Download sch103manual - university of nairobi staff profiles

UNIVERSITY OF NAIROBI
FACULTY OF SCIENCE
AND
FACULTY OF EXTERNAL STUDIES
Bachelor of Education Science (B.Ed(Sc)
PHYSICAL CHEMISTRY 1
SCH 103
(GENERAL AND PHYSICAL CHEMISTRY)
UNIVERSITY OF NAIROBI
FACULTY OF SCIENCE
AND
FACULTY OF EXTERNAL STUDIES
Bachelor of Education Science (B.Ed(Sc)
PHYSICAL CHEMISTRY 1
SCH 103
(GENERAL AND PHYSICAL CHEMISTRY)
Written By
Jacob Kithinji
Department of Chemistry
University of Nairobi.
Co-ordinated and edited By
Peter K. Nzuki
Department of Educational Studies
University of Nairobi
Reviewed By
Paul M. Shiundu
Department of Chemistry
University of Nairobi
Published by University of Nairobi
P.O. Box 30197
Nairobi
Printed by College of Biological and Physical Sciences
P.O. Box 30197
Nairobi
© University of Nairobi, 2003
2
TABLE OF CONTENTS
Introduction ………………………………………………………………………….......…... 2
Unit Objective …………………………………………………...………………......……... 3
PART 1:
THE GASEOUS STATE
Lecture 1.
Gaseous states of matter and the gas laws. …………………………….......… 5
Lecture 2:
Using the equation of the ideal gas law; Gaseous mixtures ..……………..…19
Lecture 3:
Kinetic theory of gases ………………...…………………………….............29
Lecture 4:
Behaviour of real gases ……………………...……………………..…....…..35
Lecture 5:
Determination of molar masses of gases ……………………...……......……40
PART 2:
SOLUTIONS
Lecture 6:
Classification of solutions; Properties of solutions of volatile (ideal)
liquids….......................................................................................................... 45
Lecture 7:
Colligative properties of solutions ………………...……………..………… 63
PART 3:
CHEMICAL EQUILIBRIUM
Lecture 8:
Equilibria for gaseous and heterogeneous reaction …………….......……..…79
Lecture 9:
Application of the equilibrium constant concept ………………........………92
Lecture 10:
Effects of changing the reaction conditions on chemical equilibrium ....…..101
PART 4:
IONIC EQUILIBRIUM
Lecture 11:
Acid- Base Concepts ……………………………..………...……..…..……113
Lecture 12:
Acid Base Equilibrium ……………………...……………………..…….....122
REFERENCES…………………...……………………..………………….......………….153
3
INTRODUCTION TO THE COURSE
Physical Chemistry is a branch of chemistry that establishes and develops the
principles that are used to explain and interpret the observations made in the other branches
of chemistry such as organic and inorganic chemistry.
“Physical Chemistry 1 here in referred to as General and Physical Chemistry” is a
core course in Chemistry at the undergraduate level. It is designed to train students in the
general and basic physical chemistry. The course is divided into four parts: Part one
comprises of five lectures covering the gaseous state; Part two covers classification and
properties of solutions of liquids and solutions containing non-volatile solutes. Part three
looks at the chemical equilibrium and part four covers ionic equilibrium.
It is a one year course which comprises both theoretical and practical components.
The theory component will be sent to you as a full unit, with the prescribed accompanying
text-book(s) and the practical unit. The practical sessions will be carried out at Chiromo
Campus laboratories during your residential training.
Each lecture starts with a short introduction followed by objectives. The purpose of
the introduction is to link the previous lecture with the next lecture to be learned.
Furthermore, each of the lectures has practice activities which you are expected to do in
addition to some assignments given after each section.
You are expected to mail the completed assignments according to instructions for
marking.
The material presented in this study unit, are very comprehensive. However, you are
advised to make reference to the following books: Chemistry by Chang Raymond (1991);
General Chemistry by Ebbing D.D. (1984).
It is hoped that, during your residential training those aspects of the course which you
find difficult will be explained further.
4
UNIT OBJECTIVES
At the end of this unit you should be able to:1. describe states of matter and derive the laws that govern the gaseous state.
2. describe the behavior of solutions containing volatile and non-volatile
solutes.
3. describe chemical and ionic equilibrium.
4. apply equilibrium constant to solving problems involving homogeneous as
well as heterogeneous reactions.
5
PART ONE
THE GASEOUS STATE
Introduction
Under certain conditions of pressure and temperature, most substances can exist in
any of the three states of matter: Solids, liquid or gas. Water for example, exists in the solid
state as ice, liquid state as water and in the gaseous state as steam. The physical properties of
a substance often depend on the state of the substance.
In this section, we will review the states of matter and then look at the behavior of
gases. Gases are much simpler than liquids or solids. Molecular motion in gases is totally
random, and the forces of attraction between molecules are so small that molecules move
freely and independent of each other. The behavior of a gas subjected to changes in
temperature and pressure can be described according to much simpler laws than do liquids
and solids. The laws that govern this behavior have played an important role in the
development of the atomic theory of matter and the molecular kinetic theory of gases.
Part one consists of 5 lectures as follows:Lecture 1:
Gaseous states of matter and the gas laws;
Lecture 2:
Using the equation of the ideal gas law; Gaseous mixtures;
Lecture 3:
Kinetic theory of gases;
Lecture 4:
Behaviour of real gases and
Lecture 5:
Determination of molar masses of gases
6
LECTURE 1
GASEOUS STATES OF MATTER AND THE GAS LAWS.
Introduction
As you may be aware, gases are the simplest form of matter and they fill any
container they occupy. Some of the examples of gases include helium, chlorine, carbon
dioxide, ammonia and oxygen.
Initially we shall consider only pure gases but later in the lecture we shall see that the
same idea and equations apply to mixtures of gases too.
The gas laws we will study in this lecture regards the macroscopic behaviour of
gaseous substances. They have played a major role in the development of many ideas in
chemistry. These laws are Boyle's law, Charles' law, and Avogadro's law. We shall combine
the three laws to come up with an equation called equation of state which as we shall see, will
be very useful in solving many problems involving gases.
OBJECTIVES
At the end of this lecture you should be able to:1. state the states of matter
2
define the state of gases; Pressure, temperature and amount of gas
3
state and illustrate Boyle’s law, Charles’s law, Avogadro’s law of gases
4
derive the ideal gas law
7
The state of gases
Definition
Before you are introduced to the gas laws, it is necessary to understand the concepts
of the states of matter.
List four variables that can describe a state of gas.
To describe the states of a gas, at least three of the following variables must be
specified as shown in table 1.0.
Variable
Name , Symbol
The space occupied by a sample of a gas
Volume, V
The number of molecules present in the sample
Moles, n
Pressure
Pressure, P
Temperature
Temperature, T
Table 1.1 State variable and symbols
The state of a pure gas is specified by three variables ( volume-V, moles- n and
temperature, T) and that the pressure has a quantity that depends on the value of the other
three variables.
We need to examine the various units of pressure before we develop the laws that
govern the gases.
Pressure can be defined as force per unit area.
This is the force exerted by a gas as a result of collisions that exert force which is
experienced as a steady pressure. Pressure exerted by the atmosphere is measured with a
barometer. The height of the liquid column in a mercury barometer on a typical day at sea
level is about 760 mm.
The standard SI units of pressure, the pascal is defined as 1 Newton per square meter.
This is expressed mathematically as
Pr essure 
8
Force
Area
You may have come across other units that are used to express pressure. They are related to
each other as shown in table 1.2
Unit and (Symbol)
Conversion factor
Pascal (Pa)
1NM-2 (SI unit of pressure)
Atmosphere (atm)
1.01325 x 105 Pa
Torr (torr) or mmHg
(1/760) atm
Bar (bar)
105 Pa
Table 1.2: Scientific units of pressure
Define the three laws that govern the gaseous state and use graphical method to
illustrate them.
The three laws used to describe the gaseous state are described below.
1.1
BOYLE’S LAW
Boyle investigated the pressure-volume relationship of a gas sample using an
apparatus like that shown in Figure 1.1. In Figure 1.1a, the pressure exerted on the gas is
equal to atmospheric pressure (760 mmHg). In Figure 1.1b the total applied pressure is
atmospheric pressure plus the pressure of the column of mercury (760 mmHg). Thus the
pressure doubles and the volume of the gas decreases by half. This relationship between
pressure and volume is shown pictorially in Figure 1.1a, b, and c. Boyle noticed that when
temperature is held constant, the volume (V) of a given amount (n) of a gas decreases as the
total applied pressure (P) is increased.
The results of several measurements are given in Table 1.3.
9
Figure 1.1 Boyle's Experiment.
Enclosed
gas
Enclosed
gas
b. When the pressure is doubled by adding 760 mm of
mercury, the volume is halved ( to 50 mL)
c. Tripling the pressure decreases the volume to 33 mL.
The temperature is kept constant.
100 mL
Enclosed gas
760 mm
1520 mm
a. The volume of the gas at normal atmospheric
pressure (760 mmHg) the volume is 100 mL
Hg
a
c
b
The pressure-volume data recorded from such an experiment are consistent with the
following mathematical expressions which show an inverse relationship.
V
1
P
at constant temperature (T) for a given amount of gas (n).
Where the symbol  means proportional to. To change  to an equal sign, we must
write
V  K1 x
1
P
Equation 1.0
Where K1 is a proportionality constant.
Rearranging the above equation, we obtain
PV = constant =K1
(At constant n, and T) Equation. 1.1
Equation 1.0 is an expression of Boyle’s law, which states that:“the volume of a fixed amount of gas maintained at constant temperature is inversely
proportional to the gas pressure”.
This implies that doubling the pressure, reduces the volume by half.
10
-
Gases respond readily to pressure because there is so much space between the
molecules that they can easily be confined to a smaller volume.
Boyle’s law applies to whatever the gas, argon, carbon dioxide, water vapour,
-
or any other.
Graphical representation of Boyle’s Law
Boyle’s law relation/expression can be represented as a straight line on a graph. The
equation of a straight line is represented as:y = a + bx
Thus, V  0  constant x
Where a is intercept and b is slope.
1
P
Hence it can be tested by plotting experimental values of V against
1
and seeing
P
how close the result is to a straight line. Table 1.3 shows typical pressure-volume relationship
obtained by Boyle.
P
(mmHg) 0.2500
V (L)
2.8010
0.5000
0.7500
1.0000
2.0000
3.0000
4.0000
5.0000
1.4000
0.9333
0.6998
0.3495
0.2328
0.1744
0.1394
PV
Table 1.3 Volume- pressure relationship obtained by Boyle’s law2.
-
Using the P-V data in the table calculate PV values and complete the
table. At the values of PV precisely constant?
-
What does this tell you about equation 1.0
11
Graphical representation of Boyles Law
Figure 1.2a and b shows two conventional ways of expressing Boyle’s finding
graphically. Although the individual values of pressure and volume can vary for a given
sample of gas (as long as temperature is held constant and the amount of the gas does not
change), P times V is always equal to the same constant.
Therefore, for a given sample of gas under two different sets of conditions of pressure
and volume (at constant temperature), we can write
P1 V1  K1  P2 V2
or
P1 V1  P2 V2
Equation 1.2
Where V1 and V2 are the volumes at pressure P1 and P2 respectively.
One common application of Boyle’s Law is to use equation 1.2 to predict how the
volume of a gas will be affected by a change in pressure, or how the pressure exerted by a gas
will be affected by a change in volume under constant temperature condition.
12
Example
Calculate the volume occupied by a gas when the pressure changes.
The pressure of oxygen in a 50 - L tank is 15.7 atm at 210C, what volume of
oxygen can we get from the tank at 210C if the atmospheric pressure is 1 atm?
Solution
We can write Pi and Vi for the initial pressure (15.0 atm) and initial volume
(50 – L), and Pf and Vf for the final pressure (1 atm) and final volume (to be
determined)
Thus we can write
Pf Vf = Pi Vi
Dividing both sides of the equation by Pf gives
V f  Vi x
Pi
Pf
When we substitute into this equation, we get
V f  50 Li x
15.0 atm
 750 L
1 atm
Note: In this example the pressure on the gas decrease and ratio Pi/Pf is
greater than 1, so that the gas expands to give a larger volume, that is the gas
escapes. Otherwise the volume of the container remains the same.
A sample of hydrogen gas at 0oC and 700 mmHg has a volume of 4.00L. What
would be the volume if the pressure was changed to 760 mmHg at the same
temperature?
-
Ans 3.68L
the pressure-volume product for a gas is not precisely constant. Confirm
this by looking at PV data in Table 1.3.
-
all gases follow Boyle’s law at low to moderate pressure but deviate from
this law at high pressure. The extent of deviation depends on the gas. We
will return to this point at the end of the section
1.2 CHARLES’ LAW
Boyle’s law depends on the temperature of the system remaining constant. Suppose
the temperature changes, how would this affect the volume and pressure of a gas?
13
The earliest investigation of this relationship were conducted by French scientists,
Jacques Charles and Joseph Gay-Lussac. Their studies showed that, at constant pressure, the
volume of a gas expands when heated and contracts when cooled Figure 1.3. This means that
if we plot the volume occupied by a given sample of gas at various temperatures, we get a
straight line.
Figure 1.3 When a fixed mass a gas is
heated at constant pressure (P), it
expands. Doubling the temperature
(T), doubles the volume, (V).
When the straight lines in figure 1.4 are extended from the last experimental point
towards lower temperatures, that is we extrapolate the straight lines backwards, we find that
they all intersect at a common point. This common point occurs at a temperature of –
273.150C, where the graph indicates a volume of zero. This is to say that, if the substance
remains gaseous, the volume occupied will be zero at –273.150C.
Figure 1.4 Variation of the volume of samples of O2 and CO2 at different temperatures.
14
a sample of gas shows a linear (straight-line) variation of volume with
temperature. This linear variation is obtained irrespective of the amount or type
of gas.
In 1948 Lord Kelvin realised the significance of this behaviour. He identified the
temperature –273.150C as theoretically the lowest attainable temperature and called it
absolute temperature scale , which we now call Kelvin temperature scale.
The only
difference between the absolute temperature scale and the Celsius scale is that the zero
position is shifted. A summary of the relationships between the two scales is as follows:Absolute zero:
OK = -273.150C
Freezing point of water:
273.15K = 00C
Boiling point of water:
373.15K = 1000C
The relationship between 0C and K is T(K) = t (0C) + 273.15
By convention we use T to denote absolute (Kelvin) temperature and t to indicate
temperature on the Celsius scale. The dependence of volume on temperature is given by
V  T (at a fixed pressure for a given amount of gas)
V = K2T
or
V
 K 2 where K2 is the proportionality constant.
T
This equation is known as Charles and Gay-Lussac’s Law, or simply Charles Law.
Charles’ Law states that the volume of a fixed amount of gas maintained at constant pressure
is directly proportional to the absolute temperature of the gas.
Just as we did for pressure-volume relationships at constant temperature, we can
compare volumes and temperatures at two sets of conditions for a given sample of gas at
constant pressure.
Vf
We can write
Tf
or
15
Vf
Tf
 K2 

Vf i
Ti
Vf i
Ti
Where Vi and Vf are the volume of the gas at temperatures Tf and Ti (both in Kelvin),
respectively.
Example
A quantity of gas at 100C and 1 atm pressure occupied a volume of 200cm3.
What volume will it occupy at –200C at constant pressure.
Answer
We use equation
Vf
Tf
 K2 
Vi
Tf i
Where
Initial conditions:- Vi, Ti , Vi = 200cm3, Ti = (10 + 273) K = 283 K
Final conditions:- Vf, Tf ,
Hence
Vf 
Vf = ?, Tf = (-20 + 273) = 253
Vi . T f
T fi
200cm3 x 253K

283K
 178.8 cm3
A sample of hydrogen gas occupies 3.0 L at 1250C. Calculate the temperature
at which the gas will occupy 1.5 L if the pressure remains constant.
Although gases follow Charles’ fairly well, they deviate from it at high
pressure and low temperatures – as we shall discuss in the last section.
1.3 AVOGADRO’S LAW
The work of the Italian scientist Amedeo Avogadro complemented the studies of
Boyles, Charles and Gay-Lussac. In 1811 he published a hypothesis that stated that at the
same temperature and pressure, equal volumes of different gases contain the same number of
molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas
must be proportional to the number of molecules present. That is
V  n (at constant temperature and pressure)
Hence
V = K3n
Where n represents the number of moles and K3 is the proportionality constant.
16
The above equation is the mathematical expression of Avogadro’s Law, which states
that at constant pressure and temperature, the volume of a gas is directly proportional to the
number of moles of the gas present.
One mole of any gas contains the same number of molecules ( Avogadro’s number =
6.022 x 1023) and by Avogadro’s law the gas must occupy the same volume at a given
temperature and pressure. The volume of one mole of gas is called the molar gas volume.
If we represent molar gas volume at some temperature and pressure by Vm, the
volume occupied by n moles of any gas at this temperature and pressure is V = n Vm . Where
Vm has the same value for all gases (at a fixed temperature and pressure).
-
Draw a graph to demonstrate Avogadro’ law.
-
Write an expression comparing (V, n) for two sets of conditions for a
given sample of gas at constant T and P ( As we did for V, P and V,T
relationships.
We can use the data in Table 1.4 to judge the validity of Avogadro’s law that the
moles of gaseous molecules is independent of their chemical identity.
Gas
Vol. L/mol
Ammonia
22.40
Argon
22.39
Nitrogen
22.40
Ideal gas
22.44
Table 1.4: Molar volumes of gases at 00C and 1 atm.
We have learnt from Avogadro’s law when two gases react with each other, their
reacting volumes have a simple ratio to each. If the product is a gas, its volume is related to
the volume of the reactants by a simple ratio. For example, consider the synthesis of
ammonia from molecular hydrogen and molecular nitrogen.
For reaction
3H2)g)
+
3 moles
N2(g)
2NH3(g)
1 mole
2 moles
Since at the same temperature and pressure, the volumes of gases are directly
proportional to the number of moles of the gas present, we can now write
17
3H2)g)
+
3vol.
N2(g)
2NH3(g)
1 vol.
2 vol.
The ratio of volume for molecular hydrogen and molecular nitrogen is 3:1, and that of
ammonia ( the product) and for molecular hydrogen and molecular nitrogen ( the reactants is
2:4 or 1:2. This can be summarized as
3H2)g)
+
N2(g)
2NH3(g)
Molar ratio
3 moles
1 mole
2 moles
Moleclular ratio
3 molecules
1 molecule
2 molecules
Volume ratio
3volumes
1 volume
2 volumes
At a particular temperature and pressure, 15.0g of CO2, occupies 7.16 L. What is
the volume of 12.0g of CH4 at the same temperature and pressure?
1.4 THE IDEAL GAS LAW
The ideal gas law is an example of an “equation of state.”
We have discussed three gas laws, each relates the volume occupied by a gas to another
variable:
V 
1
P
at fixed T, n
Boyle’s law
V  T
at fixed P, n
Charles’ law
V  n
at fixed T, P
Avogadro’s law
We can combine all the three expressions to form a single master equation for the behaviour
of gases:
V 
nT
P
Introducing a constant of proportionality, R,
V  R
or
nT
P
PV  nRT
Equation 1.4
Where R, is called the gas constant.
The equation PV  nRT , which describes the relationship among the four
experimental variables P,V,T and n is called the ideal gas equation
18
An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can
be completely accounted for by the ideal gas equation.
The molecules of an ideal gas do not attract or repel one another, and their volume is
negligible compared with the volume of the container.
Although there is no such thing in nature as an ideal gas, discrepancies in the
behaviour of real pressure range do not significantly affect calculations. Thus we can use the
ideal gas equation to solve many gas problems.
To use this equation, we need to know the value of the molar gas constant, R. We can
obtain it from the experimental value of the molar volume at (0oC, 1 atm). The condition 0oC
and 1 atm are called standard temperature and pressure (STP). The data for evaluation of R
are
Variable
Value
P
1 atm
V
22.414 L
T
(0+273.15) K = 273.15K
n
1 mol
From equation 1.4 we can write
PV
nT
1atm 22.414 L 

1mol  273.15 K 
 0.082057 L.atm / K .mol 
R
Recall that other units of pressure are Torr (mmHg) and Pascal. Calculate the
values of R using these units. Compare your answers with values in table 1.4
19
Where P is in
R
Units
1. Atmosphere (atm)
0.082057
L .atm / (K.mol)
2. Torr (torr)
62.37
L. Torr / (K.mol)
3. Pascal (Pa)
8.31441
J/ (K.mol)
Kg.m2/(K.mol)
dm3.kPa/(K.mol)
4. Calories
1.98719
cal/(K.mol)
Table 1.5. The gas constant R in various units
The ideal gas law includes all the information contained in Boyle’s, Charles’ and
Avogadro’s laws. Therefore, starting with the ideal gas law, we can derive any of the other
gas laws.
Review questions
1
(a)
State the following gas laws and also give their corresponding
mathematical forms:(i)
(b)
Boyle’s law
(ii) Charles’ law (iii)
Avogadro’s law
In each case and also in the form of an equation, indicate the conditions
under which the law is applicable giving the units for each quantity in the
equation.
(b)
Define absolute zero and absolute zero scale. Write the relationship
between oC and K.
20
LECTURE 2
USING THE EQUATION OF IDEAL GAS LAW
Introduction
The ideal gas law, PV = nRT we derived in the last lecture is an example of an “equation of
state”. An equation of state is a mathematical relation linking the pressure (P), volume (V)
and temperature (T) of a substance and the amount (n), present. In this lecture we are going to
use this equation to investigate ideal gas behavoiur.
OBJECTIVES
At the end of this lecture you should be able to:1.
calculate any one property once the other three properties are given
2.
derive the density relationship with pressure and temperature
3.
solve stoichiometric reactions involving gases
4.
state the Dalton’s law of partial pressure and solve problems
involving gas mixtures
1.1 Examples involving changes in P,V,T and n
Equations of state are very important because they make it possible to calculate any one the
property once we know other three as we shall see in the following examples.
Example
How many grams of oxygen are there in a 50.0L tank at 21oC when the oxygen
pressure is 15.0 atm?
Solution
In asking for the mass of oxygen we are in effect asking for moles of gas n,
because mass and moles are related. The data given in the problem are
Variables
Values
P (atm)
15
V (L)
50
T(K)
294
N (Mole
?
From the data given, we see that we can use the ideal gas law to solve for n.
21
The proper value to use for R depends on the units of V and P. These are in
litres and atmospheres, respectively, so we use
R = 0.0821 L.atm/K.mol
n
PV
RT
Substituting from the table gives
n
15.0atmx50.0L
 31.1 mol
0.0821L.atm / (K.mol)x294K
Recall that 1 mole of O2 weighs 32g,
Therefore, 31.1 moles of O2 weigh
31.1 mol x 32.0g
1mol
Example
= 994.3 g O2
Nitrogen is heated to 500 K in a vessel of constant volume. If it enters at a
pressure of 100 atm and a temperature of 300 K, what pressure would it exert
at the working temperature if it behaved as a perfect gas.
Strategy
When confronted with a problem on perfect gas, the solution can usually be
found by rearranging the equation into a form that gives the unknown quantity
in terms of the data.
In this example we are not given all the physical
properties (we are not told the amount or the volume), but this deficiency can
be overcome by writing the equation as
PV
R
nT
For both sets of conditions and noting that R is a constant, then
PV
PV
1 1
 2 2
n1T1 n 2 T2
The constant quantities (amount and volume) cancel and the data may then be
substituted into the resulting expression.
Solution
Cancellation of the amount (n1 = n2) and volume (V1 = V2)
P1 P2
T

 P2  2 x P1
T1 T2
T1
Substitution of the data gives: P2 
22
500K
x 100 atm  167 atm
300K
Calculate the pressure exerted by 1.5 g of carbon dioxide in a 1.5 litre flask at
250oC.
2.2
[Ans .098 atm]
The Density of a gas
The ideal gas/air is used to determine how other physical properties vary with
temperature and pressure. We shall use it to find the density d, or mass per unit volume of a
gas in terms of its temperature and pressure. The calculation leads to a method for measuring
the molecular weight of volatile substances.
First we express the mass of the sample in terms of the moles of molecules using the
molar mass M of the gas molecules.
Mass of sample = number of moles of sample x mass per mole = n x M
Then we use this result to express the density (d), in terms of moles
d
Mass of sample
nxM

Volume of sample
V
Finally we relate the number of moles to the pressure, using the ideal gas law in the
n
P

V RT
form
On substituting P
RT
for
n into the equation for d, we get
V
dPx
M
RT
Explain how density varies with pressure by looking at the equation for d.
The density equation shows that:(i)
The density of a gas increases as the pressure is increased at constant temperature.
(ii)
The density decreases as the temperature is raised at constant pressure. This means
that that as long as the pressure is kept constant the gas expands as we heat it
Example 1
Calculate the density of ammonia (NH3) in grams per litre (g/L) at 700 mmHg
and 50oC.
Solution
To convert the pressure to atmosphere, we write
23
P  700 mmHg x
Using the equation d 
d
PM
and
RT
1 atm
 750 

 atm
760 mmHg  760 
T = 273 + 55 = 328K, we have
atm 17.03g/mol 
= 0.634 g/l
0.0821 L.atm/K.mol 323K 
750
760
1. Calculate the density of carbon dioxide at 1.5 atm and 30oC.
2.
[Ans. 2.9 g/L]
The oil produced from eucalyptus leaves has a density of 0.4000 g/L at
200oC at a pressure of 70.0 Torr. Calculate the molecular weight of the
eucalyptus oil.
[Ans. 165 g/mol]
2.3 Stoichiometric reactions involving gases
So far, we have dealt with pure gaseous systems only. Now we can extend our
knowledge of stoichiometry to find the volume of gases obtained from reactions. The ideal
gas law allows us to calculate the number of moles (n) of gas given the volume (V) or the (V)
given (n). When the reactants and /or products are gases, we can use the relationships
between n and V to solve such problems as shown in the following example.
24
Example Hydrogen gas is prepared by reacting zinc with hydrochloric acid as per the
following chemical equation:
2HCl (aq) + Zn(s)
ZnCl(aq) + H2(g)
Calculate the amount of zinc in grams requires to prepare 2.00 L of H2 gas at 760
mmHg at 27oC.
Solution
From the specified conditions conditions of P,V and T, we can calculate the moles
of H2 produced using the ideal gas law, PV = nRT
We list the available data
Variable
Value
P
760/760 = 1atm
V
2.00 L
T
27 + 273=300K
Note we have converted pressure in mmHg to atmosphere, and degrees celsius to
Kelvin. Thus we use R= 0.0821 L.atm/K.mol.
We rearrange the ideal gas law, PV = nRT and solve for the (n)
n 
PV
RT
Substituting the data
n
1 x 2.00 = 0.08 moles
0.0821 x 300
From the balanced chemical equation we find that 1 mole of zinc produce 1 moles
of H2. Therefore, the moles of zinc used is equal to 0.08 moles.
The amount of zinc required to prepare 2.00 L of H2 is equal to 0.08 moles x 65
moles/L = 5.2 g
Ethanol (C2H5OH) burns in air
C2H5OH(l) + O2(g)
CO2(g) + H2O(l)
Balance the equation and determine the volume of air in liters at 30oC and 800
mmHg required to burn 230 g of ethanol. Assume air to be 21.0% O2 by volume.
[Ans. C2H5OH(l) + 3O2(g)
2CO2(g) + 3H2O(l) ;
25
1685 Liters]
2.4 Gas mixtures
Law of partial pressure
So far we have concentrated on the behavior of pure gaseous substances. However,
experimental studies are often based on mixture of gases. We have seen that the physical
properties of gases are largely independent of their identities. We can therefore expect a
mixture of gases to behave much like a single pure gas.
In 1801 Dalton summarised his
observations in terms of the “partial pressures” of gases.
The partial pressure of a gas in a mixture of gases is the pressure the gas would exert
if it alone occupied the container. If we have gases A and B in a container, then partial
pressures of A and B are expressed as PA and PB respectively. Dalton then described the
behaviour of mixture of gases by his ‘law of partial pressure’.
Dalton’s law of partial pressures states that the pressure of a mixture of gases is the
sum of the partial pressures of its components.
Suppose we have gases A and B in a container of volume V. If we let P T to be the
total pressure and PA , PB to be the partial pressures of the component gases in mixture, the
law of partial pressures can be written as
PT = PA + PB
Using the ideal gas law, the individual partial pressures Pi can be written as
n RT
P V  n RT or P = i
i
i
i
V
Where ni is the number of moles of component i
Therefore ,
PA =
n A RT
and PB = nB RT
V
V
Therefore, according to Dalton’s law,
PT = PA  PB 
n A RT n B RT RT


 n A  nB 
V
V
V
In general, the total pressure of a mixture of gases is given by
PT = P1 + P2 + P3 + …………..
Where P1, P2, P3 … are the partial pressures of components 1, 2, 3 …
Example
1.00g of air consists of approximately 0.76g of nitrogen and 0.24g of oxygen.
Calculate the partial pressures and the total pressure when this sample
occupies a 1.00L vessel at 20oC.
26
Solution
Strategy
Check if the given data are enough to calculate the unknown quantity using
the ideal gas equation. If there is one unknown, it may be necessary to
convert another item of data.
The data needed for use are
PN 2 = ?
nN2 = ?
V = 1.00L
T = 293 K (20oC)
PO2 = ?
n O2 = ?
V = 1.00L
T = 293 K (20oC)
The number of n N 2 and nO2 can be obtained from the masses (0.76g for N2
and 0.24g for O2 and molar masses (28.02g/mol for N2 and 32.00g/mol for
O2).
n N 2 = 0.76g N 2 x
1 mol N 2
 0.0271 mol N 2
28.02 g N 2
1 mol O
2  0.00750 mol O
n = 0.24 g O x
O
2 32.00 g O
2
2
2
Using
PA =
n A RT
V
PN2 = 0.0271 mol N 2 x
0.08206L.atm 293K
x
 0.65 atm N 2
1K.mol
1.00L
PO2 = 0.00750 mol O2 x
0.08206L.atm 293K
x
 0.18 atm O2
1K.mol
1.00L
The total pressure is the sum of these partial pressures
PT = 0.65 atm + 0.18 atm = 0.83 atm
A mixture of gases whose composition is as follows: 0.3 moles A, 0.25 moles B and
0.3 moles C have a total pressure of 2atm. Calculate the partial pressure of the gases.
[ Ans PA= 0.7; PB= 0.6; PC=0.7]
27
2.4.1 Mole fraction
The composition of a gas mixture is often described in terms of mole fraction of the
component gases. The mole fraction of a substance in a mixture is the number of moles of
the substance in the mixture, expressed as a fraction of the total number of moles in the
mixture.
The mole fraction concept applies equally to solids, liquids and gaseous mixture, but
in this lecture we will apply only to gases. For a two component system comprising of A and
B the mole fractions of A and B are XA and XB and are expressed as:XA =
nA
n A  nB
and X B =
nB
n A  nB
The mole fraction of A is zero when no A molecules are present and 1 when the
sample is pure A. If half the molecules are A, then the mole fraction of A is 0.5 as shown
below:
XA + XB =
nA
nB
n
+
 B 1
n A  nB n A  nB nT
We can relate the mole fraction to the partial pressures as follows.
Let us obtain expression for nA and nT using the ideal gas law as follows:-
PAV
nA
RT  PA
Mole fraction of A = =
nT PT V
PT
RT
or X A 
pA
PT
Therefore, when we divide each partial pressure by the total pressure, we get the mole
fractions.
From equation partial pressure of can be expressed as
PA = XA PT
Using the above equation predict the value of PA when XA =0, .5 and 1.
28
2.4.2 Collecting gases over water
Dalton’s law of partial pressure has a practical application in calculating volumes of
gases collected over water. This is a common way of collecting a-not-very soluble gas such
as oxygen that has been prepared in the laboratory. The oxygen gas collected in this way is
not pure because water vapour is also present in the same space. The total gas pressure is
equal to the sum of the pressure exerted by the oxygen gas and that of the water vapour,
which can be measured with a barometer.
PT  PO2  PH 2O
The partial pressure of water vapour above liquid water is called its vapour pressure,
which depends on the temperature. At room temperature the vapour pressure of water is only
about 20 torr. The following example shows that we can use Dalton’s law to calculate the
amount of gas collected over water.
Example
Oxygen gas generated in the decomposition of potassium chlorate and collected
over water at 20oC. The volume of the gas collected at atmospheric pressure of
755 Torr is 370mL. Calculate the mass (in grams) of oxygen gas obtained. The
pressure of the water vapour at 20oC is 17.5 Torr.
Solution
Strategy
We can calculate the number of moles using the ideal gas law once we know the
partial pressure, the volume and the temperature.
Our first step is to calculate the partial pressure of O2.
PT  PO2  PH 2O
Therefore,
PO2  755Torr  17.5Torr  737.5Torr
 737.5Torr x
1atm
760Torr
 0.97atm
From the ideal gas equation, we write
PV = nRT =
m
RT
M
Where m and M are the mass of O2 collected and the molar mass of O2,
29
respectively. Rearranging the equation, we get
m=
PVM
0.97 atm x 0.37 L x 32 g/mol
=
RT
0.0821 L.atm/K mol x 293 K
= 0.48 g of oxygen
Since , M = 32 g/mol; T= (20+273) = 293 K; V = 370 mL = 0.37 L
A piece of sodium metal undergoes complete reaction with water as follows:
2Na(S) + 2H2O(l)
2NaOH(aq) + H2(g)
The H2 gas is collected over water at 25OC. The volume of the gas is 246mL
measured at 1.00 atm. Calculate the number of grams of sodium used in the
reaction. (Vapour pressure of water at 25oC is 0.0313 atm)
30
LECTURE 3
KINETIC THEORY OF GASES
Introduction
The gas laws help us to predict the behaviour of gases in the macroscopic world
(pressure, volume, temperature changes) but do not explain the behaviour of gases in the
microscopic world (properties of molecules).
As early as in the 1850s, a number of
physicists like Ludwig Bolzmann in Germany and James Maxwell in England found that the
physical properties of gases could be explained in terms of the motion of individual
molecules as we will show in this lecture.
OBJECTIVES
At the end of this lecture you should be able to:-
3.1
1.
state the basic assumptions of the Kinetic theory of gases
2.
interpret the gas laws qualitatively
3.
derive the state equation PV = nRT using the kinetic theory of gases
4.
illustrate Maxwell distribution of molecular speeds
Postulates of Kinetic Theory
Physical theories are called postulates. They are the basic statements from which all
conclusions or predictions of a theory are deducted. The postulates are accepted as long as
the predictors from the theory agree with experiment. The kinetic theory of an ideal gas is
based on the following five postulates: 1. Gases are composed of molecules whose sizes are negligible compared to the average
distance between them.
2. Molecules move randomly in straight line in all directions and at various speeds.
3. The intermolecular forces in a gas are very weak or negligible, except when they
collide.
4. When molecules collide with one another, the collisions are elastic. In an elastic
collision, the total kinetic energy remains constant (no kinetic energy is lost).
5. The average kinetic energy of a molecule is proportional to the absolute temperature.
31
3.2 Qualitative interpretation of the gas laws
According to the kinetic theory, the pressure of a gas results from bombardment of
container walls by molecules and the pressure is determined by:(i)
the concentration of molecules (number per unit volume) which determines
the frequency of collisions.
(ii)
the average force of the molecules which determines the average force of a
collision.
We can see the meaning of the gas laws in these terms as follows: Boyle’s law: The temperature is kept constant and therefore the average kinetic energy of a
molecule is constant (postulate No. 5). Therefore the average speed and thus the average
molecular force from collision remain constant.
Suppose we raise the volume of a gas container what would happen?
This decreases the concentration of molecules and so decreases the frequency of collisions
per unit wall area. The pressure decreases.
Now consider Charles’ law. If we raise the temperature, we increase the average
molecular speed. This increases the average force of a collision. If all other factors remain
fixed, the pressure increases. For the pressure to remain constant, it is necessary for the
volume to be increased so that the number of molecules per unit volume decreases.
Avogadro’s law follows from the fact that for two different volumes of gas to have
the same temperature (same average molecular speed) and the same pressure, there must be
the same number of molecules per unit volume.
3.3 The Ideal Gas Law from Kinetic Theory
One of the key features of kinetic theory is its ability to explain the ideal gas law. To
show how we can get the ideal gas law from kinetic theory, we will first find an expression
for the pressure of a gas.
32
According to kinetic theory, the pressure of a gas P, will be proportional to both the
frequency of molecular collisions with a surface and the average force exerted by a molecule
on collision.
P  frequency of collisions x average force
The average force exerted by a molecule during a collision depends on its mass m and
its average speed u t(hat is its average momentum mu). In other words, the greater the mass
of the molecule and the faster it is moving, the greater the force exerted during collisions.
The frequency of collision is also proportional to: -
average speed u
-
the number N of molecules in the gas
and inversely proportional to
-
the gas volume V, because the larger the volume, the less often
a given molecule strikes the container walls.
Putting these factors together gives
1


P   u x x N) x mu 
V


rearranging the expression, we get
PV  Nmu2
Also, we know that average kinetic energy of a molecule of mass m and average
speed u is ½mu2,
Therefore, PV is proportional to the average kinetic energy of a molecule.
-
the average kinetic energy is also proportional to the absolute
temperature by postulate 5.
-
the number of molecules, N is proportional to the moles of
molecules, n.
We therefore get
PV  nT
We can write this as an equation by inserting a constant of proportionality R, which
we identify as the molar gas constant. Thus
PV = nRT
3.4 Distribution of Molecular Speeds and the Root Mean Square
The kinetic energy allows us to investigate motion in more detail. Suppose we have a
large number of molecules of a gas, say 1 mole, in a container. As you might expect, the
33
motion of the molecules is totally random and unpredictable. As long as we hold the
temperature constant, the average kinetic energy and the mean square speed remain
unchanged as time passes. At a given instant, how many molecules are moving at a particular
speed? The British physicist James Clerk Maxwell in 1860 showed that molecules can have
a wide range of speeds, though most are close to the average and some move at very high or
low speeds as shown in figure 3.1. The distribution of speeds of H2 molecules are shown for
0oC and 500oC.
The speed corresponding to the maximum in the curve is the most probable speed
and it increases with temperature.
Figure 3.1: Maxwell’s distribution of molecular speeds of H2 molecule for 0oC and 500oC1.
3.5 Root mean square
One of the useful results of the kinetic molecular theory is that it enables us to relate
macroscopic quantities P and V to molecular parameters such as molar mass (M) and meansquare speed according to the following equation.
According to detailed kinetic theory the total kinetic energy of a mole of any gas
equals
3
2
RT
The average kinetic energy of a molecule is therefore
1mu 2 3 RT
2
2
NA
34
where NA is the Avogadro’s number.
Rearranging the equation we get
u2  2 x
u2 
3 RT
2 mN A
3RT
M
where mNA = M (molar mass)
3RT
M
 u rms 
urms
is called the root-mean-square (rms) molecular speed (the square root of the mean square
speed). It is a type of average molecular speed, and is the speed of a molecule that has the
average molecular kinetic energy.
Example
Calculate the root-mean-square speed of helium, atom in m/s at 25oC.
Answer
We need to use equation for root-mean-square
urms 
3 RT
M
R = 8.314 J/K
T = 25 + 273 K
Mass of He = 4.003 g /mol
urms 
=
3  8.314 J / K.mol  298K 
4.003 x10  3 kg / mol
1.86 x 10 6 J / kg
Using the conversion factor
1J = 1 kgm2/S2
We get
urms  1.86 x 106 kgm2 / kg.S 2
= 1.36 x 10 3 m / s
35
(a) Calculate the root-mean-square speeds of nitrogen molecule in m/s at
25oC.
[Ans 515 m/s]
(b) Compare the root-mean-square speeds of helium atom and nitrogen
molecule.
[Ans 1360/515 = 2.64]
36
LECTURE 4
REAL GASES
Introduction
So far our discussion has assumed that molecules in the gaseous state do not exert any
force, either attractive or repulsive, on one another. Further, we have assumed that the
volume of the molecule is negligibly small compared with that of the container. A gas that
satisfies these two conditions is said to exhibit ideal behaviour. In real life however, there is
no gas that is ideal as we shall learn in this lecture.
OBJECTIVES
At the end of this lecture you should be able to:1. illustrate the deviation from the ideal behaviour
2. define corrective terms in the van der Waals equation of state
calculate pressure of gas using both ideal gas equation and the van der Waals
equation
4.1 Deviation from ideal behaviour
Under what condition will gases deviate from ideal to non ideal behaviour?
In lecture one, we found that contrary to Boyle’s law, the pressure-volume product for
O2 and 0oC was not constant, particularly at high pressures (see under PV- values in Table
1.3).
Experiments have shown that the ideal gas law describes the behaviour of real gases
quite well at moderate pressures and temperatures but not so well at high pressure and low
temperatures.
Figure 4.1 shows the PV/RT versus P plot for three real gases at a given temperature.
These plots offer a test of ideal gas behaviour. According to the ideal gas equation (for 1
mole of the gas), PV/RT equals 1, regardless of the actual gas pressure. For real gases, this is
true only at moderately low pressure ( less or equal to 5 atm).
37
Figure 4.1 plot PV/RT versus P for 1 mole of a gas at 0oC1
Significant deviations are observed as pressure increases. The attractive forces among
molecules operate at relatively short distances.
For a gas at atmospheric pressure, the
molecules are relatively far apart and these attractive forces are negligible. At high pressures,
the density of the gas increases since the molecules are much closer to one another. The
intermolecular forces can become significant enough to affect the motion of the molecules,
and the gas will no longer behave ideally.
Another way to observe the non-ideality of gases, is to lower the temperature.
Cooling a gas decreases the molecules’ average kinetic energy which in a sense deprives
molecules of the drive they need to break away from their mutual attractive forces.
4.2 Van der Waals equation
The non-ideality of gases can be dealt with mathematically by modifying PV = nRT taking
into account intermolecular forces and definite molecular volumes. A Dutch physicist J.D.
van der Waals in 1973 made simple mathematical treatment, which provides us with
interpretation of real gas behaviour at the molecular level.
38
The pressure correction term
Van der Waals suggested that the pressure exerted by an ideal gas Pideal, is related to
the experimentally measured pressure Preal, by
Where a is a constant and n and V are the number of moles and volume of the gas,
respectively.
The correction term for pressure
 an 2  represents
 2
V 
the overall effect of
intermolecular attraction towards the reduction or pressure. The quantity Pideal is the pressure
we would measure if there were no intermolecular attraction. The quantity a, then, is just a
proportionality constant in the correction term for pressure. The value of a is an expression of
how strongly molecules of a given type of gas attract one another.
The volume correction term
The volume correction concerns the volume occupied by the gas molecules. We note
that for the ideal gas law, we assume that the molecules have negligible volume. However,
each molecule does occupy a finite volume, so the effective volume of the gas becomes
(V - nb)
when n is the number of moles of the gas and b is a constant. The term nb represents
the volume occupied by n moles of the gas.
Having taken into account the corrections for pressure and volume, we can rewrite
PV=nRT as

an 2 
 P 

V 

V  nb
corrected
pressure

nRT
Equation 4.0
corrected
volume
The above equation is known as van der Waals equation. The van der Waals constants
a and b are selected for each gas to give the best possible agreement between the equation
and the actually observed behaviour.
39
Derive the units for constants a and b in equation 4.1
The table below lists the values of a and b for a number of gases.
Gas
a (atm
b
L2/mol2
(L/mol)
He
0.03412
0.02370
Ne
0.211
0.0171
H2
4.244
0.0266
N2
1.39
0.0391
CO2
3.59
0.0427
CH4
2.25
0.0428
NH3
4.17
0.0371
H2O
5.46
0.0305
Table 4.1: Van der Waals constants for some common gases2.
From table 4.1 we see that helium atom have the weakest attraction for one another,
because helium has the smallest a value. There is also a rough correlation between molecular
size and b.
Generally the larger the molecule (or atom), the greater the b is, but the
relationship between b and molecular (or atomic) size is not a simple one. The following
example compares the pressure of a gas calculated using the ideal gas equation and the van
der Waals equation.
Example
A quantity of 3.50 moles of NH3 occupies 5.20L at 47oC. Calculate the pressure
of the gas (in atm) using (a) the ideal gas equation and, (b) the van der Waals
equation
Solution
(a) We have the following data
V = 5.20L
T = (47+273)K = 320K
40
R = 0.0821 L.atm/K.mol
Which we substitute in the ideal gas equation
nRT
V

3.50mol 0.0821L.atm / K .mol 320 K 

5.20 L
 17.7 atm
P
(b) From table 4.1 we have
a = 4.17 atm.L2/mol2
b = 0.0371 L/mol
It is convenient to calculate the correction terms for van der Waals equation.
They are


4.17atm.L2 / mol 2 3.50mol 
an 2

 1.89atm
V2
5.20L 2
2
nb  3.50mol 0.0371L / mol   0.130 L
Finally, substituting in the van der Waals equation, we get
P  1.89atm5.20L  0.130L   3.50mol 0.0821L.atm / K .mol 320K 
P  16.2atm
Note that the actual pressure measured under these conditions is 16.0 atm. Thus,
the pressure calculated by the van der Waals equation (16.2atm) is closer to the
actual value than that calculated by the ideal gas equation (17.7atm)
.
1. Under what set of conditions would a gas be expected to behave most ideally?
(a) High temperature and low pressure. (b) High temperature and high
pressure.
(b) Low temperature and high pressure. (d) Low temperature and low
pressure
2. Write the van der Waals equation of a real gas. Explain clearly the meaning
of the corrective terms for pressure and volume.
41
LECTURE 5
DETERMINATION OF MOLAR MASSES OF GASES
Diffusion and Effusion
The diffusion of one gas into another contributes to the transport of pheromones
(chemical signals between animals) and perfumes through air.
It helps to keep the
composition of the atmosphere approximately constant, since gases in high concentration
diffuse away. Even in the absence of wind, diffusion helps to disperse gases leaking from
chemical plants. In this lecture we are going to discuss diffusion and effusion process, the law
that governs them and their applications.
OBJECTIVES
At the end of this lecture you should be able to:1.
define the terms diffusion and effusion
2.
state Graham’s law of diffusion and effusion
3.
calculate molar masses and separation factor of gases
Diffusion
Gaseous diffusion is the process whereby a gas spreads out through another gas to
occupy the space with uniform partial pressure. A gas or vapour having a relatively high
partial pressure will spread out towards region of lower partial pressure of that gas until the
partial pressure becomes equal everywhere in the space.
Although the rate of diffusion depends in part on the average molecular speed, the
effect of molecular collisions makes the theoretical picture a bit complicated.
From our previous lecture, we found out that the root-mean-speed of helium atom is
greater than that of nitrogen molecules. This is because a light gas will diffuse through a
certain space more quickly than will a heavier gas.
In 1932, the Scottish chemist Thomas Graham discovered that the rate of diffusion of
a gas is inversely proportional to the square root of its density. Today, we usually state
Graham’s law of diffusion in terms of its molecular weight: “under the same conditions of
42
temperature and pressure, rates of diffusion of gaseous substances are inversely proportional
to the square roots of their molar masses”.
Graham’s law of diffusion is expressed mathematically as
r1

r2
M2
M1
Equation 5.0
Where r1 and r2 are the diffusion rates of gases 1 and 2 and M1 and M2 are their molar
masses, respectively.
Theoretical Explanation of Graham Law of Effusion
Equation 5.0 is based on experimental observation, but it can also be deduced from
the kinetic theory of gases as follow.
We saw earlier that any two gases at the same temperature will have the same kinetic
energy, that is
1
1
2
2
m1u1  m2 u1
2
2
Multiplying both sides of the equation by 2 and rearranging, we get
u 12 m2

u 2 m1
Taking the square root of both sides, we get
u 
u 
rms 1

rms 2
M2
M1
Since the rate of diffusion is proportional to the root-mean-square speed, this equation
is equivalent to equation 5.0.
Effusion
Whereas diffusion is a process by which one gas mixes with another, effusion is the
process by which a gas under pressure escapes from one compartment to another by passing
through a small opening. The rate of effusion is also given by Graham’s law of effusion.
That is
r2

r1
M1
M2
43
In practice, the rate of effusion of a gas is inversely proportional to the time it takes
for the gas to effuse through a barrier – the longer the time, the slower the effusion rate. So
we can write
t1 r2
 
t 2 r1
M1
M2
Equation 5.2
Where t1 and t2 are the time of effusion for gases 1 and 2, respectively.
Calculation of molar masses of gases
It requires 60 seconds for 1.5L of an unknown gas to effuse through a porous wall
and it takes 84 seconds for the same volume of N2 gas to effuse at the same
temperature and pressure. What is the molar mass of the unknown gas?
Calculation of separation factor of gaseous isotopes
Graham’s law has practical application in the separation of isotopes using equation
5.2. We define a quantity called separation factor (S) such that
S
r1

r2
M2
M1
Thus the value of S indicates how well gases 1 and 2 can be separated from each other
in one stage effusion process. The minimum value of S is 1, which indicates that the gases
are inseparable. (We assume that M2 > M1).
Activity
Work out the separation factor for
(a) the light and heavy isotopes of hydrogen, H2 and D2
(b) two isotopes of neon 20Ne and 22Ne.
The same principle applies to the preparation of fuel rods for nuclear fission reactors.
Natural uranium consists of 99.27% uranium - 238 (which does not undergo fission) and only
0.72% uranium – 235 (which undergoes fission). A Uranium fuel rod must contain about 3%
Uranium – 235 to sustain the nuclear reaction. To increase the percentage of uranium – 235
in a sample of Uranium (a process called enrichment), one just prepares Uranium
44
hexafluoride, Uf6, a white, crystalline solid that is easily vaporised. Uf6 vapour is allowed to
pass through a series of porous membranes. Because the Uf6 molecules with the light
isotopes of uranium travel about 0.4% faster than the Uf6 molecules with the heavier isotope,
separation is achieved. To achieve complete separation of the isotopes as required for bombgrade uranium, many thousands of effusion stages are needed.
45
PART 2
SOLUTIONS OF VOLATILE LIQUIDS AND SOLUTIONS CONTAINING NONVOLATILE SOLUTES
We define a solution as a mixture of two or more chemically non-reacting substances
whose relative amounts can vary within certain limits. The mixture can be homogeneous or
heterogeneous depending upon the size of the particles which are of the molecular size 15mµ (millimicron), cannot be separated from each other by mechanical means and that
composition is uniform throughout. A solution may be gaseous, liquid or solid of variable
composition within wide limits. A solute is generally taken as the substance which dissolves
and a solvent is a substance in which dissolution takes place.
This part consists of chapter six and chapter seven. Chapter six deals with the
classification of solutions, the units that are used to express concentration and also the
properties of solutions of volatile liquids. Chapter seven deals with colligative properties
solutions containing non-volatile solutes.
46
LECTURE 6
SOLUTIONS OF VOLATILE LIQUIDS
Solutions containing many components are encountered in chemistry. We will discuss
broad classification of solutions and the units that are used to express concentration.
However, we will confine our discussions to binary ideal solution, that is, solutions
containing only two components as we develop the laws that govern the vapour composition
of volatile liquids.
.
OBJECTIVES
At the end of this lecture you should be able to:1.
classify different types of solutions
2.
express and compare concentrations of liquids given in various units
3.
define an ideal solution and state Raoult’s law
4.
illustrate liquid versus vapour composition curves at constant
temperature or pressure
5.
describe fractional distillation process and column
6.0 Types of Solutions
Table 6.0 shows examples of possible types of binary solutions.
Solution
Solvent
Examples
1. Gas
Gas
2.
3.
4.
5.
6.
7.
8.
9.
Liquid
Solid
Gas
Liquid
Solid
Gas
Liquid
Solid
Any gas mixtures, air-governed by
Dalton’s law
Ammonia in water, aerated water
Hydrogen in palladium
Vaporisation of a liquid into a gas
Water in alcohol
Liquid bezene in solid iodine
Sublimation of a solid into a gas
Sugar in water
Lead in silver, copper in nickel, etc
Gas
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
Table 6.0. Examples of possible types of binary solutions
47
6.1 Measures of concentration
To predict the effect of solutes on the physical properties of solutions, we need to know the
relative numbers of solute and solvent molecules. The different ways of measuring
composition so as to emphasize either stoichiometry or relative numbers are summarized in
the table 6.1.
Measure
Units
Note
Moles of X per litre of solution
cx
m
X
Moles per litre
(M)
Grams per litre
Moles per kg (m)
None
5. Parts per million
PPM
None
6. Mass percent
%
None
1. Molar concentration
(molarity)
2. Mass concentration
3. Molality
4. Mole fraction
Notati
on
[x]
Grams of X per litre of solution
Moles of X per kilogram of solvent
Number of moles as a fraction of the
total number of moles of solute and
solvent
Number of molecules of solute per
million of moles of solute and
solution mg/L, mg/Kg
Mass of a component, expressed as a
percentage of the total mass
The choice of concentration units is generally based on the kind of measurement that
is to be made, as we shall see later in the lecture. First, let us examine the four most common
units of concentration present: percent by mass, mole fraction, molarity and molality.
Percent by mass
Percent by mass or percent by weight or weight percent is defined as
Percent by mass of solute 
Percent
=
Mass of solute
x100%
Mass of solute  Mass of solvent
Mass of solute
x100%
mass of solution
The percent by mass has no units because it is a ratio of two similar quantities.
Example
A sample of 0.892g of potassium chloride (KCl) is dissolved in 54.5g of
water. What is the percent by mass of KCl in this solution?
48
Solution
Percent by mass of solute KCl 
Mass of solute
x100%
Mass of solvent
0.892g
x100%
0.892g + 54.6g
= 1.61%
=
Calculate the percent by mass of the of 5.50 g NaBr in 78.2g of solution.
Mole fraction (X).
The concept of mole fraction was first introduced in lecture 1. The mole fraction of a
component of a solution, say component A, is written as XA and is defined as:
Mole fraction of component A  X A 
Moles of A
Sum of moles of all components
The mole fraction has not units, since it is a ratio of two similar quantities. Recall, the
sum of the mole fraction of all the components of a solution equals 1.
Example What are the mole fractions of glucose and water in a solution containing 4.75g of
glucose, C6H12O6 dissolved in 25.2g of water?
Solution
Molecular weight of glucose  180.2 amu
 moles of glucose = 4.57g C6 H12 O6 x
1 mol C6 H12 O6
 0.0254 mol
180.2g C6 H12 O6
The moles of water in the solution are
25.2g of H 2 O x
1 mol H 2 O
18.0 g H 2 O
 1.40 mol H 2 O
Hence the total moles of components in solution are
1.40 mol + 0.0254 mol = 1.425 mol
Finally, we get
Mole fraction glucose =
49
0.0254 mol
 0.0178
1.425 mol
Mole fraction water =
1.40 mol
 0.982
1.425 mol
Note: The sum of the mole fraction = 1
A solution is prepared by mixing 62.6mL of benzene (C6H6) with 80.3mL of
toluene (C7H8). Calculate the mole fraction of these two components. The
densities are benzene 0.879g/cm3 and toluene 0.867g/cm3.
Molarity (M).
This is also called molar concentration. Molarity is the number of moles of solute in1
litre of solution
Molarity =
Moles of solute
volume of solution in liters
Molality (m): Molality is the number of moles of solute dissolved in 1 kg of solvent i.e.
Molality =
Moles of solute
Mass of solvent (kg)
Example How would we prepare a 1 molal, or 1m sodium hydroxide (NaOH) aqueous
solution?
Solution
We need to dissolve 1 mole (40g) of the substance in 1000g (1kg) of water.
Note: Depending on the nature of the solute and solvent interaction, the final
volume of the solution will be either greater or less than 1000mL.
In how many grams of water should 18.7g of ammonium nitrate (NH4NO3) be
dissolved to prepare a 0.542 m solution?
Comparison of concentration units
As mentioned, the four concentration terms have various uses. Mole fraction unit is
generally used for calculating partial pressure of gases and for dealing with vapour pressures
of solutions as we shall see later in this chapter.
Molarity has the advantage over molality in that it is easier to measure the volume of
a solution, using precisely calibrated volumetric flasks, than to weigh the solvent. On the
other hand, molality is independent of temperature. This is because the concentration is
50
expressed in number of moles of solute and mass of solvent, whereas the volume of a
solution generally increases with increasing temperature. Since the concentration is expressed
in number of moles of solute and mass of solvent, whereas the volume of a solution generally
increases with increasing temperature. A solution that is 1.0M at 25oC may become 0.97M at
45oC because of the increase in volume.
This concentration dependence on temperature can significantly affect the accuracy of
an experiment. The following examples show that various units are interconvertible when the
density of the solution is known.
Example
Calculate the molarity of a 0.396 molal glucose (C6H12O6) solution. The molar
mass of glucose is 180.2 and the density of the solution is 1.16g/mL.
Answer
In working problems of this type, the mass of the solution must be converted to
volume. Since 0.396m glucose solution contains 0.396 moles of glucose in 1 kg
of water, the total mass of the solution is
0.396 mole C6 H12 O6 x
180.2g
 1000g H 2 O solution  1071g
1 mol C6 H12 O6
From the density (1.16g.mL), the volume of the
Solution =
1071
mL
1.16
This implies that
1071
mL solution contain 0.396 moles
1.16
How many moles would be contained in 1000 mL contain ?
Molarity is =
1000 x 0.396 1000 x 0.396 x 1.16
=
= 0.429 M
1071
1071
1.16
51
Example
The density of a 2.45M aqueous methanol (CH3OH) solution is 0.976g/mL.
What is the molality of the solution? The molar mass of methanol is 32.04g.
Answer
The total mass of 1L of a 2.45M solution of methanol is
1L solution x
1000 mL soln
0.976g soln
x
= 976 of solution.
1L soln
1mL soln
Since this solution contain 2.45 moles of methanol, the mass of water in the
solution is
976 g soln - (2.45 mol CH 3 OH x
32.04 of CH 3 OH
) = 898 g H 2 O.
1 mol CH 3 OH
Now molarity of the solution can be calculated as
Molality =
2.45 mol CH 3 OH 1000 g H 2 O
x
 2.728 m
898g H 2 O
1 kg H 2 O
Calculate the molality of a 35.4 percent (by mass) aqueous solution of
phosphoric acid (H2PO4). The molar mass of phosphoric acid is 98.00g.
52
6.2 Solutions of Volatile (Ideal) Liquids
Mixtures of two liquids
The two liquids which do not react chemically when brought together give rise to three
possibilities:(i)
Completely miscible liquid pairs which dissolve in each other in all proportion
and have no saturation limit e.g water-ethyl alcohol, benzene-toluene. These pairs
are chemically similar.
Completely immisible – such liquid pairs do not dissolve at all in each other e.g.
(ii)
(iii) Partially miscible liquids dissolve in each other only to certain limits e.g. waterether, water-phenol.
The ideal solution
Just as the concept of an ideal gas as a criterion of gas behaviour has been very helpful in
helping us predict properties of gases in vapour, it will also be very helpful to set up a similar
concept in the case of solutions on the basis of certain simple laws.
For practical purposes, an ideal solution may be described as:(i)
A solution in which there is complete uniformity of cohesive forces. Thus if we
have an ideal solution of two components, A and B, the intermolecular forces
between A and A, B and B and A and B will be the same.
(ii)
There is no heat given out or absorbed in the formation of solution.
(iii)
The total volume is equal to the sum of volume of individual components.
Raoults law
If two components of a solution are volatile, they will have a natural tendency of
escaping into vapour phase and exerting a definite vapour pressure. When the solution is in
equilibrium with vapour, the total pressure above the solution will be equal to the sum of the
partial vapour pressures of the individual components in accordance with Dalton’s law of
partial pressure. If PA and PB are the partial vapour pressures of two components A and B,
the total pressure, PT = PA + PB.
The assumption made here is that vapours behave as ideal gases. The partial vapour
pressure of each component of an “ideal solution” can be easily determined from Raoult’s
law. This law states that the partial vapour pressure of any volatile constituent of a solution
53
is equal to the vapour pressure of the pure constituent multiplied by the mole fraction of that
constituent in solution over the whole concentration range at all temperatures.
Thus, if XA and XB are the mole fractions of the two components in the solution and
PAo and PBo are the vapour pressures of the pure liquids A and B, then from Raoults law, the
vapour pressure of the two constituents above the solution are:-
and
PA  PAo X A
Equation 6.1
PB  PBo X B
Equation 6.2
Therefore, the total vapour pressure P of the solution is
PT  PA  PB  PAo X A  PBo X B
XA  XB  1
Again, since
Then
Equation 6.3
XA  1 XB
Equation 6.3 can be written as
PT  PAo (1  X A )  PBo X B
PT  PBo (1  PAo ) X B  PAo
Equation 6.4
Equation 6.4 gives the total vapour pressure above a solution in terms of the vapour
pressure of the two pure constituents and the mole fraction of temperature, since PAo and are
constants, a plot of P and XB should be a straight line with the slope  PBo  PAo  and the at
X B  0 as shown by the solid line in figure 6.1
o
A
vapour pressure
P
PT  PA  PB (Total vapour pressure)
PBo
PB  X A  PBo
PA
PA  X  Po
A
A
P
B
XA=1
XB=0
0.2
0.4 0.6
0.8
mole fraction
XA=0
XB=1
Figure 6.1: Total and partial vapour pressure curves for an ideal solution over the whole
concentration range at constant temperature.
54
In figure 6.1, the dotted lines represent plots of equations 6.1 and 6.2 for partial
pressure of the two components A and B against their respective mole fractions. Each dotted
line is a straight line and passes through the origin indicating that both components are ideal
liquids. The solid line gives the total vapour pressure of the mixtures of A and B of all
compositions and is obtained by joining PAo and PBo . The total pressure at all intermediate
concentrations can be obtained by taking the sum of the partial pressures of the two
components corresponding to the particular concentration
Thus, at
X B  0.2
PT  PA  PB
This type of the plot applies to ideal solutions.
In practice it has been found that a few binary solutions e.g. benzene-toluene, ethylene
bromide-ethylene chloride, n-hexane-n-heptane, carbon tetrachloride-silicon tetrachloride,
water-methyl alcohol and chlorobenzene-bromobenzene system obey Raoults law over the
whole range at temperatures.
Example
Benzene and toluene form nearly ideal solutions.
o
g
o
If at 300k, Ptoluene  32.06 mm and Pbenzene
 103.01mmHg
(a) Compute the vapour pressure of a solution containing 0.60 mole
fraction toluene.
(b) Calculate the mole fraction of toluene in the vapour for this
composition of liquid.
Solution
(a) Partial pressure of component A is given by PA  X A PAo
Partial pressure of toluene (t) Pt  X t Pt o
X t  0.6 and
Pt o  32.06
Pt  0.6 x 32.06  19.2 mmHg
o
Partial pressure of benzene PB  X B x Pbenzene
X B  0.4 since
X t  X B  1 and PBo  103.01mmHg
 PB  0.4 x 103.01  41.2 mmHg
(b) Mole fraction of toluene Pt in the vapour phase
55
=
.(a)
Pt
19.2

 0.32
Pt  PB 19.2  41.2
Using vapour pressure values for benzene and toluene provided in the
previous example, make a plot similar to that describe by Figure 6.1.
6.3
Relationship between the composition of a solution and the composition of the
vapour above it
As we have already established, the concentration of any ideal solution is governed by
Raoult’s law and the concentration of the vapour above the solution is governed by Dalton’s
law of partial pressure. Thus, for the system as considered under ideal solution, if YB is the
mole fraction of B in the vapour above a solution of composition XB, then according to
Dalton’s Law:
PB  YB P
or
YB 
Equation 6.5
PB
P
Substituting the value of P from equation (6.2) and of P from equation (6.4), we get:
PBo X B
YB  o
PB  PAo X B  PAo


Equation 6.6
This equation gives the relationship between the composition of B in the vapour phase
and the composition of B in the solution. Equation (6.6) suggest that YB and XB are different
and will be same only when PBo  PAo i.e. the pure liquids A and B have the same vapour
pressure. Equation (6.6) can also be written as:
Y
Mole fraction of B in vapour
= B 
Mole fraction of B in solution
XB
Equation 6.7 is derived as follows:
56
1
Po
X B  Ao X A
PB
Equation 6.7

Since






or









PBo X B
= o

o
PB X B  PA X A 

YB
PBo

= o
o
X B PB X B  PA X A 

1
=

o
P
X B  Ao X A 

PB
YB 
PB
PT
From the previous activity which of the two solvents is more volatile?
If follows from equation (6) that if PAo < PBo i.e. component A is less volatile than
component B, the denominator is less than 1 and the ratio YB /X B is greater than 1. This
means that the concentration of B will be more in the vapour phase than in the solution. The
general conclusion that follows is that for a system obeying Raoult’s law, the vapour is richer
in the more volatile component.
The same result is applicable to non-ideal solutions also. With the help of equation
6.6, it is possible to calculate the composition of the vapour in equilibrium with a solution
containing known amounts of the two components. Knowing the composition of the vapour
corresponding to any composition in the solution, a vapour pressure – vapour composition
curve at any definite temperature can be constructed. Such a plot for an ideal solution is
shown in Figure 6.2.
57
Total vapour pressure
T = Constant
Liquid
P/
A
V
l
X
XA = 1
XB = 0
Figure 6.2
Y
Mole fraction
XA or XB
B
r
pou
a
V
XA = 0
XB = 1
Liquid-vapour equilibria ( temperature is constant)
The curve marked “Liquid” is known as the Liquid composition curve. It gives the total
pressure above the solution as a function of mole fraction XB. The curve marked “Vapour” is
known as the vapour composition and gives the composition of the vapour in equilibrium
with the solution having the total vapour pressure indicated on the ordinate. For example, the
composition of vapour corresponding to a solution with composition XB = B and the total
pressure P (point l on the liquid composition curve) is YB = V (point V on the vapour
composition curve). The line lV is known as the tie line because it ties together the two
phases in equilibrium.
6.4 Boiling point diagrams of binary miscible liquid mixtures
In the types of diagrams discussed above, the temperature is constant. Since the vapour
above any solution is always richer in the more volatile component than the solution, the
components of a solution can be easily separated as discussed below.
By removing the vapours above a solution, the solution can be made richer in the less
volatile component. If the vapours are condensed to form a new liquid solution, the vapours
above this new solution will again be richer in the more volatile component than the solution
from which they came. By repeating several times the process of removing the vapours from
the 'newer' solution and condensing them, it is possible, to obtain a concentration of more
volatile component in the vapour and a concentration of less volatile component in the
58
solution.
This process of concentrating the components is called isothermal fractional
distillation because the process takes place at constant temperature.
Instead of conducting distillation at constant temperature and varying pressure, the
general practice is to conduct distillation at constant pressure. The boiling point of a solution
is the temperature at which the total vapour pressure becomes equal to external pressure. The
external pressure is frequently 1 atm. Thus, a solution of two components A and B will boil
when the total pressure i.e. PA+PB becomes equal to 1 atm. Since, according to Raoult's law,
different compositions of a solution have different vapour pressures, it follows that solutions
of different compositions will boil at different temperatures. Thus a solution whose
components have higher vapour pressure will boil at a lower temperature than the solution in
which the components have lower vapour pressures. This is because solutions of higher
vapour pressures can reach the external pressure (1 atm) at relatively lower temperatures and
start boiling sooner than the solutions of lower vapour pressures.
Based upon this fact, it is possible to construct a boiling point diagram corresponding to
vapour pressure diagram for ideal solutions as shown in the Figure 6.3.
Comparing the boiling point diagrams with the vapour pressure diagrams, we find that in
boiling point diagrams, the liquid curves are below the vapour pressure curves. This can be
readily explained because at constant pressure, the vapour state exists at the higher
temperature.
The vapour pressure of pure A is the lowest pressure and that of pure B is the highest
while the vapour pressure of all possible compositions of A and B lie between the two values,
therefore, in the boiling point diagram, the boiling point of A is the highest and that of B is
the lowest while the boiling points of all possible compositions of A and B lie in between the
two values. The decrease in boiling point is regular from A and B.
59
Figure 6.3:
Types of boiling point diagram corresponding to vapour pressures
diagram.
6.4
Fractional distillation of binary miscible liquid mixtures
System which is ideal
Consider the boiling point versus composition diagram for such a system, which is
shown in Figure 6.4.
Figure 6.4 Distillation behaviour of solutions of ideal solution
Suppose we heat a solution of composition X1. The vapour pressure of the system
will increase until it reaches the point a. At this point the vapour pressure of the solution
becomes equal to the external pressure and hence the solution begins to boil. Thus t a is the
boiling point of the solution of composition X1. At this temperature, the vapours coming off
will have the composition Y1, (point a’ on the vapour composition curve). Since the vapours
are richer in the more volatile component B, (Y1 > X1.), the residual liquid will become richer
in the less volatile component A. Let it be represented by the point b on the liquidcomposition curve. A solution of composition (X2) will boil at temperature tb, which is higher
than ta. The corresponding composition of the vapours in equilibrium with solution of
composition X2 is given by Y2, point b' on the vapour composition curve. These vapours are
again richer in the component B and consequently the residual liquid will be further enriched
in the component A. The temperature of this residual liquid must again be raised until it
begins to boil. Thus we see that if the process described is continued several times, the
residue will become richer and richer in A than the original solution until pure A is obtained.
60
The boiling point will also increase continuously from ta to tA, the boiling point of the pure
component A.
Consider now the vapour phase. If the initial vapours obtained from the solution
(point a') as pointed out earlier, are condensed and redistilled, the vapours of composition Z
(represented by the point c') are obtained and the boiling point of the new solution will be tc.
This shows that the distillate becomes richer in component B than the original. When the
process is repeated several times, a stage comes when the distillate is composed of pure B
only.
Thus we see that by carrying out fractional distillation of mixtures, which are ideal, it is
possible to separate the two constituents into the pure components.
6.5 The fractionating column
Figure 6.5 shows small-scale fractional distillation apparatus used in the laboratory to
separate volatile liquids like toluene and benzene. The round-bottomed flask containing the
mixture of solvents to be separated is fitted with a long column packed with small glass beads
that provides the surface for repeated vaporization-condensation steps. At each step the
composition of the vapour in the column gets richer in the more volatile, or lower boilingpoint, component. Finally, the more volatile component rises to the top of the column and is
condensed and collected in the receiving flask The less volatile component remains in the
flask.
61
Figure 6.5 Laboratory distillation apparatus
In petroleum industry for example, the process of fractional distillation would be
extremely tedious and involve more time and labour if the separation is carried out in batches
and in discontinuous manner. Petroleum is a complex mixture of hydrocarbon compounds as
you will learn in the organic chemistry lectures. The mixture of these hydrocarbons can be
classified according to the range of their boiling points. To overcome these difficulties, a
fractionating column on a large scale is used to separate the various components of crude oil.
There are various types of fractionating columns, the most generally used in the industry
is a “bubble-cap” type of column as shown in Fig. 6.6. The column, which consists of a
number of bubble-cap plates is attached to a boiler at the bottom and to a condenser at the
top. Each plate can hold a thin layer of liquid. Liquid can flow down from one plate to the
next by a series of overflow pipes and the vapour can rise and escape through several bubble
caps. The vapours moving upward through the caps prevent the liquid descending through
them. To understand the working, the liquid mixture to be separated is boiled by heating
coils in the boiler at the bottom. The vapours after passing through the caps come into
contact and mix with the liquid at the lower plate. Some of the vapours of the less volatile
62
component will condense here and the remaining vapours, richer in the more volatile
component, will pass on through caps to upper plate. Here they will again come into contact
and mix with the liquid at upper plate.
This will cause vapours of the volatile component to condense, and vapours of the
more volatile component to move upwards. On repeating this process at every plate, the
vapours moving upwards become enriched in the more volatile component and the liquid
flowing down through the column become richer in the less volatile component. Thus, the
more volatile component can be drawn off in the pure form by condensing the vapours at the
top of the column and less volatile component from the bottom.
In order to make the separation continuous, the preheated liquid mixture is admitted
into one of the plates at an intermediate stage in the column and the two pure liquids are
drawn off from the top and bottom of the column. As the vapours from the boiler make up
the column, they come into contact with the cooler liquid moving downwards and there is a
continuous condensation-vapourization exchange between the vapour and the liquid phases at
each plate in the column. The various fraction obtained from crude oil are as shown in the
figure 6.6.
7
Figure 6.6. Diagram of bubble-cap fractionating column for separating crude oil.
63
Most solutions do not behave ideally and therefore do not follow the Raoult’s law.
They show either positive or negative deviation from the law. This aspect of nonideal behaviour will be dealt with in your second year of study.
64
LECTURE 7
COLLIGATIVE PROPERTIES OF SOLUTIONS
Introduction
In the last lecture we mainly considered physical properties of mixtures of two
components system of solutions that are volatile. In this lecture we will consider properties of
dilute solutions whose solute is not volatile. The properties of these solution, are referred to
as colligative properties which is the subject of this lecture.
OBJECTIVES
At the end of this lecture you should be able to:1.
define a colligative property
2.
illustrate the four colligative properties
3.
calculate molecular mass of an unknown compound
Definations
It has been observed that any solution containing some non-volatile solute, exhibits
the four properties, namely;
(i)
Vapour pressure lowering of the solvent,
(ii)
Boiling point elevation of the solution,
(iii)
Freezing point depression of the solution; and
(iv)
Osmotic pressure of the solution.
These four properties are collectively known as the colligative properties of solution.
A colligative property is that property which depends only upon the number of solute
particles present in the solution and in no way depends upon the chemical nature of these
particles. The solute present should not undergo any association or dissociation. If there is
association of solute particles, the number of particles actually added will become small and
correspondingly, the properties mentioned above will be lowered. Similarly, if the solute
particles are dissociated as in the case of electrolytes, the number of particles becomes larger
than the original number and the value of the colligative properties become higher.
Therefore, for such solutions in which the solute particles undergo association, or
65
dissociation, some modification is essential over the laws deduced for solutions in which the
solute particles are not associated or dissociated. While discussing these properties, it will be
assumed that the solutions are very dilute and are ideal i.e. the interactions between the solute
particles are negligibly small.
The study of these properties has proved very helpful in the determination of molecular
weight of the dissolved substances, which are non-volatile.
7.1 Vapour pressure lowering of the solvent
Whenever some non-volatile solute is added to a liquid, solvent vapour pressure of the
solvent is lowered. This can be easily explained with the help of Raoult’s Law. Consider
any solution containing some non-volatile solute and let X1 and X2 be the mole fractions of
the solvent and the solute respectively. If Po is the vapour pressure of the pure solvent and P
is the vapour pressure of the solvent above the solution, then according to Raoult’s Law.
P =Po X1
Equation 7.1
Since X1 is a positive quantity and is smaller than unity, therefore, P the vapour
pressure of the solvent above the solution is always less than Po - the vapour pressure of the
pure solvent. Hence, it can be said that whenever a non-volatile solute is added to a solvent,
the effect is the lowering of the vapour pressure of the solvent.
The decrease in vapour pressure,  P, is given by
 P = Po - P
P = Po X1,
Since from equation (7.1),
Therefore,
 Po = P - Po X1,
 Po = Po(1-X1)
But
Equation 7.2
X1 + X2 = 1 or 1-X1 = X2
Hence equation (7.2) becomes
∆P = Po X2,
Equation (7.3)
P P o  P

 X2
Po
Po
Equation (7.4)
This equation can also be written as
66
The quantity
Po  P
is known as the relative lowering of vapour pressure. It can
Po
readily be seen from equation (7.4) that the relative lowering of vapour pressure is a
colligative property as it depends only on the mole fraction of the solute.
Equation (7.4) can be utilised to calculate the molecular weight of the solute if the
vapour pressures of the pure solvent and the solution are known as follows:
Since the mole fraction of the solute, X 2 
n2
n1 + n2
where n 2 is the number of moles of solute and n1 is the number of moles of solvent. n 2 and
n1 are given by
n2 
W2
W
and n1  1
M2
M1
W2 is the weight of the solute of molecular weight M 2 and W1 , is the weight of the
solvent of molecular weight M1 . Therefore, equation (7.4) becomes
n2
W2 M 2
Po  P


o
P
n1 + n2 W1 M1  W2 M 2
Equation 7.5
Since the solution under consideration is very dilute, n 2 << n1 and n1 + n2  n1 .
P o  P  n2  W2 /M 2  W2   M 1 
  

   
 
Po
 n2  W1 M 1  W1   M 2 
Thus
Or
P o  P  W2   M 1 
  

 
Po
 W1   M 2 
Equation (7.6)
If all the quantities on the right hand side of equation 7.6 are known, M2 the molecular
weight of the solute can be calculated.
Example When 18.04g of a sugar were dissolved in 100g of water, the vapour pressure of
the later at 20oC was lowered from 17,535 mmHg to 17.216 mmHg. Calculate the
molecular weight of the sugar.
Solution
In this case, W1 is 100g and M 1 , the molecular weight of water is 18.02; W2 is
18.04 g. The values of Po and P are 17.535 and 17.226 mmHg respectively. The
67
relative lowering of the vapour pressure is obviously independent of the units used
for the vapour pressures, provided Po and P are expressed in the same units.
Hence by equation (7.6)
P o  P  W2   M 1 
  

 
W
M
Po
 1  2
17.535  17.226  20.00   18.04

 
17.535
 100   M 2




M 2  205
or
The vapour pressure of water at 20oC 17 17.0 mmHg. Calculate the vapour
pressure of a solution of 3 g of urea in 50 g of water. Molecular weight of water is
18 and that of urea is 60.
7.2 Boiling point elevation of solutions
Boiling point of a liquid is defined as the temperature at which the vapour pressure of the
liquid becomes equal to the confining pressure. It has been found that whenever some nonvolatile solute is added to a liquid, its boiling point becomes higher. The difference between
the boiling point of the solution and the boiling point of the pure solvent at any constant
pressure is known as the boiling point elevation of the solution. Boiling point elevation is a
direct consequence of vapour pressure lowering of the solvent by a non-volatile solute. We
will show mathematical relationships between the boiling point elevation and the vapour
pressure lowering.Consider the “vapour pressure-temperature” curves for the pure solvent
and solution as shown in Figure 7.1.
68
Fig. 7.2. Vapour pressure curves for solvent and solution.
In the diagram, curve AB represents the variation of vapour pressure of pure solvent
with temperature. Similarly curves CD and EF represent the variation of vapour pressure of
two solutions of different concentrations with temperature (Solution II is more concentrated
than solution 1). The curves CD and EF must be below AB at all temperatures because the
vapour pressure of a solution is always lower than that of the solvent. The boiling points of
pure solvent, solution I and solution II at the same external pressure (1 atmosphere) are To, T1
and T2 respectively.
It is evident from the diagram that the elevation of boiling point of solution I,  Tb1=
T1 – T0 is represented by distance GH and that of solution II,  Tb2= T2 – T0 is represented by
GI.
Points G, J and K represent the vapour pressure of pure solvent, solution I and
solution II respectively at the same temperature, T0. The vapour pressures corresponding to
these points are P0, P1 and P2, respectively. The distances GJ and GK represent the lowering
of vapour pressure of solution I and solution II and are given by (P0 – P2) and (P0 – P2)
respectively.
Since the solutions are dilute, their vapour pressure curves are nearly parallel to the
vapour pressure curve for pure solvent; particularly in the region of boiling point and to a
good approximation may be considered to be straight lines.
Triangles GHJ and GIK are similar. From analytical geometry,
69
GH GJ

GI
GK
GH =  Tb1 for solution I = T1 – T0
GI =  Tb2 for solution II = T2 – T0
GJ =  P1 for solution I = P0 – P1
GK =  P2 for solution II = P0 – P2
Therefore
T1  T0
P 0  P1
 0
T2  T0 P  P2
T1 P1

T2 P2
or
Dividing the numerator and denominator of the right hand side of this equation by P0, we
get,
Tb1 P1 P0

Tb2 P2 P0
This equation indicates that the boiling point elevations for the two solutions are in the
same ratio as their relative lowering of vapour pressure and hence boiling point elevation is
directly proportional to the relative lowering of vapour pressure in dilute solution, i.e.,
Tb a
Further,
P
P0
P
 X B ; mole fraction of the solute
P0
Therefore,
Tb a X B
or
Tb  AX B
where A is a constant of proportionality.
Again X B 
nB
n
 B (since the solutions are very dilute)
nA + nB nA
Where n B is the number of moles of solute and nA is the number of moles of solvent.
and
nB WB M B

nA WA M A
70
Where the terms W and M represent the weights and molecular weights of the respective
constituents.
Tb  A
Hence
WB M A
WA M B
 AM A.
WB 1
.
M B WA
Since M A is constant for a particular solvent, both constants A and M A are amalgamated
into a single constant, kb .
Thus Tb  kb
WB 1
.
M B WA
Equation 7.7
The constant kb is known as the ebullioscopic constant or boiling point elevation constant
for a solvent and signifies the boiling point elevation for one molal solution of a solute in a
solvent, i.e.,
WB
(  nB ) = 1 and WA  1 gram.
MB
Tb  kb if
If one mole of the solute is dissolved in 1000 grams of the solvent, the constant is known
as the molal elevation constant and is defined as the elevation in boiling point obtained on
dissolving one mole of a non-volatile solute in 1000 grams of the solvent. In terms of
molality, equation (7.7) can be written as
Tb  K b .
Where
The quantity
Kb 
WB 1
.
x1000
M B WA
Equation 7.8
kb
1000
WB 1
.
x1000 is known as the molality (m) of the solution.
M B WA
 Tb = Kbm
Equation 7.9
Also known as the molal boiling point elevation constant, Kb, has the units of oC/m .
The choice of molality in expressing the concentration unit here is because we are dealing
with solutions whose temperature is not kept constant. Table 7.1 lists values of Kb, for
several common solvents.
71
Solvent
Water
Benzene
Ethanol
Acetic acid
Cyclohexane
Normal freezing
point(oC)
0
5.5
-117.3
16.6
6.6
Kf
( C)/m
1.86
5.12
1.99
3.90
20.0
o
Normal boiling
Point(oC)
100
80.1
78.4
117.9
80.7
Kb
( C)/m
0.52
2.53
1.22
1.93
2.79
o
Table 7.1 Vaues of boiling points, freezing points, Kb, Kf for common solvents.
Using the boiling-point elevation constants given in figure 7.1 calculate the
boiling point benzene and water for aqueous solution whose concentrations
are:-
(a) 1 molal (b) 2 molal
(c) 3 molal
7.3 Freezing point depression of solutions
On cooling a dilute solution, a temperature is ultimately reached at which solid solvent
begins to separate from the solution. This temperature at which the precipitation of the solid
solvent begins is the freezing point of the solution. At this temperature, solid solvent and
solution are in equilibrium and consequently they must have the same vapour pressures. It is
observed that freezing point of a solution is lower than that of pure solvent. This is again a
direct consequence of the lowering of the vapour pressure of the pure solvent by the
dissolved solute. Consider the vapour pressure-temperature diagram as shown in Fig. 7.2.
Figure 7.2 Vapour pressure-temperature diagram
In the diagram, AB is the sublimation curve of the solid solvent and BC is the vapour
pressure curve of the pure liquid solvent. These two curves are intersecting at the point B.
72
The temperature T0, corresponding to this point, is the freezing point of the solvent because at
this temperature, the solid and the liquid phases are in equilibrium with each other and have
identical vapour pressures. DE and FG are the vapour pressure curves of solutions I and II
(concentration of solution II is greater than the concentration of solution I). These two curves
are intersecting the sublimation curve at points D (temperature T1) and F (temperature T2)
respectively. Therefore, T1 and T2 are the freezing points of solution I and solution II
respectively. Both T1 and T2 are lower than T0.
The freezing point depression of solution I is
T f1  T0  T1
and the freezing point depression of solution II is
T f 2  T0  T2
Let P0 is the vapour pressure of the solid and the pure liquid solvent at temperature T0
and P1, and P2 are the vapour pressures of solutions I and II respectively at the same
temperature. For dilute solutions, the freezing point depression will be small and to a good
approximation, the vapour pressure curves may be considered to be parallel straight lines in
the vicinity of the freezing point.
Since the two triangles BHD and BIF are similar, therefore,
DH BH

FI
BI
But
DH = T0 – T1 =  Tf1, the freezing point depression for solution I
FI = T0 – T2 =  Tf2, the freezing point depression for solution II
BH =P0 – P1 =  P1, vapour pressure lowering for solution I
BH =P0 – P2 =  P2, vapour pressure lowering for solution II
Therefore
T0  T1 P0  P1

T0  T2 P0  P2
or
Tf1
T f 2

P1
P2
Dividing the numerators and denominator of the right hand side of this equation by P0,
T f 1
T f 2
P1 P 0

P2 P 0
73
T f a
or
P
P0
Equation.(7.10)
i.e/, the freezing point depression of a solution is directly proportional to the vapour pressure
lowering of the solvent.
Again,
P
 X B , mole fraction of the solute and
P0
XB 
nB
n
n
W
M
 B  B  B x A.
nA + nB nA nA WA WA
All the symbols have got the same meanings as given under boiling point elevation.
Therefore, equation (7.10) becomes
WB M A
WA M B
T f a
or
T f  A
WB M A
WA M B
Equation 7.11
Where A is some proportionality constant. Again, for a given solvent, M A is constant.
Therefore, both the constants A and M A are replaced by another constant k f in the
equation (7.11).
Hence
T f  k f
WB
WA M B
Equation 7.12
The constant k f is known as the depression constant or cryoscopic constant and
depends on the nature of the solvent. If 1 mole of the solute ( i.e., WB  1 is dissolved in 1
MB
gram of the solvent ( WA  1 gram), then from equation (7.12), T f  k f i.e., the depression
constant becomes freezing point depression when 1 mole of any solute is dissolved in 1 gram
of the solvent.
But generally in dilute solutions, concentration is expressed in terms of molality m,
which is defined as the number of moles of solute dissolved per 1000 grams of the solvent.
For the solution under consideration
m  Kb.
WB
x1000
WA M B
74
Equation 8.3
In term, of molality, equation (7.13) becomes
T f  k f .
T f  K f . m
or
The quantity

m
1000
Equation 7.14
WB 1
.
x1000 is known as the molality (m) of the solution.
M B WA

Where K f  k f 1000 is known as the molal depression constant. For a molal solution,
T f  K f .
The three colligative properties i.e vapour pressure lowering, boiling point elevation and
freezing point depression can be illustrated in one phase diagram as shown in figure 7.3.
Figure 7.3. Phase diagram illustrating the colligative properties
7.4 Defination of osmosis and osmotic pressure
Osmosis is an important colligative property of solution and was first reported
by Abbe Nollet in 1748. He observed that when a solution of a solute is separated
from a pure solvent by a suitable membrane, the solvent spontaneously passed into the
solution. This passage of the solvent into solution or from more dilute to concentrated solution when the two are separated from each other by a membrane is known
as osmosis. In osmosis, only the flow of solvent takes place. If there is a movement
of solute in the opposite direction, the process is referred to as diffusion.
75
With specific examples describe the role a semi-permeable membrane.
The membrane used for osmosis is known as the semi-permeable membrane. This
because it allows only solvent molecules and not the solute molecules to pass through it
freely. There are various types of semi-permeable membranes such as animal membranes,
cellulose membrane and a film of cupric ferrocyanide, CU2[Fe(CN)6] etc. The membrane
used depends on the nature of the solvent and solute.
The phenomena of osmosis can readily be explained with the help of apparatus shown
in Figure 7.4 An inverted thistle funnel, the end of which is tied by an animal membrane, is
partially filled with a concentrated solution of sugar.
This is immersed into a beaker
containing pure water. Due to osmosis, water will begin to pass through the membrane and
the level of the sugar solution in the tube begins to rise until it reaches a definite height,
depending on the concentration of the solution. The hydrostatic pressure, set up from the,
differences in levels of the sugar solution in the tube and the surface of the, pure water is the
osmotic pressure, of the solution. This pressure can be defined as the mechanical pressure
which must be applied on a solution to, prevent the passage of solvent into the solution
through a semi-permeable membrane.
7.3 Osmosis through animal membrane.
7.5 Measurement of Osmotic Pressure
Consider the diagram as shown in Fig. 7.4.
76
7.4. Apparatus used to determine the osmotic pressure of a solution.
A is a chamber fitted at both ends with very thin bore capillary tubes. This chamber is
divided by a semi-permeable membrane into two compartments of which the right one is
filled with the solvent and the other with the solution.' At the beginning of the experiment,
levels of the liquid in the two capillaries are kept the same. Due to osmosis, solvent begins to
pass through the membrane into the solution. As a result of this, the level of the liquid in the
capillary on the solution side begins to rise and the level on the solvent side starts descending
until equilibrium is reached. The difference in capillary levels gives the osmotic pressure.
Instead of using a capillary on the solution side, a piston can also be used so that pressure can
be applied just to prevent osmosis. This pressure will then be equal to osmotic pressure.
Results of osmotic pressure measurement show that the ratio of the osmotic pressure
  to the concentration i.e.,  C , is approximately constant at constant temperatures. In
other words, the osmotic pressure of a solution is directly proportional to its concentration at
constant temperature. Further, C
1
where V is the volume of solution containing 1 mole of
V
the solute. Therefore, osmotic pressure of a solution is inversely proportional to volume at
constant temperature.

 V
1
(T = Constant)
V
= Constant
Equation 7.15
77
Experimental data shows that the osmotic pressure increases with increase in
temperature for constant concentration of solution and the ratio,  T is approximately
constant i.e.,

= Constant (At constant concentration)
T
Equation 7.16
Equations (7.15) and (7.16) are similar to Boyle's Law and Charles’s Law,
respectively for gases with the only difference that the osmotic pressure,   , is taken for the
gas pressure. Vant Hoff concluded from these results that there is a parallelism between the
properties of solution and the properties of gases. He combined equations (7.15) and (7.16) as
in the case with gases and suggested the relation,
 V  RT
Equation 7.17
where R is a constant. By substituting the value of osmotic pressure for a solution of known
concentration at a definite temperature, he calculated the value of R. This came out to be
almost identical with the familiar gas constant. Equation (7.17) is known as the Van't Hoff
Equation and holds only true for dilute solutions.
In equation (7.17), V is the volume of the solution containing 1 mole of the solute. If
n2 moles of the solutes is dissolved in volume V of the solution, equation (7.17) becomes
 V  n RT
Equation 7.18)
2
If there are two solutions having the same osmotic pressure at the same temperature, they
will have the same molar concentration. Such solutions are regarded as Isotonic Solutions).
If two solutions are of unequal pressure, the more concentrated is said to be hypertonic and
the more dilute solution is described as hypotonic. The osmotic pressure phenomenon
manifests in many applications such as:
-
Blood cells
-
Preservation of jam or jelly
-
Transportation of water upwards in plants or transpiration
Osmotic Pressure-A Colligative Property
From equation (7.18), we have
  CRT
where C is the concentration of the solution
78
This equation shows that for a definite temperature, osmotic pressure depends on the
concentration of the solute and not on the nature of solute or solvent. Therefore, osmotic
pressure of a solution is a Colligative property.
Determination of molecular weight. Osmotic pressure measurement can be employed to
obtain the molecular weight, of the solute with more degree of accuracy. This can be easily
done by putting n 2 
W2
equation (7.18) where W2 is the weight of the solute of molecular
M2
weight M 2 .
Example
A solution containing 4.0g of a polyvinyl chloride polymer in 1 litre of dioxane
was found to have an osmotic pressure of 6.4 x 10-4 atm at 27oC. Calculate the
approximate molecular weight of the polymer.
Solution
The concentration of the solution is
4.0
mole litre-1 where M is the required
M
molecular weight. Since the osmotic pressure is given in atm, R is expressed in
litre-atm R is expressed in litre-atm deg-1 mol-1, i.e., 0.082, and hence by
equation
  RTC
we have 6.4 x 10-4 = 0.082 x 300x4.0/M
or
M = 1.5x105
Molecular weight of the polymer = 1.5x105
79
PART 3
CHEMICAL EQUILIBRIUM
We have seen several examples of physical processes such as evaporation or
vaporization and dissolving that reach dynamic equilibrium, the state in which the forward
and reverse processes are continuing at equal rates. Now we shall see that the same ideas
apply to chemical reactions involving reactants and products. They also approach a dynamic
equilibrium in which the forward and reverse reactions continue at matching rates. By
making use of the responses of the equilibrium composition to changes in the pressure and
temperature, we can influence the formation of a product and hence increase the formation of
a product and hence increase its yield as we will establish later in our lessons.
This part consists of following three chapters.
Lecture 8:
The types of chemical equilibrium
Lecture 9:
Application of the equilibrium constant concept
Lecture 10:
Changing the reaction conditions
80
LECTURE 8
THE TYPES OF CHEMICAL EQUILIBRIUM
Reversible chemical reactions involve reactants and products that can be in different
phases. In this lecture we will start with descriptions, definitions and relationships that will
help us to understand the equilibrium constant concept.
OBJECTIVES
At the end of this lecture you should be able to:1.
describe dynamic equilibrium
2.
define an equilibrium constant
3.
show interrelationship between Kp and Kc
4.
illustrate heterogeneous and multiple equilibria
8.1 Description of Dynamic Equilibrium.
Let us consider a general reaction
aA  bB  cC  dD
Equation 8.1
Where A, B, C and D represent reactants and products, a, b, c and d are coefficients in
the balanced chemical equation.
Reactants A and B react to form products C and D.
However, the reverse reaction, the decomposition of C and D, also occurs. At equilibrium
the forward and reverse reactions take place at the same rate.
Figure 8.1 shows how the concentration of A, B, C and D change with time. At the start of
experiment the concentrations of C and D, which are, zero increases steadily and reaches a
constant value. On the other hand, the concentration of A and B decreases up to a certain
value when the concentration of reactants and products no longer change.
The final
composition of the reaction mixture corresponds to a dynamic equilibrium in which the
products decompose just as they are formed.
81
Figure 8.1: Concentration vs time valuation:
Note how the amounts of substances become constant at equilibrium
We show this dynamic character by writing the chemical equation of a reaction at
equilibrium with a double headed arrow ()
This signifies that the forward reaction and its reverse are continuing with equal rates.
8.2 Definition of an equilibrium constant
Guldberg and Waage studied a variety of reactions and found that in every case, the
equilibrium composition of a particular reaction could be expressed by an equilibrium
constant. They proposed the ‘law of mass action’ to summarize their conclusions. Using the
reaction given in equation 8.1, the equilibrium constant Kc can be expressed as:
 C c  D  d
Kc 
 A a  B b
Equation (8.2)
Where K c is a constant.
Notice that the products [C and D] occur in the numerator and the reactants [A and B]
in the denominator. Each concentration is raised to a power equal to the stoichiometric
coefficient in the balanced equation. The molar concentrations A, B and so on in the
definition of K c are the concentrations at equilibrium. The units of K c depend on the
stoichiometry of the reaction.
82
The law of mass action is a relation that states that the values of the equilibrium
constant K c are constant for a particular reaction at a given temperature, whatever
equilibrium concentrations are substituted.
As the following example illustrate, the
equilibrium constant expression is defined in terms of the balanced chemical equation. If the
equation is rewritten with different coefficients, then the equilibrium expression will be
changed.
Examples
Write the equilibrium constant expression K c for
(i)
CO(g) + 3H2 (g)  CH4 (g) + H2O (g)
(ii)
CH4 (g) + H2O (g)  CO(g) + 3H2 (g)
(iii)
N2 (g) + 3H2 (g)  2NH3 (g)
(iv)
½ N2 (g) + 3/2H2 (g)  NH3 (g)
(i)
Kc 
Solution
CH   H O
CO  H 
4
(ii)
2
3
CO  H 2 
3
Kc 
CH   H O
4
2
 NH 

N   H 
2
(iii)
Kc
3
3
2
(iv)
Kc 
2
2
 NH 
N   H 
3
1
2
2
3
2
2
Write equation (i) given in the previous example in reverse order and write the Kc
expression for it. What is the relationship between the two constants?
1. When the equation is written in reverse order, the expression
written for K c is inverted.
2. When the coefficients in the equation in (iii) are multiplied by ½ to
give ½N2 + 3/2H2  NH3 the equilibrium constant expression is
square root of the previous one.
That is
Kc  Kc
83
1. Write equilibrium constant expressions K c for each of the following reactions.
(a)
(b)
(i)
N (g) + Br (g)  2HBr (g)
(ii)
4HCl (g) + O2 (g)  2H2O (g) + 2Cl2 (g)
(iii)
CO (g) + 2H2 (g)  3CH3OH (g)
(iv)
3CH3OH (g)  CO (g) + 2H2 (g)
The equilibrium constant K c for the equation 2HI
(g)
 H2
(g)
+ I2
(g)
at
450oC is 1.84.
What is the value of K c for the following equation? H2 (g) + I2 (g)  2HI (g).
(.C) If you double or triple a chemical equation throughout, what happens to the
original value of the equilibrium constant?
8.3. Ways of expressing equilibrium constants
In lecture 6 we saw that the concentrations of reactants and products can be expressed
in several types of units. Also, because the reacting species are not always in the same phase,
there may be more than one way to express the equilibrium constant for the same reaction.
To begin with, we will consider reactions in which the reactants and products are in the same
phase (homogeneous). Then we will look at another type of equilibrium (heterogeneous)
reaction.
8.3.1 Homogeneous Equilibria
The term homogeneous equilibrium applies to reactions in which all reacting species
are in the same phase. An example of a homogeneous gas-phase equilibrium is the
synthesis of hydrogen iodide, which reaches the equilibrium.
H2 (g) + I2 (g)  2HI (g).
List two units in which you can express the concentration of gaseous reactants and
products.
84
The equilibrium constant K c denotes that in this form of the equilibrium constant, the
concentration of the species are expressed in moles per litre (mol/L).The concentrations of
reactants and gaseous reactions can also be expressed in terms of their partial pressures (PA).
From the ideal gas law equation, the pressure PV = nRT, we see that at constant
temperature, the pressure P of a gas is directly related to the concentration in mol/L of the
 n
gas; that is P =   RT
V 
Thus, for the equilibrium of hydrogen iodide manufacture we can write
PHI2
Kp =
PH 2 .PI 2
Where PHI , PH2 and PI2 are the equilibrium partial pressures (in atm) of HI, H2 and I2,
respectively. The subscript in Kp tells us that the equilibrium concentrations are expressed in
terms of pressure.
In general, K c is not equal to Kp, since the partial pressures of reactants and products
are not equal to their concentrations expressed in moles per litre. A simple relationship
between Kp and K c can be derived as follows:
Let us consider the following equilibrium in the gas phase.
aA(g)  bB(g)
where a and b are stoichiometric coefficients. The equilibrium constant K c is given
by
Kc 
 Bb
 Aa
PBb
and the expression for K p is K p  a
PA
where PA and PB are the partial pressures of A and B, respectively.
Assuming ideal gas behaviour,
PAV = n A RT
 PA =
n A RT
V
where V is the volume of the container in litres. Also
PBV = nB RT
85
 PB =
nB RT
V
Substituting these relations into the expression for K p , we obtain
 nB RT

V
Kp = 
 nA RT

 V
b
b

 nB 

 
   V  RT b  a
a
a

 nA 

 

V 
Now both nA/V and nB/V have the units of mol/L and can be replaced by [A] and [B],
respectively so that
Kp 
 Bb   n
RT
 Aa
 K c  RT 
n
Equation 8.3
where n  b  a = moles of gaseous products – moles of gaseous reactants.
Since pressures are usually expressed in atm, the gas constant R is given by
0.0821L.atm/K.mol, and we can write the relationship between Kp and K c as
K p = 0.0821T  K c
n
Equation 8.4
In general K p  K c except in the special case when n = 0 . In that case equation
(8.4) can be written as
K p = K c 0.0821 T   K c
0
Consider the reactions for the special case where; n = 0 , H2 (g) + I2 (g)  2HI (g);
N2 + O2  2NO ; H2 + CO2  H2 + CO. If these system was in equilibrium,
predict the effect that changes in volume would have to their equilibrium constant.
A homogeneous liquid equilibrium, is illustrated by the ionisation of acetic acid
(CH3COOH) in water:
CH 3COOH ( aq)  H 2 O(l )  CH 3COO(aq)  H 3O(aq)
The equilibrium constant is
CH COO  H O 
CH COOH  H O

-
K c' 
3
3
3
2
86
However, when the acid concentration is low  1M  and/or the equilibrium constant is
small   1 , as it is in this case, the amount of water consumed in this process is negligible


compared with the total amount present. Thus we may treat H2O as a constant and rewrite
the equilibrium constant as:
CH COO  H O 
CH COOH 

-
Kc 
3
3
3
We will extend this practice to acid-base equilibria and to solubility equilibria (to be
discussed in the lecture 11 and 12)
The following examples illustrate the procedure for writing equilibrium constant
expression and calculating equilibrium constants and equilibrium concentrations.
Example
Write expressions for Kp and K c if applicable, for the following reversible
reactions at equilibrium:
(a) HF (aq) + H2O (l)  H3O+ (aq)+ F(aq)
(b) 2NO(g)  O2 (g)  2NO2 (g)
(c) 2H2S(g) + 3O2(g)  H2O(g)+ 2SO2(g)
(d) CH3COOH (aq)+ C2H5OH(aq)  CH3COOC2H5 + H2O(l)
Solution
(a) Since there are no gases present, Kp does not apply and we have only K c'
 H O  F 
 HF  H O

K c' 

3
2
HF is a weak acid, so that the amount of water consumed in acid ionisations is
negligible compared with the total amount of water present as solvent. Thus we
can rewrite the equilibrium constant as
 H O  F 

Kc 
(b).
 NO 
 NO  O 
2
Kc 
3
 HF 
Kp 
;
2
2

2
(c).
(d).
H O2 SO 2
Kc  2 2 2 3 ;
H 2 S  O2 
P 2 NO2
P 2 NO PO2
P 2 H2O P 2 SO2
Kp  2
P H2S P3O2
The equilibrium constant K c' is given by
K c' 
 CH COOC H  H O
 CH COOH  C H OH 
3
3
87
2
5
2
2
5
Because the water produced in the reaction is negligible compared with the water
solvent, the concentration of water does not change. Thus we can write the new
equilibrium constant as
Kc 
CH COOC H 
CH COOH  C H OH 
3
2
3
Example
2
5
5
The following equilibrium process has been studied at 230oC.
2NO(g) + O2(g)  2NO2 (g)
In one experiment the concentrations of the reacting species at equilibrium are
found to be:
[NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M. Calculate
the equilibrium constant K c  of the reaction at this temperature.
Solution
The equilibrium constant is given by
 NO 
 NO  O 
2
Kc 
2
2
2
Substituting the concentrations, we find that
15.5 2
Kc 
 6.44 x 10 5
 0.0542 2  0.127 
8.4 Heterogeneous equilibria
A reversible reaction involving reactants and products that are in different phases
leads to a heterogeneous equilibrium. For example, when calcium carbonate is heated in a
closed vessel, the following equilibrium is attained:
CaCO3(s)  CaO(s) + CO2 (g)
The two solids and one gas constitute three separate phases. At equilibrium, we might
expect the equilibrium constant to be
K c' 
 CaO  CO 
 CaCO 
2
Equation 8.5
3
But how do we express the concentration of a solid substance? Because both CaCO3
and CaO are pure solids, their “concentrations” do not change as the reaction proceeds. This
88
follows from the fact that the molar concentration of a pure solid (or a pure liquid) is a
constant at a given temperature; it does not depend on the quantity of the substance present.
From the above discussion we can express the equilibrium constant for the
decomposition of CaCO3 in a different way. Rearranging equation (8.5), we obtain
 CaCO  K  CO
 
 CaO
3
'
c
2
Equation 8.6
Since both [CaCO3] and [CaO] are constants and K c' is an equilibrium constant, all the
terms on the left-hand side of the equation are constants. We can simplify the equation by
writing
 CaCO  K
3
CaO
'
c
 K c  CO2 
where K c , the “new” equilibrium constant, is now conveniently expressed in terms of
a single concentration, that of CO2. Keep in mind that the value of K c does not depend on
how much CaCO3 and CaO are present, as long as some of each is present at equilibrium).
Alternatively, we can express the equilibrium constant as
K p  PCO2
Equation 8.7
The equilibrium constant in this case is numerically equal to the pressure of the CO 2
gas, an easily measurable quantity.
What has been said about pure solids also applied to liquids. Thus if a pure liquid is a
reactant or a product in a reaction, we can treat its concentration as constant and omit it from
the equilibrium constant expression.
Reactions involving heterogeneous equilibria are shown in the following examples.
Examples
Write the equilibrium constant expression K c and K p if applicable, for each of
the following heterogeneous systems:
(a) (NH4)2Se (s)  2NH3(g)+ H2Se(g)
(b) AgCl(s)  Ag+(aq) + Cl-1(aq)
(c) P4(s) + 6Cl2(g)  4PCl3(l)
Solutions
(a)
The equilibrium constant is given by
K c' 
89
NH 3 2 H 2 Se
NH 3 2 Se
However, since (NH4)2 Se is a solid, we write the new equilibrium constant as
K c   NH 3   H 2 Se
2
where
K c  K c' H 2 Se .
Alternatively, we can express the equilibrium
constant K p in terms of the partial pressures of NH3 and H2Se:
2
K p  PNH
PH 2 Se
3
(b)
 Ag  Cl 
 AgCl 

+
K c' 

 
K c  Ag+ Cl 
Again, we have incorporated [AgCl] into K c because AgCl is a solid.
 PCl 
 P  Cl 
4
K 
'
c
3
4
6
2
Since pure solids and pure liquids do not appear in the equilibrium constant
expression, we write
Kc 
1
Cl2 6
Alternatively, we can express the equilibrium constant in terms of the pressure of
Cl2;
Kp 
1
PCl6 2
90
Example
Consider the following heterogeneous equilibrium:
CaCO3(s)  CaO(s) + CO2(g)
At 800oC, the pressure of CO2 is 0.236 atm. Calculate (a) K p and (b) K c for the
reaction at this temperature.
Solution
(a)
Using Equation (8.7), we write K p  PCO2  0.236
(b)
From Equation (8.4), we know K p  K c  0.0821T 
(c)
T = (800 + 273) = 1073K and ∆n = 1, so we substitute these in the
equation
n
and obtain 0.236 = Kc(0.0821 x 1073)
K c  2.68 x103
8.41 Multiple equilibria
The reactions we have considered so far are all relatively simple. Let us now consider
a more complicated situation in which the product molecules in one equilibrium system are
involved in a second equilibrium process:
A + B  C + D E + F
This equation can be broken into to steps as follows:
A+BC+D
C+DE+F
The products formed in the first reaction, C and D, react further to form products E and F. At
equilibrium we can write two separate equilibrium constants:
K c' 
and
K c" 
C D
 A B
 E  F 
C D
The overall reaction is given by the sum of the two reaction
Overall reaction:
A +B  C + D
K c'
C +D  E + F
K c"
A +B  E + F
Kc
and the equilibrium constant K c for the overall reaction is
91
Kc 
 E  F 
 A B
We obtain this same expression if we take the product of the expressions for K c' and K c"
K c' K c" 
C D  E  F   E  F 
x

 A B C D  A B
Therefore we arrive at K c = K c' K c"
Equation 8.8
We can now make an important statement about multiple equilbria: If a reaction can be
expressed as the sum of two or more reactions, the equilibrium constant for the overall
reaction is given by the product of the equilibrium constants of the individual reactions.
Among the many known examples of multiple equilibria is the ionisation of diprotic acids in
aqueous solution. The following equilibrium constants have been determined for carbonic
acid (H2CO3) at 25oC:
H 2 CO3 aq   H + aq  + CO3- aq  : K c' 
HCO3- aq   H + aq  + HCO32- aq  : K c" 
H CO   4.2 x10
+
3
H 2 CO3 
H HCO   4.8x10
H CO 
+
23
2

3
The overall reaction is the sum of these two reactions
H 2 CO3 aq   2 H + aq  + CO32 aq 
and the corresponding equilibrium constant is given by
 H   HCO 

 H CO 
+ 2
Kc
23
2
Using Equation (8.8) we arrive at
92
7
3
11
K c = K c' K c"
= 4.2 x 10 -7  4.8 x 10 -11 
 2.0 x 10 17
93
LECTURE 9
APPLICATION OF THE EQUILIBRIUM CONSTANT CONCEPT
Introduction
In the last lecture we described how a chemical reaction reaches equilibrium and how
an equilibrium constant for a given reaction can be calculated from known equilibrium
concentrations at a constant temperature. In this lecture we shall explore the uses of the
equilibrium constant.
OBJECTIVES
At the end of this lecture you should be able to:1.
determine the qualitative interpretation of the equilibrium constant
2.
predict the direction of a reaction
3.
calculate the equilibrium concentrations
9.1 Qualitative interpretation an equilibrium constant
By looking at the magnitude of K c , we can tell whether a particular equilibrium reaction
favours products or reactants. Let us consider three cases to get a sense of whether reactants
or products are favoured.
Consider the K c for a reaction
aA  bB
 cC  dD
reactants
products
Kc 
at ToC
C c Dd
Ac Bb
Case I
If K c is large compared to 1, this means that the numerator (product concentrations) is
larger than the denominator (reactants concentrations). This means that at this temperature
the reaction favours the formation of products.
Case II
If K c is small compared to 1, the reactants are favoured at equilibrium.
94
Case III
If K c is neither large nor small between 0.1 and 10, neither reactants nor products are
favoured at equilibrium.
The equilibrium constant K c for the reaction: N2 (g) + O2 (g)  2NO
Equals 4.6 x 10-31 at 25oC. Does the equilibrium mixture contain mostly reactants
or products? If [N2] = [O2] = 0.4M at equilibrium, what is the equilibrium
concentration of NO at 25oC?
9.2 Predicting the direction of a reaction
The equilibrium constant K c for the reaction CO
(g)
+ H2O
(g)
 H2 (g) + CO2 is 5.1 at
850oC. Suppose that in a certain experiment, we initially placed certain amounts of the
reactants and products. Would the reaction go towards the right or towards the left? That is,
would the mixture form more H2
(g)
and CO2 in going towards equilibrium or would it form
more CO and H2O? To answer this question, we substitute the initial concentrations into the
equilibrium constant expression. We write
 H   CO 
CO  H O
2 o
o
 Qc
2 o
2
Equation 9.1
o
where the subscript o indicates initial concentrations.
This new constant is called
reaction quotient (Qc). The reaction quotient, Qc, is an expression that has the same form as
the equilibrium constant expression, but whose concentration values are not necessarily those
at equilibrium.
To determine in which direction a reaction will proceed to achieve
equilibrium, we compare the values of Qc and Kc. The three possible cases are as follows:-
(i)
Qc > Kc, the ratio of initial concentrations of product to reactant is too large. To
reach equilibrium, the value of Qc must decrease to equal Kc. That is, the
numerator of the expression Qc = [H2]o [CO2]o will have to decrease. Thus,
products must be converted to reactants. The system proceeds from right to left.
(ii)
Qc = Kc, the initial concentration of reactants and products are equilibrium
concentrations. The system is already at equilibrium.
(iii)
Qc < Kc, the ratio of initial concentrations of product to reactants is too small. To
reach equilibrium the value of Qc must increase to equal Kc.
95
That is, the
numerator of Qc equation 9.1. [H2]o[CO2] will increase, and the denominator
[CO]o [H2O]o will decrease. Thus, reactants must be converted to products. The
system proceeds from left to right.
Example
At the start of a reaction, there is 1.50 mol N2, 3.50 mol H2 and 0.55 mol NH3 in
a 60.0 L reaction vessel at 350oC. If the reaction equilibrium constant (Kc) for
the reaction
N2 (g) + 3H2 (g)  2NH3(g)
is 0.510 at 350oC.
Determine whether the reaction is at equilibrium and if not, predict the direction
of reaction.
Solution
The composition of the gas has been given in terms of moles. We convert these
to molar concentrations by dividing by the volume (60.0L).
N 2 o  1.50 mol  0.025
H 2 o  3.50 mol  0.058
60.0 L
NH 3 o

60.0 L
0.55 mol
 .0092
60.0 L
We substitute these concentration into the reaction quotient gives
Q
NH 3 02
N 2 o H 2 3o

0.092 2
0.0250.0583
   17.227
Since Qc is greater than Kc = 0.550, the reaction will go from right to left as it
approaches equilibrium. Therefore ammonia will dissociate.
96
Example
At the start of a reaction, there is 0.249 mol N2, 3.21 x 10-2 mol H2 and 6.42 x
10-4 mol NH3 in a 3.50-L reaction vessel at 200oC. If the equilibrium constant
N2 (g) + 3H2 (g)  2NH3(g) is 0.65 at this
(Kc) for the reaction:
temperature, decide whether the system is at equilibrium. If it is not, predict
which way the net reaction will proceed.
Solution
The initial concentrations of the reacting species are:
N 

H 
3.21 x 10 -2 mol

 9.17 x 10  3 M
3.50 L
2 o
2 o
 NH 
3 o
0.249 mol
 0.0711 M
3.50 L
6.42 x 10 -4 mol

 1.83 x 10  4 M
3.50 L
Next we write
2
NH 3 02

1.83 x 10  4 

N 2 o H 2 3o 0.0711 9.17 x 10  3 3
 0.611  Qc
Since Qc is smaller than Kc = 0.65, the system is not at equilibrium. The net
result will be an increase in the concentration of NH3 and a decrease in the
concentrations of N2 and H2. That is, the net reaction will proceed from left to
right until equilibrium is reached.
9.3
Calculating equilibrium concentrations
If we know the equilibrium constant for a particular reaction, we can calculate the
concentrations in the equilibrium mixture from the initial concentrations. Depending on the
information given, the calculation may be straightforward or complex. In the most common
situation, only the initial reactant concentrations are given. Let us consider the following
system, which has an equilibrium constant (Kc) of 24.0 at a certain temperature:
A

B
Suppose that A is initially present at a concentration of 0.850 mol/L. How do we calculate
the concentrations of A and B at equilibrium? From the stoichiometry of the reaction, we see
that for every mole of A converted, 1 mole of B is formed. Let x be the equilibrium
concentration of B in mol/L; therefore, the equilibrium concentration of A must be (0.850 – x)
mol/L. It is useful to summarize the changes in concentration as follows:
97

A
Initial:
B
0.850 M
0M
-x M
+xM
Change:
Equilibrium: (0.850 - x) M
xM
A positive (+) change represents an increase and a negative (-) change indicates a decrease in
concentration at equilibrium. Next we set up the equilibrium constant expression.
Kc 
B
A
 24.0 
x
0.850  x
Solving for x
x  0.816
Having solved for x, we can now calculate the equilibrium concentrations A and B as follows:
[A] = (0.850 – 0.816) M = 0.034 M
[B] = 0.816 M
We summarize our approach to solving equilibrium constant problems as follows:1.
Express the equilibrium concentrations of all species in terms of the initial
concentrations and a single unknown x, which represents the change in
concentration.
2.
Write the equilibrium constant expression in terms of the equilibrium
concentrations. Knowing the value of the equilibrium constant, solve for x.
3.
Having solved for x, calculate the equilibrium concentrations of al species.
The following examples illustrate the application of the three-step approach.
Example
A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainlesssteel flask at 430oC. Calculate the concentrations of H2, I2 and HI at equilibrium.
98
The equilibrium constant Kc for the reaction H2 (g) + I2 (g)  2HI (g) is 54.3 at this
temperature.
Solution
Step 1: The stoichiometry of the reaction is 1 mol H2 reacting with 1 mol I2 to
yield 2 mol HI. Let x be the depletion in concentration (mol/L) of either H2 or I2
at equilibrium. It follows that the equilibrium concentration of HI must be 2x.
We summarize the changes in concentrations as follows:
I2

0.500 M
-xM
(0.500 - x) M
H2
+
Initial:
0.500 M
Change:
-x M
Equilibrium: (0.500 - x) M
2HI
0.000 M
+ 2x M
+ 2x M
Step 2: The equilibrium constant is given by
Kc 
 HI  2
H  I 
2
2
 2x 2
54.3 
 0.500 - x 0.500  x
Substituting
Taking the square root of both sides, we get
7.37 
2x
0.500 - x
x  0.393 M
Step 3: At equilibrium, the concentrations are
[H2] = (0.500 - 0.393) M = 0.107 M
[I2] = (0.500 - 0.393) M = 0.107 M
[HI] = 2 x 0.393 M = 0.786 M
99
Example
For the same reaction and temperature as I Example 14.9, suppose that the initial
concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M,
respectively. Calculate the concentrations of these species at equilibrium.
Solution
Step 1:
Let x be the depletion in concentration (mol/L) for H2 and I2 at
equilibrium. From the stoichiometry of the reaction if follows that the increase
in concentration for HI must be 2x. Next we write
I2

0.00414 M
-xM
(0.00414 -x) M
H2
+
Initial:
0.00623 M
Change:
-x M
Equilibrium: (0.00623 - x) M
2HI
0.0224 M
+ 2x M
(0.0224+2x M
Step 2: The equilibrium constant is
Kc 
 HI  2
H  I 
2
Substituting, we get
2
 0.0224  2 x  2
54.3 
0.00623 - x 0.00414  x 
It is not possible to solve this equation by the square root shortcut, as the starting
concentrations [H2] and [I2] are unequal. Instead, we must first carry out the
multiplications
54.3 (2.58 x 10-5 – 0.0104x + x2) = 5.02 x 10-4 + 0.0896 + 4x2
Collecting like terms together, we get
50.3x2 – 0.654x + 8.98 x 10-4 = 0
This is a quadratic equation of the form ax2 + bx + c = 0. The solution for a
quadratic equation of this kind:
x
 b  b 2  4ac
2a
Here we have a = 50.3, b = -0.654, and c = 8.98 x 10-4, so that
x
0.654 
 0.6542
x = 0.0114 M or

 450.3 8.98 x10 4
2 x 50.3

x = 0.00156 M
The first solution is physically impossible since the amounts of H2 and I2 reacted
would be more than those originally present. The second solution gives the
100
correct answer. Note that in solving equations of this type, one answer is always
physically impossible, so the choice of which value to use for x is easy to make.
Step 3: At equilibrium, the concentrations are
[H2] = (0.00623 - 0.00156) M = 0.00467 M
[I2] = (0.00623 - 0.00156) M = 0.00258 M
[HI] = (0.0224 + 2 x 0.00156) M = 0.0255 M
In some cases simplifications are possible that greatly reduce the mathematical
difficulties in equilibrium constant calculations. The following example shows the conditions
under which valid approximations can be made.
Example The equilibrium constant (Kp) for the following reaction is found to be 4.31 x 10-4
at 200oC.
N2 (g) + 3H2 (g)  2NH3(g)
In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2
in a constant volume vessel at 200oC. Calculate the pressures of all species when
equilibrium is reached.
Solution
Step 1: Since initially there is no NH3 present, we conclude that the reaction must
proceed from left to right to establish equilibrium. In examples 14.9 and 14.10 we
used molarities in calculations. However, it is equally valid to use pressures for a
gas-phase reaction at constant temperature and volume. The ideal gas equation
 n
P    RT shows that the pressure of a gas is directly proportional to the
V 
number of moles per litre of volume.
Thus we can represent the change needed to establish equilibrium in terms of
pressures. We summarize the changes in pressures as follows:
Initial:
Change:
Equilibrium:
N2
+
0.862 atm
-x atm
(0.862 - x) atm
101
3H2

0.373 atm
- 3x atm
(0.373 - 3x) atm
2NH3
0 atm
+ 2x atm
2x atm
Step 2: The equilibrium constant is given by
P 2 NH 3
Kp 
PN2 P3 H 2
Substituting, we get
4.31 x 10
4
2

2x

0.862- x0.373  3x 3
Multiplying and collecting like terms together would give an equation with terms
containing x4, x3, x2, and x, which would be very difficult to solve. Fortunately,
we can greatly simplify the above equation by realizing that, because Kp is a small
number, the amount of NH3 formed at equilibrium must also be very small. Thus
we make the following approximations: 0.862 – x  0.862 and 0.373 – 3x  0.373.
The equilibrium constant expression can now be written as
 2x 2
4.31x10 
0.862 0.373 3
4
Solving for x gives
x = 2.20 x 10-3 atm
Step 3: At equilibrium pressures are
PN 2 = (0.862 – 2.20 x 10-3) atm = 0.860 atm
PH2 = (0.373 – 3 x 2.20 x 10-3) atm = 0.366 atm
PNH3 = 2(2.20 x 10-3) atm = 4.40 x 10-3 atm
102
LECTURE 10
EFFECTS OF CHANGING THE REACTION CONDITIONS ON CHEMICAL
EQUILIBRIUM
Introduction
Getting the maximum amount of product from a reaction depends on the proper
selection of reaction conditions. By changing these conditions, we can increase or decrease
the yield of a product.
There are several ways in which we can alter the equilibrium
composition of a gaseous reaction mixture and possibly increase the yield of product as we
will show in this lecture.
OBJECTIVES
At the end of this lecture you should be able to:1.
State LeChatelier’s Principle
2.
determine the direction of a reaction upon:- addition of a catalyst,
removal of product or addition of a reactant, or change of the pressure
and temperature to reaction mixture
3.
choose the optimum conditions of reaction
10.1 ADDING A CATALYST
A catalyst is a substance that increases the rate of a reaction but is not itself
consumed. The significance of a catalyst can be seen in the reaction of sulfur dioxide with
oxygen to give sulfur trioxide.
2SO2(g) + O2(g)  2SO3(g)
The equilibrium constant K c , for this reaction is 1.7 x 1O26, which indicates that for
all practical purposes the reaction should go almost completely to products. Yet, when sulfur
is burned in air or oxygen, it forms predominantly SO2 and very little S03. Oxidation of SO2
to S03 is simply too slow to give a significant amount of product. However, the rate of the
reaction is appreciable in the presence of a platinum or divanadium pentoxide catalyst. The
oxidation of SO2 in the presence of a catalyst is the main step in the contact process for the
103
industrial production of sulfuric acid, H2SO4. The acid is prepared by reacting SO3 with
water.
It is important to understand that a catalyst has no effect on the equilibrium
composition of a reaction mixture. A catalyst merely speeds up the attainment of equilibrium.
A catalyst increases both the forward and reverse reaction rates. For example, suppose we
mix 2.00 mol SO2 and 1.00 mol O2 in a 100.0-L vessel. In the absence of a catalyst, these
substances appear unreactive. Much later, if we analyze the mixture, we find essentially the
same amounts of SO2 and O2. But when a catalyst is added, the rates of both forward and
reverse reactions are very much increased. As a result, the reaction mixture comes to
equilibrium in a short time. The amounts of SO2, O2, and SO3 can be calculated from the
equilibrium constant. We find that the mixture is mostly SO3 (2.00 mol), with only 1.7 x 10-8
mol SO2 and 8.4 x 10-9 mol O2.
A catalyst is useful for a reaction, such as 2SO2 + O2  2 SO3, that is normally slow
but has a large equilibrium constant. However, if the reaction has an exceedingly small
equilibrium constant, a catalyst is of little help. The reaction
N2(g) + O2(g)  2NO(g)
has been considered for the industrial production of nitric acid (NO reacts with O2 and H2O to
give nitric acid). At 25oC, however, the equilibrium constant K c , equals 4.6 x 10-31.
An equilibrium mixture would contain an extremely small concentration of NO. We cannot
expect a catalyst to give a significant yield at this temperature; a catalyst merely speeds up
the attainment of equilibrium. The equilibrium constant increases as the temperature is
raised, so that at 2000oC, air (which is a mixture of N2 and O2) forms about 0.4% NO at
equilibrium. An industrial plant was set up in Norway in 1905 to prepare nitrate fertilizers
using this reaction. The plant was eventually made obsolete by the Ostwald process for
making nitric acid in which NO is prepared by the oxidation of ammonia. This latter reaction
is more economical than the direct reaction of N2 and O2, in part because the equilibrium
constant is larger at moderate temperatures.
The Ostwald process is an interesting example of how a catalyst can affect the product
obtained from a mixture by making one reaction much faster than other possible reactions.
Ammonia burns in oxygen to give nitrogen.
4NH 3( g )  3O 2 ( g )  2N 2 ( g )  6H 2 O ( g )
However, if a hot platinum wire is inserted into a mixture of NH3 and O2, the reaction is
different. This time the product is nitric oxide, NO.
104
4NH 3( g )  5O 2 ( g )
pt
 4NO
(g)
 6H 2 O ( g )
True equilibrium actually favors the oxidation of NH3 to N2 rather than NO. We can see this
by
noting
that
the
equilibrium
2NO( g )  N 2 ( g ) + O 2( g ) is 2.2 x1030 .
constant
Kc ,
for
the
reaction
Therefore any NO that forms should eventually
dissociate to the elements. However, this dissociation of NO is extremely slow at room
temperature. The platinum catalyst speeds up the formation of NO from NH3 and has no
effect on its rate of dissociation.
Table 10.1 lists the catalysts used in some important gaseous reactions. Note that
carbon monoxide and hydrogen react to give various products, depending on the catalyst.
Mixtures of CO and H2 can be made from many organic, or carbon containing, substances by
reacting with steam. For example,
CH 4( g )  H 2 O ( g )  CO ( g )  3H 2 ( g )
natural gas
C s   H 2 O ( g )  CO ( g )  H 2 ( g )
coal
REACTION
N 2  3H 2  2NH 3
4NH 3  5O 2  4NO  6H 2 O
CATALYST
Fe with K2O and
Al2O3
Pt - Rh mixture
2SO 2  O 2  2SO 3
Pt, V2O5
C2 H 4  H 2  C2 H 6
Ni, Pt, Pd
CO  H 2 O  CO 2 + H 2
ZnO – CuO mixture
CO  3H 2  CH 4 + H 2 O
Ni
8CO  17H 2  C 8 H 18 + 8H 2 O
Fe-Co mixtures
CO  2H 2  CH 3 OH
CH 4  2S 2  CS 2 + 2H 2 S
ZnO-Cr2O3 mixture
Al2O3
105
COMMERCIAL APPLICATION
Haber process for ammonia
Ostwald process for nitric acid
(NO reacts with O2 and H2O to
give HNO3)
Contact process for sulphuric acid
(SO3 reacts with H2O to give
H2SO4)
Hydrogenation of compounds with
C=C (such as unsaturated
vegetable oils)
Water-gas-shift reaction (to
produce H2 or synthesis gas with
higher concentration of H2)
Catalytic methanation (preparation
of substitute natural gas)
Fischer-Tropsch process
(preparation of synthetic gasoline)
Industrial synthesis of methanol
Industrial synthesis of methanol
2CO  O 2  2CO 2
C 7 H 16


 11O 2  7CO 2  8H 2 O 
Pt, Pd, V2O5
Automobile catalytic converter (to
educe pollutants in exhaust gases)
Table 10.1 Catalysts used in some gaseous reactions2
In the presence of ZnO and Cr2O3, carbon monoxide and hydrogen give methanol,
CH3OH. Methanol is the starting material for formaldehyde, CH2O, which is used to make
plastics. In the Fischer-Tropsch process, developed in Germany in 1940, an iron-cobalt
catalyst is used to convert mixtures of CO and H2 to liquid hydrocarbons for synthetic
gasoline. In each case, the catalyst determines the product by allowing one reaction to attain
equilibrium much faster than all others.
Study of the catalysts and reaction conditions for the various reactions of CO and H2
has become an active research area. These reactions make it possible to convert a variety of
organic materials, including coal and organic wastes, to gaseous and liquid fuels and all of
the industrial chemicals presently obtained from petroleum.
10.2 REMOVING PRODUCTS OR ADDING REACTANTS
One way to increase the yield of a desired product is to change concentrations in a
reaction mixture by removing a product or adding a reactant. Consider the methanation
reaction,
CO( g )  3H 2 ( g )  CH 4( g )  H 2 O ( g )
If we place 1.000 mol CO and 3.000 mol H2 in a 10.00-L reaction vessel, the
equilibrium composition at 1200 K is 0.613 mol CO, 1.839 mol H2, 0.387 mol CH4, and
0.387 mol H2O. Can we alter this composition by removing or adding one of the substances
to improve the yield of methane?
To answer this question, we can apply LeChatelier's principle, which states that if a
system in chemical equilibrium is altered by the change of some condition', chemical reaction
occurs to shift the equilibrium composition in a way that acts to reduce the effect of that
change of conditions. Suppose we remove a substance from or add one to the equilibrium
mixture in order to alter the concentration of the substance. Chemical reaction then occurs to
partially restore the initial concentration of the removed or added substance. (However, if the
concentration of the substance cannot be changed, as in the case of a pure solid or liquid
reactant or product, changes in amount will have no effect on the equilibrium.)
106
For example, suppose that water vapor is removed from the reaction vessel containing
the equilibrium mixture for methanation. LeChatelier's principle predicts the net chemical
change will occur to partially reinstate the original concentration of water vapor. This means
that the methanation reaction momentarily goes in the forward direction,
CO( g )  3H 2 ( g )  CH 4( g )  H 2 O ( g )
until equilibrium is re-established. Going in the forward direction, the concentrations of both
water vapor and methane increase.
A practical way to remove water vapor in this reaction might be to cool the reaction mixture
quickly to condense the water. Liquid water could be removed and the gases reheated until
equilibrium was again established. The concentration of water vapor would build up again as
the concentration of methane increased. Table 10.2 lists the amounts of each substance at
each stage of this process. Note how the yield of methane has been improved.
STAGE OF PROCESS
Mol CO
Mol H2
Mol CH4
Mol H2O
Original reaction mixture
0.613
1.839
0.387
0.387
After removing water
0.613
1.839
0.387
0
When equilibrium is re-established
0.491
1.473
0.509
0.122
Table 10.2 The effect of removing water vapor from a methanation Mixture2
It is often useful to add an excess of a cheap reactant in order to force the reaction
toward more products. In this way, the more expensive reactant is made to react to a greater
extent than it would otherwise.
Consider the process of ammonia synthesis:
N 2( g )  3H 2 ( g )  2NH 3 ( g )
If we wished to convert as much hydrogen to ammonia as possible, we might increase
the concentration of nitrogen. To understand the effect of this, first suppose that a mixture of
nitrogen, hydrogen, and ammonia is at equilibrium. If nitrogen is now added to this mixture,
the equilibrium is disturbed. According to LeChatelier's principle, the reaction will now go
in the direction that will use up some of the added nitrogen. Consequently, adding more
nitrogen has the effect of converting a greater quantity of hydrogen to ammonia
107
We can look at these situations in terms of the reaction quotient, Q. Consider the
methanation reaction, in which
Q
 CH   H O
CO  H 
4
2
3
2
If the reaction mixture is at equilibrium, Q = K c . Suppose we remove some H2O from this
equilibrium mixture. Now Q = K c and from what we said in lecture 9, we know the reaction
will proceed in the forward direction to restore equilibrium.
We can now summarize the conclusions from this section:
When more reactant is added to, or some product is removed from, an equilibrium mixture,
thereby changing the concentration of reactant or product, net reaction occurs left to right
(that is, in the forward direction) to give a new equilibrium, and more products are produced.
Example Predict the direction of reaction when H2 is removed from a mixture in which the
following equilibrium is established:
H 2( g )  I 2 ( g )  2HI ( g )
Solution
When H2 is removed from the reaction mixture, the reaction goes in the reverse
direction (more HI dissociates to H2 and I2) to partially restore the H2 that was
removed.
Consider each of the following equilibria that are disturbed in the manner
indicated. Predict the direction of reaction.
(a)
CaCO 3 ( s )  CaO (s)  CO 2 ( s )
The equilibrium
is disturbed by increasing the pressure (that is, concentration) of carbon dioxide.
(b)
The equilibrium
2Fe ( s )  3H 2 O (g)  Fe 2 O 3 (s)  3H 2 ( s )
is disturbed by increasing the concentration of hydrogen.
A reaction whose equilibrium constant is extremely small remains almost
completely as reactants and cannot be shifted to products by adding an excess of
108
one reactant. For example, the reaction
CO ( g )  2H 2 O ( g )  CH 4( g )  O 2 ( g )
has an equilibrium constant K c equal to 10-140. This value is so small that the
equilibrium mixture practically consists only of carbon dioxide and water. Adding
more carbon dioxide to the reaction vessel has no appreciable effect
10.3 CHANGING THE PRESSURE AND TEMPERATURE
The optimum conditions for catalytic methanation involve moderately elevated
temperatures and normal to moderately high pressures.
230  450 o C
CO
 3H
CH
H O

(g)
2 (g)
4( g )
2 (g)
1 - 100 atm
Let us see whether we can gain insight into why these might be the optimum conditions for
the reaction.
Effect of Pressure Change
A pressure change obtained by changing the volume can affect the yield of product in
a gaseous reaction if the reaction involves a change in total moles of gas. The methanation
reaction, CO + 3H2  CH4 + H2O, is an example of a reaction involving changes in moles of
gas. When the reaction goes in the forward direction, four moles of reactant gas (CO + 3H 2)
become two moles of product gas (CH4 + H2O).
To see the effect of such a pressure change, consider what happens when an
equilibrium mixture from the methanation reaction is compressed to one-half of its original
volume at a fixed temperature. The total pressure is doubled (PV = constant at a fixed
temperature, according to Boyle's law, so halving V requires that P double). Because the
partial pressures and therefore the concentrations of reactants and products have changed, the
mixture is no longer at equilibrium. The direction in which the reaction goes to again attain
equilibrium can be predicted by applying LeChatelier's principle. Reaction should go in the
109
forward direction, because then the moles of gas decrease, and the pressure (which is
proportional to moles of gas) decreases. In this way, the, initial pressure increase is partially
reduced.
We can see the quantitative effect of the pressure change by solving the equilibrium
problem. Recall that if we put 1.000 mol CO and 3.000 mol H2 into a 10.00-L vessel, the
equilibrium composition at 1200 K, where K c , equals 3.92, is 0.613 mol CO, 1.839 mol H2,
0.387 mol CH4, and 0.381 mol H20. Suppose the volume of the reaction gases is halved so
that the initial concentrations are doubled. The temperature remains at 1200 K, so K, is still
3.92. But if we solve for the new equilibrium composition, we find 0.495 mol CO, 1.485 mol
H2, 0.505 mol CH4, and 0.505 mol H2O ‘Note that the amount of CH4 has increased from 0.
3 87 mol to 0. 505 mol. We conclude that high pressure of reaction gases favors high yields
of rnethane.
In order to decide the direction of reaction when the pressure of the reaction mixture
is increased (say, by decreasing the volume), we ignore liquids and solids. These are not
much affected by pressure changes because they are nearly incompressible. Consider the
reaction
C( s )  CO2 (g)  2CO(g)
The moles of gas increase when the reaction goes in the forward direction (1 mol CO 2
goes to 2 mol CO). Therefore, when we increase the pressure of the reaction mixture by
decreasing its volume, the reaction goes in the reverse direction. The moles of gas decrease,
and the initial increase of pressure is partially reduced, as Le Chatelier's principle leads us to
expect.
It is important to note that an increase or a decrease in pressure of a gaseous reaction
must result in changes of partial pressures of substances in the chemical equation if it is to
have an effect on the equilibrium composition.
Only changes in partial pressures, or
concentrations, of these substances can affect the reaction quotient. Consider the effect of
increasing the pressure in the methanation reaction by adding helium gas. Although the total
pressure increases, the partial pressures of CO, H2, CH4, and H2O do not change. Thus the
equilibrium composition is not affected. However, changing the pressure by changing the
volume of this system changes the partial pressures of all gases, so the equilibrium
composition is affected.
110
We can summarize these conclusions. If the pressure is increased by decreasing the
volume of a reaction mixture, the reaction shifts in the direction of fewer moles of gas. The
following example illustrates applications of this statement.
Example
Look at each of the following equations and decide whether an increase of
pressure obtained by decreasing the volume will increase, decrease, or have no
effect on the amounts of products.
(a)
(c)
SolSoSolution
CO ( g )  Cl ( g )  COCl 2( g )
(b)
2H 2S (g)  2H 2 ( g )  S 2 (g)
Cgraphite   O 2 ( g )  2CO 2( g ) )
(a) Reaction decreases the number of molecules of gas (from two to one).
According
to LeChatelier's principle, an increase of pressure increases the
amount of product
(b) Reaction increases the number of molecules of gas (from two to three); hence,
an increase of pressure decreases the amounts of products.
(c) Reaction does not change the number of molecules of gas. (We ignore the
change in volume due to consumption of solid carbon because the change in
volume of a solid is insignificant. Look only at gas volumes when deciding the
effect of pressure change on equilibrium composition.) Pressure change has no
effect.
Can you increase the amount of product in each of the following equations by
increasing the pressure? Explain.
(a) CO 2 ( g )  H 2 ( g )  CO ( g )  H 2 O ( g )
(b) 4Cu ( s )  2Cu 2 O ( s )  O 2 ( s )
(c) 2SO 2 ( s )  O 2( s )  2SO 3 ( s )
Effect of Temperature Change
Temperature has a profound effect on most reactions. In the first place, reaction rates
usually increase with an increase in temperature, meaning that equilibrium is reached sooner.
Many gaseous reactions are sluggish or have low rates at room temperature but speed up
enough at higher temperature to become commercially feasible processes.
111
Second, equilibrium constants vary with temperature.
Table 10.3 gives values
of K c for methanation at various temperatures. Note that K c equals 4.9 x 1027 at 298 K. Thus
an equilibrium mixture at room temperature is mostly methane and water.
TEMPERATURE
Kc
(K)
298
4.9 x 1027
800
1.38 x 105
1000
2.54 x 102
1200
3.92
Table 10.3: Equilibrium Constant for Methanation at Different Temperatures2
Whether we should raise or lower the temperature of a reaction mixture to increase
the equilibrium amount of product can be shown by LeChatelier's principle. Again, consider
the methanation reaction,
COg   3H 2 g   CH 4 ( g )  H 2 O g   H o  206.2kJ
The value of Ho shows this reaction to be quite exothermic. Thus, as products are
formed, considerable heat is released. According to LeChatelier's principle, as the
temperature is raised, the reaction shifts to form more reactants, thereby absorbing heat and
attempting to counter the increase in temperature.
CO  g  3H 2 g   CH 4( g )  H 2 O g   heat
Thus, we would predict the equilibrium constant to be larger for lower temperatures, in
agreement with the values of K c , given in Table 10.3
The conclusions from Le Chatelier's principle regarding temperature effects on an
equilibrium can be summarized this way. For an endothermic reaction (H positive), the
amounts of products are increased at equilibrium by an increase in temperature ( K c is larger
112
at higher T).
For an exothermic reaction (H negative), the amounts of products are
increased at equilibrium by a decrease in temperature ( K c is larger at lower T).
Example Carbon monoxide is formed when carbon dioxide reacts with solid carbon
(graphite).
CO 2  g  Cgraphite  2CO ( g ) ; H o  172.5kJ
Is a high or low temperature more favourable to the formation of carbon
monoxide?
Solution
The reaction absorbs heat in the forward direction.
Heat + CO2(g)  C(graphite)  2CO(g)
As the temperature is raised, reaction occurs in the forward direction, using heat
and thereby tending to lower the temperature. Thus high temperature is more
favourable to the formation of carbon monoxide. This is why combustions of
carbon and organic materials can produce significant amounts of carbon
monoxide.
Consider the possibility of converting carbon dioxide to carbon monoxide by the
endothermic reaction;
CO 2  g  H 2 g   CO ( g )  H 2 O g 
Is a high or a low temperature more favourable to the production of carbon
monoxide? Explain.
10.4
Choosing the Optimum Conditions for Reaction
We are now in a position to understand the optimum conditions for the methanation
reaction. We will consider the temperature, pressure catalytic factors.
COg   3H 2 g   CH 4 ( g )  H 2 O g   H o  206.2kJ
(a)
K c = 4.9 x 1027 at 298 K.
Because the reaction is exothermic, low temperatures should favor high
yields of methane; that is, the equilibrium constant is large for low
temperature. However, gaseous reactions are often very slow at room
temperature. In practice, the methanation reaction is run at moderately
elevated temperatures (230-450oC) in the presence of a nickel catalyst,
113
where the rate of reaction is sufficiently fast but the equilibrium constant is
not too small.
(b)
Because the methanation reaction involves a decrease in moles of gas, the
yield of methane should increase as the pressure increases. However, the
equilibrium constant is large at the usual operating temperatures, so very
high pressures are not needed to obtain economical yields of methane.
Pressures of 1-100 atm are usual for this reaction.
As another example, consider the Haber process for the synthesis of ammonia.
N 2  g   3 H 2 g 
(a)
Fe catalyst

2NH 3( g ) ; H o  918
. kJ  206.2kJ
Because the reaction is exothermic, the equilibrium constant is larger for lower
temperatures. But the reaction proceeds too slowly at room temperature to be
practical, even in the presence of the best available catalysts The optimum choice
of temperature, found experimentally to be about 450'C, is a compromise between
an increased rate of reaction at higher temperature and an increased yield of
ammonia at lower temperature.
(b)
Because the formation of ammonia decreases the moles of gases, the yield of
product is improved by high pressures. The equilibrium constant Kc, is only 0.
159 at 450oC, so higher pressures (up to 1000 atm) are required for an economical
yield of ammonia.
Consider the reaction
2CO 2 g   2CO ( g )  O2( g ) ; H o  566kJ
Predict the optimum temperature and pressure conditions that would give high
yield of carbon monoxide.
114
PART 4
IONIC EQUILIBRIUM
Introduction
As you are probably aware, two of the most important concepts in chemistry are those
of ‘acid’ and ‘base’. Acids and bases are produced in huge amounts by industry. Sulphuric
acid for example is manufactured in great tonnage than any other chemical and is used to
make fertiliser from phosphates used in agricultural practices. Bases too are produced on an
enormous scale. Ammonia ranks fourth in annual tonnage among industrial chemicals.
In this section we discuss more fully what the term acids and bases mean. We also
see how to describe the behaviour of acids and bases in terms of equilibrium constants which
we have already covered in our previous lectures.
LECTURE 11
ACID-BASE CONCEPTS
OBJECTIVES
At the end of this lecture you should be able to:1.
illustrate Arrhenius and Bronsted-Lowry Concepts
2.
show the ionisation of water
3.
define pH and illustrate a pH Scale
In your earlier encounter with acid, you described it as a compound that contains
hydrogen and releases hydrogen ions in water. A broader view of acids and bases is obtained
if we adopt the Bronsted–Lowry definitions in which an acid is proton donor and a base is
any proton acceptor.
Our understanding of acids and bases can be enriched by combining the Bronsted–
Lowry definitions with what we have learnt about chemical equilibria. In particular, we shall
see that we can set up a scale of strengths for acids and bases using equilibrium constants and
115
that we can use that scale to make quantitative predictions about reactions involving acids
and bases.
11.1 The Bronsted–Lowry Theory of Acids and Bases
Bronsted–Lowry theory defines an acid as any molecule or ion that acts as a proton
donor denoted as (HA) and a base as any molecule or ion that acts as a proton acceptor
denoted as B.
The proton donation from an acid to water is given as:
HA(aq) + H2O (l)  H3O+(aq) + A-(aq)
Equilibrium established is represented:
HA(aq) + H2O (l)  H3O+(aq) + A-(aq)
An H2O molecule that has accepted a proton from a proton donor to become a
hydronium ion, H3O+.
A proton donation from water to a base in aqueous solution is
H2O(l) + B(aq)  BH+(aq) + OH-(aq)
Equilibrium established is represented as:
H2O(l) + B(aq)  BH+(aq) + OH-(aq)
Acid-base neutralization in the Bronsted - Lowry theory can be summarized as:H3O+(aq) + OH-(aq  2H2O(l)
Equilibrium established:
H3O+(aq) + OH-(aq  2H2O(l)
Conjugate Acids and Bases
Consider the reactions
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
In the forward reaction, acetic acid, CH3COOH donates a proton and hence is a
Bronsted acid. In the reverse reaction, the acetate CH3COO- accepts a proton from hydronium
ion. This makes CH3COO- a Bronsted base, or conjugate of CH3COOH.
The conjugate base of a Bronsted acid is the Bronsted base formed when the acid has
donated a proton.
This can be expressed as:
CH3COOH(aq) +
acid1
+
H2O(l) 
base2
116
CH3COO-(aq) + H3O+(aq)
base1
+
acid2
Conjugate in this context means related.
Likewise if we consider the Bronsted
equilibrium involving ammonia, we can write:
NH3(aq)
base1
+
+
H2O(l) 
acid2
NH4+(aq) +
acid1
+
OHbase2
From the above two examples we therefore state that every Bronsted acid has its conjugate
base, and every Bronsted base has its conjugate acid.
We can therefore write equilibrium constant aqueous ammonia as follows:-
 NH  OH 

 NH   H O

Kc

4
3
2
For acetic acid as
CH3COOH
acid1
+
H2O 
base2
CH3CO2base1
H3O+
acid2
CH CO   H O 


Kc
+
3
2

3
 CH COOH   H O
3
2
Identify conjugate acids and bases in ; (a) H2SO4 + H2O  HSO3- + H3O+ in
water
(b) H2CO3 + H2O  H3O+ + HCO3- in water
11.2 Autoionization of water
As we know, water is a unique solvent. One of its special properties is its ability to act
as an acid and as a base.
H2O
acid1
+
H2O 
base2
H3O+
acid2
+
OHbase1
Water undergoes ionisation to a small extent thus it is a very weak electrolyte and
therefore a poor electrical conductor.
H2O(l)  H+(aq) + OH-(aq)
From this equation, we can write the equilibrium constant for autoionization of water
as
117
 H  OH 
 H O

Kc 

2
Since a very small fraction of water molecules are ionised, the concentration of water,
[H2O], remain unchanged. Therefore, rearranging the equation above we get
Kc  H2O  K w   H   OH  
Kw is the ionic product constant. At 25oC, [H] = [OH-] = 1.0 x 10-7 M, thus
Kw = [H] [OH-] = (1.0 x 10-7 ) x (1.0 x 10-7 ) = 1.0 x 10-14 at 25oC
Predict the acidity of the solution when;
(i)
 H  = OH 


(ii)
 H  > OH 
(iii)
  



A solution is said to be neutral when H  = OH 
 H  < OH 


A solution is said to be acidic when there is an excess of  H  
A solution is said to be basic when there is an excess of OH  
In the laboratory we can adjust either the concentration of  H   or OH   ions in
solution. For example if we adjust the  H   to 1.0 x 10-5M, then OH   changes to
 K  10. x10
OH    H   10. x10

14
w

6
 10
. x10 8 M
Note that K w is an equilibrium constant, its value changes with temperature.
For example at 41oC, K w = 3.8 x 10-14
As we shall find out later most calculation, discussions with equation involving acids and
bases we use the logarithm of K w . We therefore use the quantity “ pK w ” defined as
pK a   log K w
So that


pK w   log 1.0x10 14  14.00
118
The value of pK w is the order of magnitude of K w . This is the origin of the p (for
power of 10 in pK w .
11.3 pH scale – A measure of acidity
We know that whether an aqueous solution is acidic, neutral, or basic depends on the
hydronium (H3O+) or (H+) concentration. We can quantitatively describe the acidity by
giving the hydrogen ion concentration. Because their concentrations are often very small
numbers, a Danish biochemist, Soren Sorensen in 1909 proposed a more practical measure
called pH, which is defined as the negative logarithm of the molar hydrogen ion
concentration.
The table below summarises the three types of solutions
Types of solution
[H+]
pH
Neutral
10 x 10-7
=7
Acidic
Greater than (>)10 x 10-7
<7
Basic
Less than (<)10 x 10-7
>7
Table 11.0 Types of solutions
Example What is the pH of a sample of gastric juice (digestive juice in the stomach) whose
hydrogen ion concentration is 0.047 M?
Solution
We have that pH = - log [H+] = - log 0.047 = - (-1.33)
=1.33
To get this answer using an electronic calculator with a log key, we simply enter
[H+] press log and change the sign of the result.
A scale similar to the pH scale can be devised using the negative logarithm of the
hydroxide ion concentration, OH   as pOH is defined as
pOH   logOH   a measure of hydroxide ion concentration.
Now consider the ion product of constant for water:
119
 H OH   K


w
 10
. x1014
Taking negative logarithm of both sides, we obtain
 




 log H   log OH    log K w   log 1.0 x10 14  14
 


 log H   log OH   pK w  14
From the definition of pH and pOH we obtain
pH + pOH = 14.00
For example, we can use this equation to calculate the pH of a basic solution as
follows: First, we calculate pOH = - log OH   and then use pH = 14.00 - pOH .
Example Tusker beer has a pH of 4.50. What is the hydrogen-ion concentration of tusker?
Solution
The hydrogen-ion concentration is
pH = -log10[H+] = 4.50
pH = log10[H+] = -4.50
[H+] = 10 -4.5
= 3.16 x 10-5
The pH of a basic solution is 10.50. What is the hydroxyl-ion concentration of the
solution? (Answer 3.16 x 10-4).
Measurement of pH
In the laboratory, the pH of a solution is measured with a pH meter. This instrument
consists of special designed electrodes that are dipped into the solution.
A voltage, which
depends on the pH, is generated between the electrodes and is read on a meter calibrated
directly in pH scale. A less precise method of pH measurement is acid-base indicators.
120
Phenolphthalein is an example of an acid-base indicator that is used to measure pH
because it changes colour within a small pH range. The colour change of an indicator
involves establishment of an equilibrium between an acid form and a base form, which have
different colours. In case of the phenolphthalein indicator the acid form is colourless and the
base form is pink. When a base is added to an acidic solution of phenolphthalein, the colour
changes to pink.
11.4 Degree of ionisation
Depending on the nature of an acid, some or all of its molecules may ionise when the
acid is dissolved in water. Percent ionisation is defined as
Percent ionisation 
ionised acid concentration at equilibrium
x100%
Initial concentration of acid
Consider two aqueous solutions containing acids HA and HB at the same
concentrations. If HA is a stronger acid than HB, which of the solutions will have a higher
pH. Because HA is stronger, it can transfer a proton to water more readily than HB can.
Therefore, the percent ionisation is greater for HA than for HB. At equilibrium, the solution
containing HA will have a higher concentration of H+ ions and a lower pH than the one
containing HB.
Strong acids such as HCl, HNO3 and H2SO4 may be assumed to be 100% ionised in
water.
HCl (aq) + H2O(l)  H3O+(aq)+ Cl-(aq)
Most acids, however are classified as weak acids although the strength can vary
greatly. The limiting ionisation of weak acids is related to the equilibrium constant for
ionisation, a relationship we will study in the next lecture.
Much of what we have said about acids also applies to bases. The strength of base
refers to its ability to accept a proton from an acid. NaOH, KOH and Ba(OH)2 are strong
bases.
NaOH (aq)  Na+(aq)+ OH-(aq)
On the other hand, ammonia is a weak base.
NH3 (aq) + H2O(l)  NH4+(aq) + OH-(aq)
The following example shows calculations of pH for solutions containing a strong acid and a
strong base.
121
Example
Answer
Calculate the pH of
(a)
2.0 x 10-3 M HCl
(a)
Since HNO3 is a strong acid, it is completely ionised in solution.
(b)
0.010 M Ba (OH)2
HCl(aq)  H+(aq) + Cl-(aq)
The concentrations of all the species (HCl, H+ and Cl-) before and after
ionisation can be represented as follows: Initial:
Change:
Final
HCl(aq)
2.0 x 10-3
-2.0 x 10-3
0.0M

H+(aq) +
0.0 M
+2.0 x 10-3 M
2.0 x 10-3 M
Cl-(aq)
0.0 M
+2.0 x 10-3
2.0 x 10-3
A positive (+) change represents an increase and a negative (-) change
indicates a decrease in concentration. Thus
[H+]
=
2.0 x 10-3M
pH
=
- log (2.0 x 10-3)
=
2.70
(b)
Ba(OH)2 is a strong base, each Ba(OH) 2 produces two ions:
Ba(OH)2 (aq)

Ba2+(aq)
+
2OH-(aq)
The change in the concentration of all the species can be represented as
follows: Initial:
Change:
Final
Thus
Ba(OH)2 (aq)
0.010 M
-0.010 M
0.00 M

Ba2+(aq)
0.00 M
+0.010 M
.0 0 M
[OH-] =
0.020 M
pOH
=
- log [0.020] = 1.70
=
14.00 - pOH =
+
2OH-(aq)
0.00 M
+2(0.020 M)
0.020 M
Therefore
pH
122
14.00 - 1.70
=
12.30
123
LECTURE 12
ACID-BASE EQUILIBRIA
Many known substances are weak acids or bases and their reactions with water do not
go to completion. To discuss such acid-base reactions, we need to look at the equilibria
involved and be able to calculate the concentrations of species in a reaction.
At the end of this lecture you should be able to:1.
describe the behaviour of solutions of a weak acid, weak
base or salt
2.
describe the common ion effect
3.
define and describe a buffer solution
4.
explain the functioning of acid-base indicators
5.
solve problems involving acids and bases and their salts
12.1 Weak acids and acid ionisation constants
Consider a weak monoprotic acid HA. The Bronsted equilibrium for HA in water
HA(aq) + H2O(l)  A-(aq) + H3O-(aq)
A H O 


Kc
or
 H O K
2

3
HA H 2O
 A  H O 



3
c
 HA
We can treat the concentration of H2O as a constant and combine it with K c to obtain
a new constant called the acid ionisation constant K a .
Hence
Ka   H2 O Kc
 A  H O 


Ka

3
 HA
We can also write the acid ionisation in the form
HA(aq)  H+(aq) + A-(aq).
124
HA might be acetic acid, in which case we would have
Ka
CH CO   H 

CH COOH 

2
3

3
This is the result we got for HA, with [H+] written in place of [H3O+].
It is convenient to define the quantity
pK a   log K a
Then, for acetic acid’


pK a   log 1.8 x 10 5  4.74
At a given temperature, the strength of the acid HA is measured quantitatively by the
magnitude of K a .
The larger the K a , the stronger the acid, that is the greater the
concentration of H+ ions at equilibrium due to its ionisation.
An example of an acid with lower ionisation equilibrium than acetic acid is
hydrocyanic acid (HCN), which has K a  4.9 x10 10 and pK a   log 4.9 x10 10   9.31
Table 12.1 lists some examples of weak acids and their constants.
Name of acid
Formula
Ka
pK a
Conjugate
Kb
pK b
base
Hydrofluoric acid HF
7.1 x 10-4
3.15
F-
1.4 x10-11
10.85
Formic acid
HCOOH
1.7 x 10-4
3.77
HCOO-
5.9 x10-11
10.23
Benzoic acid
C6H5COOH 6.5 x 10-5
4.19
C6H5COO-
1.5 x10-10
9.81
Acetic acid
CH3COOH
1.8 x 10-5
4.74
CH3COO-
5.6 x10-10
9.26
Phenol
C6H5OH
1.3 x 10-10
9.89
C6H5O-
7.7 x 10-5
4.11
Table 12.1 below lists a number of weak acids, their K a and pK a in order of decreasing acid
strength1.
Note that, the smaller the K a is, the larger the pK a becomes.
We can calculate K a from the initial concentration of the acid and the pH of the solution,
and we can use K a and the initial concentration of the acid to calculate equilibrium
concentrations of all the species and pH of solution.
In calculating the equilibrium
concentrations in a weak acid, we follow essentially the same procedure outlined in lecture 9;
125
the systems may be different, but the calculations are based on the same principle, the law of
mass action. The three basic steps are:
1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
2. Write the acid ionization constant in terms of the equilibrium concentrations.
Knowing the value of K a , we can solve for x.
3. Having solved for x, calculate the equilibrium concentrations of all species and/ or the
pH of the solution.
Unless otherwise stated, we will assume that the temperature is 25oC for all such
calculations.
The following three examples illustrate the calculations using the above procedure.
Example
Calculate the concentrations of the non-ionized acid and of the ions of a 0.100 M
formic acid (HCOOH) solution at equilibrium.
Solution
Step 1: Table 12.1 shows that HCOOH is a weak acid. Since it is monoprotic,
one HCOOH molecule ionises to give one H+ ion and one HCOO- anion. Let x
be equilibrium concentration of H+ and HCOO- ions in mol/L.
Then the
equilibrium concentration of HCOOH must be (0.100 – x) mol/L or (0.100 – x)
M. We can now summarize the changes in concentrations as follows: Initial:
Change:
Final
HCOOH(aq)

0.100 M
-x M
(0.100 - x) M
H+(aq)
0.000 M
+xM
xM
+
HCOO-(aq)
0.000 M
+xM
xM
Step: 2 According to Table 12.1
 H   HCOO   17. x10


Ka


 HCOOH 
4
x2
 1.7 x10 4
0.100 - x
This equation can be rewritten as
x2 + 1.7 x 10-4 x – 1.7 x 10-5 = 0
Which fits the quadratic equation ax2 + bx + c = 0.
To avoid solving a quadratic equation we can often apply a simplifying
126
approximation to this type of problem. Since HCOOH is a weak acid, the extent
of its ionisation must be small. Therefore x is small compared with 0.100. As a
general rule, if the quantity x that is subtracted from the original concentration of
the acid (0.100 M in this case) is equal to or less than 5 percent of the original
concentration, we can assume 0.100 – x  0.100. This removes x from the
denominator of the equilibrium constant expression and we avoid a quadratic
equation. If x is more than 5 percent of the original value, then we must solve the
quadratic equation. Normally we can apply the approximation in cases where K a
is small (equal or less than 1 x 10-4) and the initial concentration of the acid is
high (equal to or greater than 0.1 M). In case of doubt, we can always solve for x
by the approximate method and then check the validity of our approximation.
Assuming that 0.100 – x  0.100, then
x2
x2

 1.7 x10  4
0.100- x 0.100
x 2  1.7 x10 5
Taking the square root of both sides, we obtain
x = 4.1 x 10-3 M
Step 3: At equilibrium, therefore
[H+]
=
4.1 x 10-3 M
[HCOO-]
=
4.1 x 10-3 M
[HCOOH]
=
(0.100 – 0.0041) M = 0.096 M
Example
Calculate the pH of a 0.050 M nitrous acid (HNO2) solution.
Solution
Step 1: From Table 16.1 we see that HNO2 is a weak acid. Letting x be the
equilibrium concentration H+ and NO2- ions in mol/L, we summarize.
Initial:
Change:
Final
HNO2(aq)

0.050 M
-x M
(0.050 - x) M
H+(aq)
0.00 M
+x M
xM
+
Step: 2 According to Table 12.1
 H   NO   4.5x10
 HNO 

Ka 

2
2
127
4
NO2-(aq)
0.00 M
+x M
xM
x2
 4.5x10 4
0.050 - x
Applying the approximation 0.050 – x  0.050, we obtain
x2
x2

 4.5 x10  4
0.050- x 0.050
x 2  2.3 x10 5
Taking the square root of both sides gives
x = 4.8 x 10-3 M
Step 3: At equilibrium
and
[H+]
=
4.6 x 10-3 M
pH
=
- log (4.6 x 10-3) = 2.34
Example The pH of a 0.100 M solution of a weak monoprotic acid HA is 2.85. What is
the K a of the acid?
Solution
In this case we are given the pH, from which we can obtain the equilibrium
concentrations, and we are asked to calculate the acid ionisation constant. We can
follow the same basic steps.
Step 1: First we need to calculate the hydrogen ion concentration from the pH
value
pH
=
- log [H+]
2.85
=
- log [H+]
Taking the antilog of both sides, we get
[H+] = 1.4 x 10-3M
Next we summarize the changes:
Initial:
Change:
Final
HA (aq)

0.100 M
-0.0014 M
(0.100 - 0.0014) M
H+(aq)
0.000 M
+0.0014 M
0.0014 M
+
A-(aq)
0.000 M
+0.0014 M
0.0014 M
Step 2: The acid ionisation constant is given by
 H   HA  0.00140.0014  2.0x10


Ka
HA
128
0100
.  0.0014
5
Percent Ionization
In, the last lecture we saw that the percent ionisation can also be used to compare the
strengths of acids. For monoprotic acids, the concentration of acid that undergoes ionisation
is equal to the concentration of the H+ ions and the concentration of the conjugate base at
equilibrium. Therefore, we can write the percent ionisation of a monoprotic acid as
% ionization =
hydrogen ion concentration at equilibrium
x 100%
initial cocentration of acid
=
equilibrium concentration of conjugate base
x 100%
initial cocentration of acid
The following example compares the percent ionisation of two weak monoprotic acids.
Example
Calculate the percent ionization of (a) a 0.60 M hydrofluoric acid (HF) solution
and (b) a 0.60 M hydrocyanic acid (HCN) solution
Solution
(a)
HF
Step 1: Let x be the concentrations of H+ and F- ions at equilibrium in mol/L.
We summarize:

Initial
HF(aq)
0.60 M
Change
Equilibrium
-xM
(0.60 -x) M
H+(aq) +
0.00 M
F-(aq)
0.00
+xM
xM
+xM
xM
M
Step 2: From Table 12.1
 H   F   7.1 x 10


Ka

 HF 
4
x2
 7.1 x 10  4
0.60  x
Recall the assumptions we made while solving quadratic equations,
We can assuming that 0.60 - x 0.60, then
129
x2
x2

 7.1 x 10  4
0.60  x 0.60
x 2  4.3 x 10  4
x  0.021 M
Because HF is monoprotic, the concentration of hydrogen ions produced by the
acid at equilibrium is equal to that of the F- ions. Thus
% ionization =
0.021 M
x 100%
0.60 M
= 3.5%
(b)
HCN
Step 1: Let x be the concentrations of H+ and CN- ions at equilibrium in mol/L.
We summarize:
Initial
Change
Equilibrium
HCN(aq)
0.60 M
-xM
(0.60 -x) M

H+(aq) +
0.00 M
+xM
xM
CN-(aq)
0.00 M
+xM
xM
Step 2: Referring to Table 12.1, we obtain
 H  CN   4..9 x 10

Ka 

 HCN 
x2
 4..9 x 10-10
0.60  x
We assume that 0.60 – x  0.60:
130
-10
x2
x2

 4.9 x 10 10
0.60  x 0.60
x 2  2.9 x 10 10
x  17
. x 10  5 M
Like HF, HCN is monoprotic. Therefore
% ionization =
1.7 x 10 -5 M
x 100%
0.60 M
= 0.0028 %
Thus, at the same concentration, HF ionises to a much greater extent than does
HCN
The extent to which a weak acid ionises depends on the initial concentration of the
acid.
The more dilute the solution, the greater the percent ionisation (Figure 12.1). I
qualitative terms, when an acid is diluted, initially the number of particles (nonionized acid
molecules plus ions) per unit volume is reduced. According to Le Chatelier’s principle, to
counteract this “stress” (that is, the dilution), the equilibrium shifts from un-ionized acid to
H+ and its conjugate base to produce more particles (ions).
131
The following example shows that the percent ionisation, unlike the equilibrium
constant, depends on the initial acid concentration.
Example Compare the percent ionisation of HF at 0.60 M and at 0.00060 M.
Solution
In the previous example, we found the percent ionisation at 0.60 M HF to be 3.5
percent. In order to compare we must find the ionisation of 0.00060 M HF.
Step 1: Let x be the concentrations in mol/L of H+ and F- at equilibrium. We
summarize the changes:
HF(aq)
H+(aq) +
F-(aq)

Initial
0.00060 M
0.00 M
0.00 M
Change
-xM
+xM
+xM
Equilibrium (0.00060 -x) M
xM
xM
Step 2:
x2
 71
. x 10 -4
0.00060  x
Since the concentration of the acid is very low the ionisation constant is
fairly large, the approximation method is not applicable. We express the equation
in quadratic form and then substitute in the quadratic formula:
x2 + 7.1 x 10-4x – 4.3 x 10-7=0
x
 7.1 x 10  4 
 7.1 x 10 
4 2
 4 1   4.3x10  7 
21
x  3.9 x 10 -4 M
Therefore
3.9 x 10 -4 M
% ionization =
x 100%
0.00060 M
= 65 %
132
The large percent ionisation here shows clearly that the approximation 0.00060 –
x  0.00060 would not be valid. It also confirms our prediction that the extent of
acid ionisation increases with dilution.
12.2 Weak bases and base ionisation constants
Strong bases such as the hydroxides of alkali metals and of the alkaline earth metals other
than beryllium are completely ionised in water:
NaOH (aq)

Na+(aq)
+
OH-(aq)
KOH (aq)

K+(aq)
+
OH-(aq)
Ba(OH)2 (aq)

Ba2+(aq)
+
2OH-(aq)
Recall that OH- is the strongest base that can exist in aqueous solutions. Weak bases
are treated like weak acids. When ammonia dissolves in water, it undergoes the reaction
NH3(aq) + H2O(l)

NH4+(aq)
+
OH-(aq)
The production of hydroxide ions in this base ionization reaction means that, in this
solution at 25oC, [OH-] > [H+], and therefore pH > 7. Because, compared with the total
concentration of water, very few water molecules are consumed by the reaction, we can treat
[H20] as a constant. Thus we can write the equilibrium constant as
K H 2 O  K b 
NH OH   1.8 x 10

4

NH3 
5
where K b the equilibrium constant for base ionization, is called the base ionisation constant.
Table 12. lists a number of common weak bases and their ionization constants.
Name of base Formula
K b*
Conjugate acid
Ka
Methylamine
CH3NH2
4.4 x 10-4
CH3NH3
2.3 x 10-11
Ammonia
NH3
1.8 x 10-5
NH4
5.6 x 10-10
Urea
N2H4CO
1.5 x 10-14 H2NCONH3
0.67
Table 12.2: Ionisation constants of some common weak bases at 25oC1
The following example shows how to calculate the pH of ammonia solution.
133
Example What is the pH of a 0.400 M ammonia solution?
Solution
The procedure is essentially the same as the one we use for weak acids.
Step 1: Let x be the concentrations in mol/L of NH4+ and OH- ions at equilibrium.
Next we summarize:
Initial
NH3(aq) + H2O(l)
0.400 M
Change
Equilibrium
-xM
(0.400 -x) M

NH4+(aq)
0.000 M
+
OH-(aq)
0.000
M
+xM
xM
+xM
xM
Step 2:
Kb 
NH OH   1.8 x 10

4

NH3 
5
x2
1.8 x 10  5
0.400  x
Applying the approximation 0.400 – x ≈ 0.400 gives
Kb 
x2
x2

 1.8 x 10 5
0.400  x 0.400
x 2  7.2 x 10 6
x  2.7 x 10 3 M
Step 3: At equilibrium,
[OH-] = 2.7 x 10-3 M. Thus
pH
=
- log (2.7 x 10-3)
=
2.57
= 14 – pOH = 14.0 – 2.57
pH
=
11.43
Note that we omitted the contribution of [OH-] from water.
134
12.3 The relationship between conjugate acid-base ionisation constants
An important relationship between the acid ionisation constant and the ionisation
constant of its conjugate base can be derived as follows, using acetic acid as an example:

CH3COOH (aq)
H+(aq) +
 H  CH COO 

 CH COOH 

Ka
CH3COO- (aq)

3
3
The conjugate base CH3COO-, reacts with water according to the equation
CH3COO- (aq) + H2O (l) 
OH-(aq)
CH3COOH (aq) +
and we can write the base ionisation constant as
 CH COOH  OH 

CH COO 

Kb
3

3
The product of these two ionisation constants is given by
 H  CH COO  x  CH COOH  OH 

 CH COOH 
CH COO 

Ka Kb


3
3

3
3
  OH 
= H
=

Kw
This result may seem strange at first, but it can be understood by realising that the
sum of reactions (1) and (2) below is simply the autoionization of water.

(1)
CH3COOH (aq)
(2)
CH3COO- (aq) + H2O (l) 
(3)
+ H2O (l) 
H+(aq) +
CH3COO- aq)
Ka
CH3COOH (aq) +
OH-(aq)
Kb
H+ (aq)
OH-(aq)
Kw
+
This example illustrates one of the rules we learned about chemical equilibria. When
two reactions are added to give a third reaction, the equilibrium constant for the third reaction
is the product of the equilibrium constants for the added reactions. Thus, for any conjugate
acid-base pair it is always true that
135
Ka Kb  Kw
Equation
12.1
Expressing equation (12.1) in the following ways
Ka 
Kw
Kb
and
Kb 
Kw
Ka
enables us to draw an important conclusion: The stronger the acid (the larger K a ) the weaker
its conjugate base (the smaller K b ), and vice versa (see Tables 12. 1 and 12.2). We can use
Equation (12. 1) to calculate the K b of the conjugate base (CH3COO-) of CH3COOH as
follows. We find the K a value of CH3COOH in Table 12.1 and write
Kb 

Kw
Ka
1.0 x 10 14
1.8 x 10  5
 5.6 x 10 10
16.4 Diprotic and Polyprotic Acids
The treatment of diprotic and polyprotic acids is more involved than that of
monoprotic acids because these substances may yield more than one hydrogen ion per
molecule. These acids ionize in a stepwise manner; that is, they lose one proton at a time.
An ionization constant expression can be written for each step of ionization. Consequently,
two or more equilibrium constant expressions must often be used to calculate the
concentrations of species in the acid solution. For example, for H2CO3 we write
H 2 CO3 aq   H  aq   HCO32 aq  : K a1 
HCO3 aq   H  aq  CO32 aq  : K a2 
H HCO  


3
H 2 CO3 
H CO  
HCO 

2
3

3
2.4 x 10 8
2.1 x 10 4
Note that the conjugate base in the first step of ionization becomes the acid in the
second step of ionization. For a given acid, the first ionization constant is much larger than
136
the second ionization constant, and so on. This trend seems logical when we realize that it is
easier to remove a H+ ion from a neutral molecule than to remove another H+ from a
negatively charged ion derived from the molecule.
Calculation involving ionization constants of diprotic or polyprotic acids is more complex
than the calculation for a monoprotic acid because it involves more than one stage of
ionization.
Phosphoric acid (H3PO4) is an important polyprotic acid with three ionizable
hydrogen atoms:
H  H PO   7.5 x 10

H PO 

H 3 PO 4 aq  H aq  H 2 PO aq


4
K a1
3
H 2 PO aq  H aq  HPO

4

2
4
K a2
HPO
aq  H aq 

PO
3
4
aq
H  PO   4.8 x 10

HPO 

K a3
-8

4
2
2
4
2
4
2
3
4
H  H PO   6.2 x 10

H PO 

aq

4
2
3
4
-13
2
4
We see that phosphoric acid is a weak polyprotic acid and that its ionization constants
decrease rapidly. Thus we can predict that, in a solution containing phosphoric acid, the
concentration of the nonionized acid is the highest and the only other species - ions. Present
in significant concentrations are H+ and H 2 PO 4 ions.
16.5 Acid-Base Properties of Salts
As defined earlier, a salt is an ionic compound formed by the reaction between an acid
and a base. Salts are strong electrolytes that completely dissociate in water. The term salt
hydrolysis describes the reaction of an anion or a cation of a salt, or both, with water. Salt
hydrolysis usually affects the pH of a solution. In this section we discuss various the types of
salt hydrolysis.
Salts That Produce Neutral Solutions
When NaNO3, formed by the reaction between NaOH and HNO3, dissolves in water, it
dissociates as follows:
137
H2 O
NaNO3 aq  Na  aq  NO 3 aq
As we saw in the last lecture, the hydrated Na+ ion neither donates nor accepts H+ ions. The
NO3- ion is the conjugate base of the strong acid HNO3, and it has no affinity for H+ ions.
Consequently, a solution containing Na+ and NO3- is neutral, with a pH of 7
Salts That Produce Basic Solutions
The dissociation of sodium acetate (CH3COONa) in water is given by
H 2O
CH 3COONa s   Na  aq   CH3 COO- aq 
The hydrated Na+ ion has no acidic or basic properties. The acetate ion CH3COhowever, is the conjugate base of the weak acid CH3COOH and therefore has affinity for H+
ions. The hydrolysis reaction is given by
CH 3COO  aq   CH 3COOH aq   OH - aq 
Because this reaction produces OH- ions, the sodium acetate solution will be basic.
The equilibrium constant for this hydrolysis reaction is identical to the base ionization
constant expression for CH3COO-, so we write
 CH COOH  OH 

CH COO 

Kb
3

3
The value of Kb can be obtained from Equation (1 2. I):
K w 1.0 x 10 14
10
Kb 

 5  5.6 x 10
K a 1.8 x 10
where K a is the acid ionization of CH3COOH.
Since each CH3COO- ion that hydrolyses produces one OH- ion, the concentration of OH- at
equilibrium is the same as the concentration of CH3COO- that hydrolysed. We can define the
percent hydrolysis as
138
CH COO 
CH COO 

% hydrolysis =
3
hyrolyzed

3
OH 
x 100%
initial

=
equilibrium
CH COO 

3
x 100%
initial
The CH3COONa example shows that the solution of a salt derived from a strong base
and a weak acid is basic. A calculation based on the hydrolysis of CH3COONa is illustrated
in the following example..
Example Calculate the pH of a 0.15 M solution of sodium acetate (CH3COONa). What is
the percent hydrolysis?
Solution
We note that CH3COONa is the salt formed from a weak acid (CH3COOH) and a
strong base (NaOH). Consequently, only the anion (CH3COO-) will hydrolyse.
The initial dissociation of the salt is
H 2O

CH COONas  CH3COO aq   Na   aq 
3
0.15 M
0.15 M
We can now treat the hydrolysis of CH3COO-. The acetate ion acts as a weak
Bronsted base.
Step 1: Let x be the concentrations of CH3COO- and OH- ions in mol/L. We
summarize the changes:
Initial
Change
Equilibrium
CH3COO- (aq) 
0.15 M
-xM
(0.15 -x) M
Step 2:
139
CH3COOH (aq)
0.00 M
+xM
xM
+
OH-(aq)
0.00 M
+xM
xM
Kb 
CH 3COOH  OH    5.6 x 10 10
CH COO 

3
x 
2
0.15  x
 5.6 x 10 10
Since K b is very small and the initial concentration of the base is large, we can
apply the approximation 0.15 – x  0.15:
x2
x2

 5.6 x 10 10
015
.  x 015
.
x  9.2 x 10  6 M
Step 3: At equilibrium,
[OH-] =
9.2 x 10-6 M. Thus
pOH
=
- log (9.2 x 10-6)
=
5.04
=
14.00 – 5.04
=
8.96
pH
Thus the solution is basic, as we would expect. The percent hydrolysis is given
by
% hydrolsis =
7.5x 10 -6 M
x 100%
0.10 M
= 0.0075 %
The results shows that only a very small portion of the anion undergoes
hydrolysis. The small percent hydrolysis also justifies the approximation that
0.15 - x  0.15.
140
Salts that produce acidic solutions
When a salt derived from a strong acid and a weak base dissolves in water, the
solution becomes acidic. For example, consider the process
H 2O
NH 4 Cl s   NH 4 aq   Cl - aq 
The Cl- ion has no affinity for H+. The ammonium ion NH4- is the weak conjugate acid of the
weak base NH3 and ionises as follows:
NH 4 aq  H 2 O l   NH3 aq  H3O  aq
NH 4 aq  NH3 aq  H  aq
or simply
Since this reaction produces H+ ions, the pH of the solution decreases. As you can see, the
hydrolysis of the NH4 ion is the same as the ionization of the NH4 acid. The equilibrium
constant (or ionisation constant) for this process is given by
Ka 
NH3  H    K w
NH 

4
Kb
=
1.0 x10 14
= 5.6 x 10 -10
5
1.8 x10
The following example deals with the pH change and percent hydrolysis of an NH4Cl
solution.
Example What are the pH and percent hydrolysis of a 0.10 M NH4Cl solution?
We note that CH3COONa is the salt formed from a weak acid (CH3COOH) and a
Solution
strong base (NaOH). Consequently, only the anion (CH3COO-) will hydrolyse.
The initial dissociation of the salt is
H 2O
NH 4 s   NH 
4
0.10 M

aq   Cl   aq 
0.10 M
We can now treat the hydrolysis of the cation as the ionisation of the acid.
141
Step 1:
We represent the hydrolysis of the cation NH 4 , and let x be the
equilibrium concentration of NH3 andH+ ions in mol/L:
Initial
Change
Equilibrium
NH 4 (aq)
0.10 M
-xM
(0.10 -x) M

NH3 (aq)
0.00 M
+xM
xM
+
H+(aq)
0.00 M
+xM
xM
Step 2: From Table 12.2 we obtain the K a for NH 4
Ka 
NH3  H    5.6 x 10 -10
NH 

4
x2
 5.6 x 10 10
010
. x
Applying the approximation 0.10 - x  0.10, we get:
x2
x2

 5.6 x 10 10
0.10  x 0.10
x  7.5 x 10 6 M
Thus the pH is given by
pH
=
=
- log (7.5 x 10-6)
5.12
The percent hydrolysis is
7.5x 10 -6 M
% hydrolsis =
x 100%
0.10 M
= 0.0075 %
We see that the extent of hydrolysis is very small in a 0.10 M ammonium
chloride solution and that the approximation 0.10 - x  0.10 was justified.
142
Salts in Which Both the Cation and the Anion Hydrolyze
So far we have considered salts in which only one ion undergoes hydrolysis. For salts
derived from a weak acid and a weak base, both the cation and the anion hydrolyse.
However, whether a solution containing such a salt is acidic, basic, or neutral depends on the
relative strengths of the weak acid and the weak base. Since the mathematics associated with
this type of system is rather involving, we limit ourselves to making qualitative predictions
about these solutions. We consider three situations.

K b > K a . If Kb for the anion is greater than K a for the cation, then the solution must
be basic because the anion will hydrolyse to a greater extent than the cation. At
equilibrium, there will be more OH- ions than H+ ions.

Kb< Ka.
Conversely, if K b of the anion is smaller than K a of the cation, the
solution will be acidic because cation hydrolysis will be more extensive than anion
hydrolysis.

K b  K a . If K a is approximately equal to K b , the solution will be nearly neutral.
Type of salt
Examples
Cation from strong base; anion from strong acid
NaCl, KI, KNO3,
RbBr, BaCl2
CH3COONA,
KNO2
NH4Cl, NH4NO3
NH4NO2,
CH3COONAH4,
NH4CN
Cation from strong base; anion from weak acid
Cation from weak base; anion from strong acid
Cation from weak base; anion from weak acid
Small, highly charged cation; anion from strong
acid
AlCl3, Fe(NO3)3
Ions that
undergo
hydrolysis
None
pH of solution
Anion
>7
Cation
Anion and
cation
<7
< 7 if K b < K a
 7 if K b  K a
< 7 if K b > K a
<7
Hydrated
cation
 7
Table 12.3 summarizes the behaviour in aqueous solution of the salts discussed in this
section1.
12.6 The Common Ion Effect
Our discussion of acid-base ionization and salt hydrolysis so far has been limited to
solutions containing a single solute. What happens when two different compounds are
143
dissolved? If both sodium acetate and acetic acid are in solution, they both dissociate and
ionize to produce CH3COO- ions:
H 2O

CH 3 COON s  CH 3COO aq  Na  aq
 

CH3COOH (aq)
CH3COO- (aq) +
H+(aq)
CH3COONa is a strong electrolyte, so it dissociates completely in solution, but CH3COOH, a
weak acid, ionizes only slightly. According to Le Chatelier's principle, the addition of
CH3COO- ions from CH3COONa to a solution of CH3COOH will suppress the ionization of
CH3COOH (that is, shift the equilibrium from right to left), thereby decreasing the hydrogen
ion concentration. Thus a solution containing both CH3COOH and CH3COONa will be less
acidic than a solution containing only CH3COOH at the same concentration. The shift in
equilibrium of the acetic acid ionization is caused by the additional acetate ions from the salt.
The CH3COO- ion is
called the common ion because it is supplied by both CH3COOH and CH3COONa,
The shift in equilibrium caused by the addition of a compound having an ion
in common with the dissolved substance is called the common ion effect. The common ion
effect plays an important role in determining the pH of a solution and the solubility of a
slightly soluble salt. We will deal with the latter in Chapter 17. Here we will study the
common ion effect as it relates to the pH of a solution. Keep in mind that despite its
distinctive name, the common ion effect is simply a special case of Le Chatelier's principle.
Let us consider the pH of a solution containing a weak acid HA and a soluble salt of the weak
acid, such as NaA. We start by writing
HA(aq) + H2O(l) 
H3O+(aq) + A-(aq)
HA(aq) 
or simply
H+(aq) + A-(aq)
The ionization constant Ka is given by
Ka 
144
H A 
+

HA
Rearranging the equation gives
H+ 
K a  HA 
A 

Taking the negative logarithm of both sides, we obtain
- log  H +
- log  H
or
+


= - log K a  log
 HA 
A 
A 


= - log K a  log
A 
 HA 

so
pH = pK a  log
where
pK a   log K a
 HA 
This equation is called the Henderson-Hasselbalch equation. In a more general for, it can be
expressed as
pH = pK a  log
conjugate base
acid 
In our example, HA is the acid and A- is the conjugate base. Thus, if we know Ka the
concentrations of the acid and the salt of the acid, we can calculate the pH of the solution. It
is important to remember that the Henderson-Hasselbalch equation is derived from the
equilibrium constant expression. It is valid regardless of the source of the conjugate base
(that is, whether it comes from the acid alone or is supplied by both the acid and its salt).
In solving problems that involve the common ion effect, we are usually given-,1
starting concentrations of a weak acid HA and of its salt, such as NaA. As long as 1
concentrations of these species are reasonably high (≥ 0.1 M), we can neglect ionization of
the acid and the hydrolysis of the salt. Thus we can use the starting concentrations as the
equilibrium concentrations in Henderson-Hasselbalch equation.
The following example deals with calculation of the pH of a solution containing a common
ion.
145
Example
(a)
Calculate the pH of a solution containing 0.20 M CH3COOH and
0.30 M CH3COONa
(b)
What would be the pH of the 0.20 M CH3COOH solution if no
were present?
Solution
(a)
Sodium acetate is a strong electrolyte, so it dissociates
completely in solution:
H 2O
CH 3COONa s  CH 3COO
0.30 M

aq   Na   aq 
0.30 M
The equilibrium concentrations of both the acid and the conjugate
base are assmed to be
the same as the starting concentrations; that is,
[CH3COOH]
=
0.20 M
and
[CH3COO-] = 0.30 M
This is a valid assumption because 91) CH3COOH is a weak acid
and the extent of hydrolysis of the CH3COO- ion is very small. and
(2) the presence of CH3COO- ions further suppresses the ionisation
of CH3COOH, and the presence of CH3COOH further suppresses
the hydrolysis of the CH3COO- ions. We have
146
H+ 
K a  HA 
A 

1.8 x 10  0.20
-5

0.30
= 1.2 x 10 5 M
pH = - log H + 
= - log 1.2 x 10 5 
Thus
= 4.92
Alternatively, we can calculate the pH of the solution by using the
Henderson-Hasselbalch equation. In this case we need to calculate pK a
of the acid first.
pK a   log K a
  log 1.8 x 10 -5 
 4.74
We can calculate the pH of the solution by substituting the value pK a and
the concentrations of the acid and its conjugate base in HendersonHasselbalch:
CH COO 
log
CH COOH
-
pH = -
3
3
= 4.74 + log
=
(b)
0.30 M
0.20 M
4.92
Following the procedure in the previous examples, we find that
the pH of a 0.20 M CH3COOH solution is
147
H 
= 1.9 x 10 3 M
+
pH
= - log 1.9 x 10  3 
= 2.72
Thus, without the common ion effect, the pH of a 0.20 M
CH3COOH solution is 2.72 considerably lower than 4.92, the pH in the
presence of CH3COONa, as calculated in (a).
The presence of the
common ion CH3COO- clearly suppresses the ionisation of the acid
CH3COOH.
The common ion effect also operates in a solution containing a weak base (for
example, NH3) and a salt of the base (for example, NH4C1). At equilibrium
NH 4

(aq)
H+(aq)
NH3 (aq) +
 NH  H 

 NH 

Ka
3

4
We can derive the Henderson-Hasselbalch equation for this system as follows. Rearranging
the above equation we obtain
 
 H    NH 
K a NH 4

3
Taking the negative logarithm of both sides gives
NH 
- log  H  = - log K  log
 NH 
 NH 
- log  H  = - log K + log
NH 
 NH 
pH = pK  log
NH 
+
4
+
a
3
3
+
a
+
4
3
a
+
4
A solution containing both NH3 and its salt NH4C1 is less basic than a solution
containing only NH3 at the same concentration. The common ion NH+ suppresses the
ionization of NH3 in the solution containing both the base and the salt.
148
16.7 Buffer Solutions
A buffer solution is a solution of
(1) a weak acid or base and
(2) its salt; both components must be present.
The solution has the ability to resist changes in pH upon the addition of small
amounts of either acid or base. Buffers are very important to chemical and biological
systems. The pH in the human body varies greatly from one fluid to another; for example,
the pH of blood is about 7.4, whereas the gastric juice in our stomachs has a pH of about 1.5.
These pH values, which are crucial for functioning of enzymes and the balance of osmotic
pressure, are maintained by buffers in most cases.
A buffer solution must contain an acid to react with any OH- ions that may be added
to it and a base to react with any added H+ ions. Furthermore, the acid and the base
components of the buffer must not consume each other in a neutralization reaction. These
requirements are satisfied by an acid-base conjugate pair (a weak acid and its conjugate base
or a weak base and its conjugate acid).
One of the simplest buffers is the acetic acid-sodium acetate system. A solution
containing these two substances has the ability to neutralize either added acid or added base
as follows. If a base is added to the buffer system, the OH- ions will be neutralized by the
acid in the buffer:
CH3COOH(aq) + OH-(aq)
→
CH3COO-(aq) + H2O(l)
If an acid is added, the H+ ions will be consumed by the conjugate base in the buffer,
according to the equation
CH3COO-(aq) +
→
H+(aq)
CH3COOH(aq)
Here the acetate ions are provided by the dissociation of the sodium acetate:
H2O
CH 3 COONa s  CH 3 COO  aq  Na  aq
As you can see, the two reactions that characterize this buffer system are identical to
those for the common ion effect we described earlier. In general, a buffer system can be
represented as salt/acid or conjugate base/acid. Thus the sodium acetate acetic acid buffer
system can be written as CH3COONa/CH3COOH or CH3COO-/CH3COOH
149
Which of the following solutions are buffer systems? (a) KH2PO4/H3PO4 (b)
NaCLO4/HClO4, (c) C5H5N/ C5H5N (C5H5N is pyridine; its Kb is given in Table
16.2). Explain your answer.
(a)
H3PO4 is a weak acid and its conjugate base, H2PO4-, is a weak base. Therefore, this
is a buffer system.
(b)
Because HClO4 is a strong acid, its conjugate base ClO4, and extremely weak base.
This means that the ClO4- ion will not combine with a H+ ion in solution to form
HClO4. Thus the system cannot act as a buffer system.
(c)
As Table 12.2 shows, C5H5N is a weak base and its conjugate acid C5H5NH (the
cation of the salt (C5H5NHCL), is a weak acid. Therefore, this is a buffer system.
The action of a buffer solution is demonstrated by the following example.
Example (a)
Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0
M CH3COONa.
(b)
What is the pH of the buffer system the addition of 0.10 mole of gaseous
HCl to 1 L of the solution? Assume that the volume of the solution does
not change when the HCl is added.
Solution
(a)
The pH of the buffer system before the addition of HCl can be calculated
according to the procedure described in our earlier examples. Assuming
negligible ionisation of the acetic acid and hydrolysis of the acetate ions,
we have, at equilibrium,
[CH3COOH] = 1.0 M
150
and
[CH3COO-] = 1.0 M
H   CH COO  = 1.8 x 10

 CH COOH 

+
3
Ka
-5
3
H  
+
K a  CH 3 COOH 
CH COO 

3
1.8 x 10 10. 
-5
=
1.0
= 1.8 x 10 -5 M
pH = - log (1.8 x 10-5)
= 4.74
Thus when the concentration of the acid and the conjugate base are the
same, the pH of the buffer is equal to the pKa of the acid.
(b)
After the addition of HCl, complete ionisation of HCl acid occurs.
HCl(aq)
→
H+(aq)
+
0.10 mol
0.10 mol
Cl-(aq)
0.10 mol
Originally, there were 1.0 mol CH3COOH and 1.0 mol CH3COOpresenting 1 L of the solution. After neutralization of the HCl acid CH3COO-,
which we write as
CH3COO- (aq)
+
0.10 mol
→
H+(aq)
0.10 mol
CH3COOH (aq)
0.10 mol
the number of moles of acetic acid and acetate ions present are
CH3COOH:
(1.0 + 0.1) mol = 1.1 mol
CH3COO-:
(1.0 - 0.1) mol = 0.90 mol
Next we calculate the hydrogen ion concentration:
151
H  
+
K a  CH 3 COOH 
CH COO 

3
1.8 x 10 11. 
-5
=
0.90
= 2.2 x 10 -5 M
The pH of the solution becomes
pH = - log (2.2 x 10-5)
= 4.66
Note that since the volume of the solution is the same for both species, we
replaced the ratio of their molar concentrations by the ratio of the number of
moles present, that is, (1.1 mol/L) (0.90 mol/L) = 1.1 mol/0.90 mol).
We see that in this buffer solution there is a decrease in pH of 0.08 (becoming more
acidic) as a result of the addition of HCI. We can also compare the change in H+ ion
concentrations as follows:
Before addition of HCI:
[H+] = 1.8 x 10-5 M
After addition of HCI:
[H+] = 2.2 x 10-5 M
Thus the H+ ion concentration increases by a factor of
2.2 x 10 -5 M
1.8 x 10 -5 - 5 M
= 1.2
To appreciate the effectiveness of the CH3COONa/CH3COOH buffer, let us find out
what would happen if 0.10 mol HCl were added to 1 L of water, and compare it with the
result in the preceding example. In this case
Before addition of HCI:
[H+] = 1.0 x 10-7 M
After addition of HCI:
[H+] = 0.10 M
Thus, as a result of the addition of HCI, the H' ion concentration increase by a factor of
0.10 M
1.0 x 10 - 7 M
152
= 1.0 x 10 6
amounting to a million fold increase! This comparison shows that a properly chosen buffer
solution can maintain a fairly constant H' ion concentration, or pH.
Preparing a Buffer Solution with a Specific pH
Now a question arises: How do we prepare a buffer solution with a specific pH. HendersonHasselbalch equation indicates that if the molar concentrations of the acid and its conj base
are approximately equal, then
log
 conjugate base  0
acid 
pH  pK a
Thus to prepare a buffer solution, we choose a weak acid whose pKa is equal to or
close to the described pH. This choice not only gives the correct pH value of the buffer
system, but also ensures that we have comparable amounts of the acid and its conjugate base
present, both are prerequisites for the buffer system to function effectively. The following
example shows this approach.
Example
Describe how you would prepare a “phosphate buffer” at a pH of about 7.40.
Solution
We write three stages of ionization of phosphoric acid as follows. The Ka values
are obtained from Table 16.2, and then pKa values are found by applying pH = logKa.
H 3 PO 4 aq  H  aq  H 2 PO 4 aq
K a1  7.5 x 10 3 ; pK a1  212
.
H 2 PO 4 aq  H  aq  HPO 24 aq
K a2  6.2 x 10-8 ; pK a2  7.21
HPO 24 aq  H  aq  PO 34 aq
K a3  48
. x 10-13 ; pK a3  12.32

The most suitable of the three buffer systems is HPO 2
4 / H 2 PO 4 , because the
153
pKa of the acid H 2 PO 4 is closest to the desired pH. From the HendersonHasselbalch equation we write
pH = pK a + log
 conjugate base
7.40 = 7.21 + log
acid 
HPO 
H PO 
24
2
log

4
HPO   019
.
H PO 
24
2

4
Taking the antilog, we obtain
HPO   15.
H PO 
24
2

4
Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve
disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate
(NaH2PO4) in molar ration of 1.5:1.0 in water. For example, we could dissolve
1.5 moles of Na2HPO4 and 1.0 mole of NaH2PO4 in enough water to make up a
1-L solution.
REFERENCES
1. General Chemistry by Ebbing D.D. (1984)
2. Chemistry by Chang Raymond (1991)
154