any weak acid and its conjugate
base, in the same solution. In equilibrium, they maintain a relatively
constant pH when small amounts of acid or base are added.
I.e. The addition of a small amount of base
produces more acetate ions. The very small
change in the acid-base conjugate pair
ration and the complete consumption of the
OH- explains why the pH change is very
slight
The addition of a small amount of acid
produces more acetic acid. The very small
change in the acid-base conjugate pair ratio
and the complete consumption of the H3O+
explains why the pH change is very slight
+
Buffer Example: Blood Plasma

Blood plasma has a remarkable buffering ability, as shown by
the following table.

This is very useful, as a change of more than 0.4 pH units, can
be lethal. If the blood were not buffered, the acid absorbed
from a glass of orange juice would likely be fatal.
+ Buffering Capacity

The limit of the ability of a buffer to maintain a pH level.

When one of the entities of the conjugate acid-base pair
reacts with an added reagent and is completely consumed,
the buffering fails and the pH changes dramatically.
All of the CH3COOH(aq) is used up, OHadditions will now cause the pH to
drastically increase
All of the CH3COO-(aq) is used up,
H3O+ additions will now cause the pH
to drastically decrease
+
Practice

Pg. 766 #16-21
+
Modelling Dynamic Equilibrium

Mini Investigation pg. 678
Volume of
Cylinder #1
Volume of
Cylinder #2
25.0
0.0
20.0
5.0
17.0
8.0
14.0
11.0
11.0
14.0
8.0
17.0
5.0
20.0
2.0
23.0
2.0
23.0
2.0
23.0
2.0
23.0
Assume large straw
transfers 5 mL each time
and the smaller straw
transfers 2 mL each time