192 CHEMISTRY

UNIT 7

EQUILIBRIUM

Chemical equilibria are important in numerous biological

and environmental processes. For example, equilibria

involving O

molecules and the protein hemoglobin play a

crucial role in the transport and delivery of O

from our

lungs to our muscles. Similar equilibria involving CO

molecules and hemoglobin account for the toxicity of CO.

When a liquid evaporates in a closed container,

molecules with relatively higher kinetic energy escape the

liquid surface into the vapour phase and number of liquid

molecules from the vapour phase strike the liquid surface

and are retained in the liquid phase. It gives rise to a constant

vapour pressure because of an equilibrium in which the

number of molecules leaving the liquid equals the number

returning to liquid from the vapour. We say that the system

has reached equilibrium state at this stage. However, this

is not static equilibrium and there is a lot of activity at the

boundary between the liquid and the vapour. Thus, at

equilibrium, the rate of evaporation is equal to the rate of

condensation. It may be represented by

O (l)



O (vap)

The double half arrows indicate that the processes in

both the directions are going on simultaneously. The mixture

of reactants and products in the equilibrium state is called

an equilibrium mixture.

Equilibrium can be established for both physical

processes and chemical reactions. The reaction may be fast

or slow depending on the experimental conditions and the

nature of the reactants. When the reactants in a closed vessel

at a particular temperature react to give products, the

concentrations of the reactants keep on decreasing, while

those of products keep on increasing for some time after

which there is no change in the concentrations of either of

the reactants or products. This stage of the system is the

dynamic equilibrium and the rates of the forward and

After studying this unit you will be

able to

• identify dynamic nature of

equilibrium involved in physical

and chemical processes;

• state the law of equilibrium;

• explain characteristics of

equilibria involved in physical

and chemical processes;

• write expressions for

equilibrium constants;

• establish a relationship between

and K

;

• explain various factors that

affect the equilibrium state of a

reaction;

• classify substances as acids or

bases according to Arrhenius,

Bronsted-Lowry and Lewis

concepts;

• classify acids and bases as weak

or strong in terms of their

ionization constants;

• explain the dependence of

degree of ionization on

concentration of the electrolyte

and that of the common ion;

• describe pH scale for

representing hydrogen ion

concentration;

• explain ionisation of water and

its duel role as acid and base;

• describe ionic product (K

) and

for water;

• appreciate use of buffer

solutions;

• calculate solubility product

constant.

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193EQUILIBRIUM

reverse reactions become equal. It is due to

this dynamic equilibrium stage that there is

no change in the concentrations of various

species in the reaction mixture. Based on the

extent to which the reactions proceed to reach

the state of chemical equilibrium, these may

be classified in three groups.

(i) The reactions that proceed nearly to

completion and only negligible

concentrations of the reactants are left. In

some cases, it may not be even possible to

detect these experimentally.

(ii) The reactions in which only small amounts

of products are formed and most of the

reactants remain unchanged at

equilibrium stage.

(iii) The reactions in which the concentrations

of the reactants and products are

comparable, when the system is in

equilibrium.

The extent of a reaction in equilibrium

varies with the experimental conditions such

as concentrations of reactants, temperature,

etc. Optimisation of the operational conditions

is very important in industry and laboratory

so that equilibrium is favorable in the

direction of the desired product. Some

important aspects of equilibrium involving

physical and chemical processes are dealt in

this unit along with the equilibrium involving

ions in aqueous solutions which is called as

ionic equilibrium.

7.1 EQUILIBRIUM IN PHYSICAL

PROCESSES

The characteristics of system at equilibrium

are better understood if we examine some

physical processes. The most familiar

examples are phase transformation

processes, e.g.,

solid

liquid

liquid gas

solid gas

7.1.1 Solid-Liquid Equilibrium

Ice and water kept in a perfectly insulated

thermos flask (no exchange of heat between

its contents and the surroundings) at 273K

and the atmospheric pressure are in

equilibrium state and the system shows

interesting characteristic features. We observe

that the mass of ice and water do not change

with time and the temperature remains

constant. However, the equilibrium is not

static. The intense activity can be noticed at

the boundary between ice and water.

Molecules from the liquid water collide against

ice and adhere to it and some molecules of ice

escape into liquid phase. There is no change

of mass of ice and water, as the rates of transfer

of molecules from ice into water and of reverse

transfer from water into ice are equal at

atmospheric pressure and 273 K.

It is obvious that ice and water are in

equilibrium only at particular temperature

and pressure. For any pure substance at

atmospheric pressure, the temperature at

which the solid and liquid phases are at

equilibrium is called the normal melting point

or normal freezing point of the substance.

The system here is in dynamic equilibrium and

we can infer the following:

(i) Both the opposing processes occur

simultaneously.

(ii) Both the processes occur at the same rate

so that the amount of ice and water

remains constant.

7.1.2 Liquid-Vapour Equilibrium

This equilibrium can be better understood if

we consider the example of a transparent box

carrying a U-tube with mercury (manometer).

Drying agent like anhydrous calcium chloride

(or phosphorus penta-oxide) is placed for a

few hours in the box. After removing the

drying agent by tilting the box on one side, a

watch glass (or petri dish) containing water is

quickly placed inside the box. It will be

observed that the mercury level in the right

limb of the manometer slowly increases and

finally attains a constant value, that is, the

pressure inside the box increases and reaches

a constant value. Also the volume of water in

the watch glass decreases (Fig. 7.1). Initially

there was no water vapour (or very less) inside

the box. As water evaporated the pressure in

the box increased due to addition of water

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194 CHEMISTRY

molecules into the gaseous phase inside the

box. The rate of evaporation is constant.

However, the rate of increase in pressure

decreases with time due to condensation of

vapour into water. Finally it leads to an

equilibrium condition when there is no net

evaporation. This implies that the number of

water molecules from the gaseous state into

the liquid state also increases till the

equilibrium is attained i.e.,

rate of evaporation= rate of condensation

O(l) H

O (vap)

At equilibrium the pressure exerted by the

water molecules at a given temperature

remains constant and is called the equilibrium

vapour pressure of water (or just vapour

pressure of water); vapour pressure of water

increases with temperature. If the above

experiment is repeated with methyl alcohol,

acetone and ether, it is observed that different

liquids have different equilibrium vapour

pressures at the same temperature, and the

liquid which has a higher vapour pressure is

more volatile and has a lower boiling point.

If we expose three watch glasses

containing separately 1mL each of acetone,

ethyl alcohol, and water to atmosphere and

repeat the experiment with different volumes

of the liquids in a warmer room, it is observed

that in all such cases the liquid eventually

disappears and the time taken for complete

evaporation depends on (i) the nature of the

liquid, (ii) the amount of the liquid and (iii) the

temperature. When the watch glass is open to

the atmosphere, the rate of evaporation

remains constant but the molecules are

dispersed into large volume of the room. As a

consequence the rate of condensation from

vapour to liquid state is much less than the

rate of evaporation. These are open systems

and it is not possible to reach equilibrium in

an open system.

Water and water vapour are in equilibrium

position at atmospheric pressure (1.013 bar)

and at 100°C in a closed vessel. The boiling

point of water is 100°C at 1.013 bar pressure.

For any pure liquid at one atmospheric

pressure (1.013 bar), the temperature at

which the liquid and vapours are at

equilibrium is called normal boiling point of

the liquid. Boiling point of the liquid depends

on the atmospheric pressure. It depends on

the altitude of the place; at high altitude the

boiling point decreases.

7.1.3 Solid – Vapour Equilibrium

Let us now consider the systems where solids

sublime to vapour phase. If we place solid iodine

in a closed vessel, after sometime the vessel gets

filled up with violet vapour and the intensity of

colour increases with time. After certain time the

intensity of colour becomes constant and at this

stage equilibrium is attained. Hence solid iodine

sublimes to give iodine vapour and the iodine

vapour condenses to give solid iodine. The

equilibrium can be represented as,

(solid)



(vapour)

Other examples showing this kind of

equilibrium are,

Camphor (solid)



Camphor (vapour)

Cl (solid)



Cl (vapour)

Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature

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195EQUILIBRIUM

7.1.4 Equilibrium Involving Dissolution of

Solid or Gases in Liquids

Solids in liquids

We know from our experience that we can

dissolve only a limited amount of salt or sugar

in a given amount of water at room

temperature. If we make a thick sugar syrup

solution by dissolving sugar at a higher

temperature, sugar crystals separate out if we

cool the syrup to the room temperature. We

call it a saturated solution when no more of

solute can be dissolved in it at a given

temperature. The concentration of the solute

in a saturated solution depends upon the

temperature. In a saturated solution, a

dynamic equilibrium exits between the solute

molecules in the solid state and in the solution:

Sugar (solution) Sugar (solid), and

the rate of dissolution of sugar = rate of

crystallisation of sugar.

Equality of the two rates and dynamic

nature of equilibrium has been confirmed with

the help of radioactive sugar. If we drop some

radioactive sugar into saturated solution of

non-radioactive sugar, then after some time

radioactivity is observed both in the solution

and in the solid sugar. Initially there were no

radioactive sugar molecules in the solution

but due to dynamic nature of equilibrium,

there is exchange between the radioactive and

non-radioactive sugar molecules between the

two phases. The ratio of the radioactive to non-

radioactive molecules in the solution increases

till it attains a constant value.

Gases in liquids

When a soda water bottle is opened, some of

the carbon dioxide gas dissolved in it fizzes

out rapidly. The phenomenon arises due to

difference in solubility of carbon dioxide at

different pressures. There is equilibrium

between the molecules in the gaseous state

and the molecules dissolved in the liquid

under pressure i.e.,

(gas)

(in solution)

This equilibrium is governed by Henry’s

law, which states that the mass of a gas

dissolved in a given mass of a solvent at

any temperature is proportional to the

pressure of the gas above the solvent. This

amount decreases with increase of

temperature. The soda water bottle is sealed

under pressure of gas when its solubility in

water is high. As soon as the bottle is opened,

some of the dissolved carbon dioxide gas

escapes to reach a new equilibrium condition

required for the lower pressure, namely its

partial pressure in the atmosphere. This is how

the soda water in bottle when left open to the

air for some time, turns ‘flat’. It can be

generalised that:

(i) For solid liquid equilibrium, there is

only one temperature (melting point) at

1 atm (1.013 bar) at which the two phases

can coexist. If there is no exchange of heat

with the surroundings, the mass of the two

phases remains constant.

(ii) For liquid vapour equilibrium, the

vapour pressure is constant at a given

temperature.

(iii) For dissolution of solids in liquids, the

solubility is constant at a given

temperature.

(iv) For dissolution of gases in liquids, the

concentration of a gas in liquid is

proportional to the pressure

(concentration) of the gas over the liquid.

These observations are summarised in

Table 7.1

Liquid Vapour p

constant at given

O (l) H

O (g) temperature

Solid Liquid Melting point is fixed at

O (s) H

O (l) constant pressure

Solute(s) Solute Concentration of solute

(solution) in solution is constant

Sugar(s) Sugar at a given temperature

(solution)

Gas(g) Gas (aq) [gas(aq)]/[gas(g)] is

constant at a given

temperature

(g) CO

(aq) [CO

(aq)]/[CO

(g)] is

constant at a given

temperature

Table 7.1 Some Features of Physical

Equilibria

Process Conclusion

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196 CHEMISTRY

7.1.5 General Characteristics of Equilibria

Involving Physical Processes

For the physical processes discussed above,

following characteristics are common to the

system at equilibrium:

(i) Equilibrium is possible only in a closed

system at a given temperature.

(ii) Both the opposing processes occur at the

same rate and there is a dynamic but

stable condition.

(iii) All measurable properties of the system

remain constant.

(iv) When equilibrium is attained for a physical

process, it is characterised by constant

value of one of its parameters at a given

temperature. Table 7.1 lists such

quantities.

(v) The magnitude of such quantities at any

stage indicates the extent to which the

physical process has proceeded before

reaching equilibrium.

7.2 EQUILIBRIUM IN CHEMICAL

PROCESSES – DYNAMIC

EQUILIBRIUM

Analogous to the physical systems chemical

reactions also attain a state of equilibrium.

These reactions can occur both in forward

and backward directions. When the rates of

the forward and reverse reactions become

equal, the concentrations of the reactants

and the products remain constant. This is

the stage of chemical equilibrium. This

equilibrium is dynamic in nature as it

consists of a forward reaction in which the

reactants give product(s) and reverse

reaction in which product(s) gives the

original reactants.

For a better comprehension, let us

consider a general case of a reversible reaction,

A + B

C + D

With passage of time, there is

accumulation of the products C and D and

depletion of the reactants A and B (Fig. 7.2).

This leads to a decrease in the rate of forward

reaction and an increase in he rate of the

reverse reaction,

Fig. 7.2 Attainment of chemical equilibrium.

Eventually, the two reactions occur at the

same rate and the system reaches a state of

equilibrium.

Similarly, the reaction can reach the state of

equilibrium even if we start with only C and D;

that is, no A and B being present initially, as the

equilibrium can be reached from either direction.

The dynamic nature of chemical

equilibrium can be demonstrated in the

synthesis of ammonia by Haber’s process. In

a series of experiments, Haber started with

known amounts of dinitrogen and dihydrogen

maintained at high temperature and pressure

and at regular intervals determined the

amount of ammonia present. He was

successful in determining also the

concentration of unreacted dihydrogen and

dinitrogen. Fig. 7.4 (page 191) shows that after

a certain time the composition of the mixture

remains the same even though some of the

reactants are still present. This constancy in

composition indicates that the reaction has

reached equilibrium. In order to understand

the dynamic nature of the reaction, synthesis

of ammonia is carried out with exactly the

same starting conditions (of partial pressure

and temperature) but using D

(deuterium)

in place of H

. The reaction mixtures starting

either with H

or D

reach equilibrium with

the same composition, except that D

and ND

are present instead of H

and NH

. After

equilibrium is attained, these two mixtures

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Dynamic Equilibrium – A Student’s Activity

Equilibrium whether in a physical or in a chemical system, is always of dynamic

nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible

in a school laboratory. However this concept can be easily comprehended by performing

the following activity. The activity can be performed in a group of 5 or 6 students.

Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes

each of 30 cm length. Diameter of the tubes may be same or different in the range of

3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this

purpose add a crystal of potassium permanganate to water) and keep second cylinder

(number 2) empty.

Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder

1, close its upper tip with a finger and transfer the coloured water contained in its

lower portion to cylinder 2. Using second tube, kept in 2

cylinder, transfer the coloured

water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring

coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you

notice that the level of coloured water in both the cylinders becomes constant.

If you continue intertransferring coloured solution between the cylinders, there will

not be any further change in the levels of coloured water in two cylinders. If we take

analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the

two cylinders, we can say the process of transfer, which continues even after the constancy

of level, is indicative of dynamic nature of the process. If we repeat the experiment taking

two tubes of different diameters we find that at equilibrium the level of coloured water in

two cylinders is different. How far diameters are responsible for change in levels in two

cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning.

Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the

equilibrium is attained.

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198 CHEMISTRY

2NH

(g) N

(g) + 3H

(g)

Similarly let us consider the reaction,

(g) + I

(g) 2HI(g). If we start with equal

initial concentration of H

and I

, the reaction

proceeds in the forward direction and the

concentration of H

and I

decreases while that

of HI increases, until all of these become

constant at equilibrium (Fig. 7.5). We can also

start with HI alone and make the reaction to

proceed in the reverse direction; the

concentration of HI will decrease and

concentration of H

and I

will increase until

they all become constant when equilibrium is

reached (Fig.7.5). If total number of H and I

atoms are same in a given volume, the same

equilibrium mixture is obtained whether we

start it from pure reactants or pure product.

, N

, NH

and D

, N

, ND

) are mixed

together and left for a while. Later, when this

mixture is analysed, it is found that the

concentration of ammonia is just the same as

before. However, when this mixture is

analysed by a mass spectrometer, it is found

that ammonia and all deuterium containing

forms of ammonia (NH

, NH

D, NHD

and ND

)

and dihydrogen and its deutrated forms

, HD and D

) are present. Thus one can

conclude that scrambling of H and D atoms

in the molecules must result from a

continuation of the forward and reverse

reactions in the mixture. If the reaction had

simply stopped when they reached

equilibrium, then there would have been no

mixing of isotopes in this way.

Use of isotope (deuterium) in the formation

of ammonia clearly indicates that chemical

reactions reach a state of dynamic

equilibrium in which the rates of forward

and reverse reactions are equal and there

is no net change in composition.

Equilibrium can be attained from both

sides, whether we start reaction by taking,

(g) and N

(g) and get NH

(g) or by taking

(g) and decomposing it into N

(g) and

(g).

(g) + 3H

(g)

2NH

(g)

Fig 7.4 Depiction of equilibrium for the reaction

Fig.7.5 Chemical equilibrium in the reaction

(g) + I

(g)



2HI(g) can be attained

from either direction

7.3 LAW OF CHEMICAL EQUILIBRIUM

AND EQUILIBRIUM CONSTANT

A mixture of reactants and products in the

equilibrium state is called an equilibrium

mixture. In this section we shall address a

number of important questions about the

composition of equilibrium mixtures: What is

the relationship between the concentrations of

reactants and products in an equilibrium

mixture? How can we determine equilibrium

concentrations from initial concentrations?

N g H g NH g

2 2 3

3 2

( )

( ) ( )



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199EQUILIBRIUM

What factors can be exploited to alter the

composition of an equilibrium mixture? The

last question in particular is important when

choosing conditions for synthesis of industrial

chemicals such as H

, NH

, CaO etc.

To answer these questions, let us consider

a general reversible reaction:

A + B

C + D

where A and B are the reactants, C and D are

the products in the balanced chemical

equation. On the basis of experimental studies

of many reversible reactions, the Norwegian

chemists Cato Maximillian Guldberg and Peter

Waage proposed in 1864 that the

concentrations in an equilibrium mixture are

related by the following equilibrium

equation,

[ ][ ]

C D

A B

(7.1) where K

is the equilibrium constant

and the expression on the right side is called

the equilibrium constant expression.

The equilibrium equation is also known as

the law of mass action because in the early

days of chemistry, concentration was called

“active mass”. In order to appreciate their work

better, let us consider reaction between

gaseous H

and I

carried out in a sealed vessel

at 731K.

(g) + I

(g) 2HI(g)

1 mol 1 mol 2 mol

Six sets of experiments with varying initial

conditions were performed, starting with only

gaseous H

and I

in a sealed reaction vessel

in first four experiments (1, 2, 3 and 4) and

only HI in other two experiments (5 and 6).

Experiment 1, 2, 3 and 4 were performed

taking different concentrations of H

and / or

, and with time it was observed that intensity

of the purple colour remained constant and

equilibrium was attained. Similarly, for

experiments 5 and 6, the equilibrium was

attained from the opposite direction.

Data obtained from all six sets of

experiments are given in Table 7.2.

It is evident from the experiments 1, 2, 3

and 4 that number of moles of dihydrogen

reacted = number of moles of iodine reacted =

½ (number of moles of HI formed). Also,

experiments 5 and 6 indicate that,

(g)]

= [I

(g)]

Knowing the above facts, in order to

establish a relationship between

concentrations of the reactants and products,

several combinations can be tried. Let us

consider the simple expression,

[HI(g)]

/ [H

(g)]

It can be seen from Table 7.3 that if we

put the equilibrium concentrations of the

reactants and products, the above expression

Table 7.2 Initial and Equilibrium Concentrations of H

, I

and HI

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200 CHEMISTRY

is far from constant. However, if we consider

the expression,

[HI(g)]

/ [H

(g)]

we find that this expression gives constant

value (as shown in Table 7.3) in all the six

cases. It can be seen that in this expression

the power of the concentration for reactants

and products are actually the stoichiometric

coefficients in the equation for the chemical

reaction. Thus, for the reaction H

(g) + I

(g)

2HI(g), following equation 7.1, the equilibrium

constant K

is written as,

= [HI(g)]

/ [H

(g)]

(7.2)

Generally the subscript ‘eq’ (used for

equilibrium) is omitted from the concentration

terms. It is taken for granted that the

concentrations in the expression for K

are

equilibrium values. We, therefore, write,

= [HI(g)]

/ [H

(g)] [I

(g)] (7.3)

The subscript ‘c’ indicates that K

expressed in concentrations of mol L

–1

At a given temperature, the product of

concentrations of the reaction products

raised to the respective stoichiometric

coefficient in the balanced chemical

equation divided by the product of

concentrations of the reactants raised to

their individual stoichiometric coefficients

has a constant value. This is known as

the Equilibrium Law or Law of Chemical

Equilibrium.

The equilibrium constant for a general

reaction,

a A + b B

c C + d D

is expressed as,

= [C]

[D]

/ [A]

[B]

(7.4)

where [A], [B], [C] and [D] are the equilibrium

concentrations of the reactants and products.

Equilibrium constant for the reaction,

4NH

(g) + 5O

(g)

4NO(g) + 6H

O(g) is written

= [NO]

/ [NH

]

Molar concentration of different species is

indicated by enclosing these in square bracket

and, as mentioned above, it is implied that these

are equilibrium concentrations. While writing

expression for equilibrium constant, symbol for

phases (s, l, g) are generally ignored.

Let us write equilibrium constant for the

reaction, H

(g) + I

(g) 2HI(g) (7.5)

as, K

= [HI]

/ [H

] [I

] = x

(7.6)

The equilibrium constant for the reverse

reaction, 2HI(g) H

(g) + I

(g), at the same

temperature is,

′

= [H

] [I

] / [HI]

= 1/ x = 1 / K

(7.7)

Thus, K

′

= 1 / K

(7.8)

Equilibrium constant for the reverse

reaction is the inverse of the equilibrium

constant for the reaction in the forward

direction.

If we change the stoichiometric coefficients

in a chemical equation by multiplying

throughout by a factor then we must make

sure that the expression for equilibrium

constant also reflects that change. For

example, if the reaction (7.5) is written as,

½ H

(g) + ½ I

(g)

HI(g) (7.9)

the equilibrium constant for the above reaction

is given by

K″

= [HI] / [H

]

1/2

]

1/2

{[HI]

/ [H

][I

]}

1/2

= x

1/2

= K

1/2

(7.10)

On multiplying the equation (7.5) by n, we get

Table 7.3 Expression Involving the

Equilibrium Concentration of

Reactants

(g) + I

(g) 2HI(g)

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201EQUILIBRIUM

(g) + nI

(g) D 2nHI(g) (7.11)

Therefore, equilibrium constant for the

reaction is equal to K

. These findings are

summarised in Table 7.4. It should be noted

that because the equilibrium constants K

and

′

have different numerical values, it is

important to specify the form of the balanced

chemical equation when quoting the value of

an equilibrium constant.

800K. What will be K

for the reaction

(g)

+ O

(g)

2NO(g)

Solution

For the reaction equilibrium constant,

can be written as,

2.8 10 M

-3

[ ]

[ ][ ]

( )

− −

2 2

3 3

3 0 10 4 2 10

N O

M M. .

= 0.622

7.4 HOMOGENEOUS EQUILIBRIA

In a homogeneous system, all the reactants

and products are in the same phase. For

example, in the gaseous reaction,

(g) + 3H

(g) 2NH

(g), reactants and

products are in the homogeneous phase.

Similarly, for the reactions,

COOC

(aq) + H

O (l) CH

COOH (aq)

+ C

OH (aq)

and, Fe

(aq) + SCN

–

(aq) Fe(SCN)

(aq)

all the reactants and products are in

homogeneous solution phase. We shall now

consider equilibrium constant for some

homogeneous reactions.

7.4.1 Equilibrium Constant in Gaseous

Systems

So far we have expressed equilibrium constant

of the reactions in terms of molar

concentration of the reactants and products,

and used symbol, K

for it. For reactions

involving gases, however, it is usually more

convenient to express the equilibrium

constant in terms of partial pressure.

The ideal gas equation is written as,

pV = nRT

⇒ p

T= R

Here, p is the pressure in Pa, n is the number

of moles of the gas, V is the volume in m

and

T is the temperature in Kelvin

Problem 7.1

The following concentrations were

obtained for the formation of NH

from N

and H

at equilibrium at 500K.

] = 1.5 × 10

–2

M. [H

] = 3.0 ×10

–2

M and

[NH

] = 1.2 ×10

–2

M. Calculate equilibrium

constant.

Solution

The equilibrium constant for the reaction,

(g) + 3H

(g) 2NH

(g) can be written

as,

NH g

N g H g

2 2

1 2 10

1 5 10

( )









( )









( )









( )

−

. 33 0 10

. ×

( )

−

= 0.106 × 10

= 1.06 × 10

Problem 7.2

At equilibrium, the concentrations of

=3.0 × 10

–3

M, O

= 4.2 × 10

–3

M and

NO= 2.8 × 10

–3

M in a sealed vessel at

Table 7.4 Relations between Equilibrium

Constants for a General Reaction

and its Multiples.

Chemical equation Equilibrium

constant

a A + b B c C + dD K

c C + d D a A + b B K

′

=(1/K

)

na A + nb B

ncC + ndD K

′″

= (K

)

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202 CHEMISTRY

Therefore,

n/V is concentration expressed in mol/m

If concentration c, is in mol/L or mol/dm

and p is in bar then

p = cRT,

We can also write p = [gas]RT.

Here, R= 0.0831 bar litre/mol K

At constant temperature, the pressure of

the gas is proportional to its concentration i.e.,

p ∝ [gas]

For reaction in equilibrium

(g) + I

(g)

2HI(g)

We can write either

HI g

H g I g

(

)









( )









( )









2 2

or K

p p

H I

(

)

( )( )

2 2

(7.12)

Further, since p T

HI g R=

( )









(

)









(

)









I g R

H g R

p T

Therefore,

p p

( )

( )( )

( )









[ ]

(

)









(

)









H I

g g

2 2

HI R

H R I. RRT

H I

g g

( )









( )









( )









2 2

(7.13)

In this example, K

= K

i.e., both

equilibrium constants are equal. However, this

is not always the case. For example in reaction

(g) + 3H

(g) 2NH

(g)

( )

( )( )

p p

N H

2 2

( )









[ ]

( )









( )









(

)

NH g R

N g R g R

2 2

T H T.

( )









[ ]

( )









( )









(

)

−

NH g R

N g g

2 2

K T

or K K T

p c

(

)

−

(7.14)

Similarly, for a general reaction

a A + b B c C + d D

p p

c d c d

a b

( )( )

[ ] [ ]

(

)

[ ] [ ]

( )

(

)

( )

C D R

A B R

[ ] [ ]

( )

(

)

− +

(

)

C D

A B

c d

a b

c d a b

[ ] [ ]

( )

C D

A B

R R

c d

a b

T K T

n n∆ ∆

(7.15)

where ∆n = (number of moles of gaseous

products) – (number of moles of gaseous

reactants) in the balanced chemical equation.

It is necessary that while calculating the value

of K

, pressure should be expressed in bar

because standard state for pressure is 1 bar.

We know from Unit 1 that :

1pascal, Pa=1Nm

–2

, and 1bar = 10

values for a few selected reactions at

different temperatures are given in Table 7.5

Table 7.5 Equilibrium Constants, K

for a

Few Selected Reactions

Problem 7.3

PCl

, PCl

and Cl

are at equilibrium at

500 K and having concentration 1.59M

PCl

, 1.59M Cl

and 1.41 M PCl

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203EQUILIBRIUM

Calculate K

for the reaction,

PCl

+ Cl

Solution

The equilibrium constant K

for the above

reaction can be written as,

PCl Cl

PCl

[ ]

( )

(

)

3 2

1 59

1 41

1 79

Problem 7.4

The value of K

= 4.24 at 800K for the reaction,

CO (g) + H

O (g)

(g) + H

(g)

Calculate equilibrium concentrations of

, H

, CO and H

O at 800 K, if only CO

and H

O are present initially at

concentrations of 0.10M each.

Solution

For the reaction,

CO (g) + H

O (g) CO

(g) + H

(g)

Initial concentration:

0.1M 0.1M 0 0

Let x mole per litre of each of the product

be formed.

At equilibrium:

(0.1-x) M (0.1-x) M x M x M

where x is the amount of CO

and H

equilibrium.

Hence, equilibrium constant can be

written as,

= x

/(0.1-x)

= 4.24

= 4.24(0.01 + x

-0.2x)

= 0.0424 + 4.24x

-0.848x

3.24x

– 0.848x + 0.0424 = 0

a = 3.24, b = – 0.848, c = 0.0424

(for quadratic equation ax

+ bx + c = 0,

b b ac

− ± −

(

)

x = 0.848±√(0.848)

– 4(3.24)(0.0424)/

(3.24×2)

x = (0.848 ± 0.4118)/ 6.48

= (0.848 – 0.4118)/6.48 = 0.067

= (0.848 + 0.4118)/6.48 = 0.194

the value 0.194 should be neglected

because it will give concentration of the

reactant which is more than initial

concentration.

Hence the equilibrium concentrations are,

[CO

] = [H

] = x = 0.067 M

[CO] = [H

O] = 0.1 – 0.067 = 0.033 M

Problem 7.5

For the equilibrium,

2NOCl(g) 2NO(g) + Cl

(g)

the value of the equilibrium constant, K

is 3.75 × 10

–6

at 1069 K. Calculate the K

for the reaction at this temperature?

Solution

We know that,

= K

(RT)

∆n

For the above reaction,

∆n = (2+1) – 2 = 1

= 3.75 ×10

–6

(0.0831 × 1069)

= 0.033

7.5 HETEROGENEOUS EQUILIBRIA

Equilibrium in a system having more than one

phase is called heterogeneous equilibrium.

The equilibrium between water vapour and

liquid water in a closed container is an

example of heterogeneous equilibrium.

O(l) H

O(g)

In this example, there is a gas phase and a

liquid phase. In the same way, equilibrium

between a solid and its saturated solution,

Ca(OH)

(s) + (aq) Ca

(aq) + 2OH

–

(aq)

is a heterogeneous equilibrium.

Heterogeneous equilibria often involve pure

solids or liquids. We can simplify equilibrium

expressions for the heterogeneous equilibria

involving a pure liquid or a pure solid, as the

molar concentration of a pure solid or liquid

is constant (i.e., independent of the amount

present). In other words if a substance ‘X’ is

involved, then [X(s)] and [X(l)] are constant,

whatever the amount of ‘X’ is taken. Contrary

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204 CHEMISTRY

to this, [X(g)] and [X(aq)] will vary as the

amount of X in a given volume varies. Let us

take thermal dissociation of calcium carbonate

which is an interesting and important example

of heterogeneous chemical equilibrium.

CaCO

(s)

CaO (s) + CO

(g) (7.16)

On the basis of the stoichiometric equation,

we can write,

( )









(

)









( )









CaO s CO g

CaCO s

Since [CaCO

(s)] and [CaO(s)] are both

constant, therefore modified equilibrium

constant for the thermal decomposition of

calcium carbonate will be

K´

= [CO

(g)] (7.17)

or K

= p

(7.18)

This shows that at a particular

temperature, there is a constant concentration

or pressure of CO

in equilibrium with CaO(s)

and CaCO

(s). Experimentally it has been

found that at 1100 K, the pressure of CO

equilibrium with CaCO

(s) and CaO(s), is

2.0 ×10

Pa. Therefore, equilibrium constant

at 1100K for the above reaction is:

= P

= 2 × 10

Pa/10

Pa = 2.00

Similarly, in the equilibrium between

nickel, carbon monoxide and nickel carbonyl

(used in the purification of nickel),

Ni (s) + 4 CO (g) Ni(CO)

(g),

the equilibrium constant is written as

( )









[ ]

Ni CO

It must be remembered that for the

existence of heterogeneous equilibrium pure

solids or liquids must also be present

(however small the amount may be) at

equilibrium, but their concentrations or

partial pressures do not appear in the

expression of the equilibrium constant. In the

reaction,

O(s) + 2HNO

(aq) 2AgNO

(aq) +H

O(l)

AgNO

[ ]

HNO

Problem 7.6

The value of K

for the reaction,

(g) + C (s) 2CO (g)

is 3.0 at 1000 K. If initially P

= 0.48 bar

and P

= 0 bar and pure graphite is

present, calculate the equilibrium partial

pressures of CO and CO

Solution

For the reaction,

let ‘x’ be the decrease in pressure of CO

then

(g) + C(s) 2CO(g)

Initial

pressure: 0.48 bar 0

Units of Equilibrium Constant

The value of equilibrium constant K

can

be calculated by substituting the

concentration terms in mol/L and for K

partial pressure is substituted in Pa, kPa,

bar or atm. This results in units of

equilibrium constant based on molarity or

pressure, unless the exponents of both the

numerator and denominator are same.

For the reactions,

(g) + I

(g) 2HI, K

and K

have no unit.

(g) 2NO

(g), K

has unit mol/L and

has unit bar

Equilibrium constants can also be

expressed as dimensionless quantities if

the standard state of reactants and

products are specified. For a pure gas, the

standard state is 1bar. Therefore a pressure

of 4 bar in standard state can be expressed

as 4 bar/1 bar = 4, which is a

dimensionless number. Standard state (c

)

for a solute is 1 molar solution and all

concentrations can be measured with

respect to it. The numerical value of

equilibrium constant depends on the

standard state chosen. Thus, in this

system both K

and K

are dimensionless

quantities but have different numerical

values due to different standard states.

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205EQUILIBRIUM

At equilibrium:

(0.48 – x)bar 2x bar

= (2x)

/(0.48 – x) = 3

= 3(0.48 – x)

= 1.44 – x

+ 3x – 1.44 = 0

a = 4, b = 3, c = –1.44

b b ac

− ± −

( )

= [–3 ± √(3)

– 4(4)(–1.44)]/2 × 4

= (–3 ± 5.66)/8

= (–3 + 5.66)/ 8 (as value of x cannot be

negative hence we neglect that value)

x = 2.66/8 = 0.33

The equilibrium partial pressures are,

= 2x = 2 × 0.33 = 0.66 bar

= 0.48 – x = 0.48 – 0.33 = 0.15 bar

5. The equilibrium constant K for a reaction

is related to the equilibrium constant of the

corresponding reaction, whose equation is

obtained by multiplying or dividing the

equation for the original reaction by a small

integer.

Let us consider applications of equilibrium

constant to:

• predict the extent of a reaction on the basis

of its magnitude,

• predict the direction of the reaction, and

• calculate equilibrium concentrations.

7.6.1 Predicting the Extent of a Reaction

The numerical value of the equilibrium

constant for a reaction indicates the extent of

the reaction. But it is important to note that

an equilibrium constant does not give any

information about the rate at which the

equilibrium is reached. The magnitude of K

or K

is directly proportional to the

concentrations of products (as these appear

in the numerator of equilibrium constant

expression) and inversely proportional to the

concentrations of the reactants (these appear

in the denominator). This implies that a high

value of K is suggestive of a high concentration

of products and vice-versa.

We can make the following generalisations

concerning the composition of

equilibrium mixtures:

• If K

> 10

, products predominate over

reactants, i.e., if K

is very large, the reaction

proceeds nearly to completion. Consider

the following examples:

(a) The reaction of H

with O

at 500 K has a

very large equilibrium constant,

= 2.4 × 10

(b) H

(g) + Cl

(g) 2HCl(g) at 300K has

= 4.0 × 10

(g) + Br

(g) 2HBr (g) at 300 K,

= 5.4 × 10

• If K

< 10

–3

, reactants predominate over

products, i.e., if K

is very small, the reaction

proceeds rarely. Consider the following

examples:

7.6 APPLICATIONS OF EQUILIBRIUM

CONSTANTS

Before considering the applications of

equilibrium constants, let us summarise the

important features of equilibrium constants as

follows:

1. Expression for equilibrium constant is

applicable only when concentrations of the

reactants and products have attained

constant value at equilibrium state.

2. The value of equilibrium constant is

independent of initial concentrations of the

reactants and products.

3. Equilibrium constant is temperature

dependent having one unique value for a

particular reaction represented by a

balanced equation at a given temperature.

4. The equilibrium constant for the reverse

reaction is equal to the inverse of the

equilibrium constant for the forward

reaction.

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206 CHEMISTRY

(a) The decomposition of H

O into H

and O

at 500 K has a very small equilibrium

constant, K

= 4.1 × 10

–48

(b) N

(g) + O

(g)

2NO(g),

at 298 K has K

= 4.8 ×10

–31

• If K

is in the range of 10

–3

to 10

appreciable concentrations of both

reactants and products are present.

Consider the following examples:

(a) For reaction of H

with I

to give HI,

= 57.0 at 700K.

(b) Also, gas phase decomposition of N

is another reaction with a value

of K

= 4.64 × 10

–3

at 25°C which is neither

too small nor too large. Hence,

equilibrium mixtures contain appreciable

concentrations of both N

and NO

These generarlisations are illustrated in

Fig. 7.6

= K

, the reaction mixture is already

at equilibrium.

Consider the gaseous reaction of H

with I

(g) + I

(g)

2HI(g); K

= 57.0 at 700 K.

Suppose we have molar concentrations

]

=0.10M, [I

]

= 0.20 M and [HI]

= 0.40 M.

(the subscript t on the concentration symbols

means that the concentrations were measured

at some arbitrary time t, not necessarily at

equilibrium).

Thus, the reaction quotient, Q

at this stage

of the reaction is given by,

= [HI]

/ [H

]

= (0.40)

/ (0.10)×(0.20)

= 8.0

Now, in this case, Q

(8.0) does not equal

(57.0), so the mixture of H

(g), I

(g) and HI(g)

is not at equilibrium; that is, more H

(g) and

(g) will react to form more HI(g) and their

concentrations will decrease till Q

= K

The reaction quotient, Q

is useful in

predicting the direction of reaction by

comparing the values of Q

and K

Thus, we can make the following

generalisations concerning the direction of the

reaction (Fig. 7.7) :

Fig.7.6 Dependence of extent of reaction on K

7.6.2 Predicting the Direction of the

Reaction

The equilibrium constant helps in predicting

the direction in which a given reaction will

proceed at any stage. For this purpose, we

calculate the reaction quotient Q. The

reaction quotient, Q (Q

with molar

concentrations and Q

with partial pressures)

is defined in the same way as the equilibrium

constant K

except that the concentrations in

are not necessarily equilibrium values.

For a general reaction:

a A + b B c C + d D (7.19)

= [C]

[D]

/ [A]

[B]

(7.20)

Then,

If Q

> K

, the reaction will proceed in the

direction of reactants (reverse reaction).

If Q

< K

, the reaction will proceed in the

direction of the products (forward reaction).

Fig. 7.7 Predicting the direction of the reaction

• If Q

< K

, net reaction goes from left to right

• If Q

> K

, net reaction goes from right to

left.

• If Q

= K

, no net reaction occurs.

Problem 7.7

The value of K

for the reaction

2A B + C is 2 × 10

–3

. At a given time,

the composition of reaction mixture is

[A] = [B] = [C] = 3 × 10

–4

M. In which

direction the reaction will proceed?

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207EQUILIBRIUM

The total pressure at equilbrium was

found to be 9.15 bar. Calculate K

, K

and

partial pressure at equilibrium.

Solution

We know pV = nRT

Total volume (V ) = 1 L

Molecular mass of N

= 92 g

Number of moles = 13.8g/92 g = 0.15

of the gas (n)

Gas constant (R) = 0.083 bar L mol

–1

Temperature (T ) = 400 K

pV = nRT

p × 1L = 0.15 mol × 0.083 bar L mol

–1

× 400 K

p = 4.98 bar

2NO

Initial pressure: 4.98 bar 0

At equilibrium: (4.98 – x) bar 2x bar

Hence,

total

at equilibrium = p

+ p

9.15 = (4.98 – x) + 2x

9.15 = 4.98 + x

x = 9.15 – 4.98 = 4.17 bar

Partial pressures at equilibrium are,

= 4.98 – 4.17 = 0.81bar

= 2x = 2 × 4.17 = 8.34 bar

K p p

p NO N O

2 2 4

( )

= (8.34)

/0.81 = 85.87

= K

(RT)

∆n

85.87 = K

(0.083 × 400)

= 2.586 = 2.6

Problem 7.9

3.00 mol of PCl

kept in 1L closed reaction

vessel was allowed to attain equilibrium

at 380K. Calculate composition of the

mixture at equilibrium. K

= 1.80

Solution

PCl

+ Cl

Initial

concentration: 3.0 0 0

Solution

For the reaction the reaction quotient Q

is given by,

= [B][C]/ [A]

as [A] = [B] = [C] = 3 × 10

–4

= (3 ×10

–4

)(3 × 10

–4

) / (3 ×10

–4

)

= 1

as Q

> K

so the reaction will proceed in

the reverse direction.

7.6.3 Calculating Equilibrium

Concentrations

In case of a problem in which we know the

initial concentrations but do not know any of

the equilibrium concentrations, the following

three steps shall be followed:

Step 1. Write the balanced equation for the

reaction.

Step 2. Under the balanced equation, make a

table that lists for each substance involved in

the reaction:

(a) the initial concentration,

(b) the change in concentration on going to

equilibrium, and

In constructing the table, define x as the

concentration (mol/L) of one of the substances

that reacts on going to equilibrium, then use

the stoichiometry of the reaction to determine

the concentrations of the other substances in

terms of x.

Step 3. Substitute the equilibrium

concentrations into the equilibrium equation

for the reaction and solve for x. If you are to

solve a quadratic equation choose the

mathematical solution that makes chemical

sense.

Step 4. Calculate the equilibrium

concentrations from the calculated value of x.

Step 5. Check your results by substituting

them into the equilibrium equation.

Problem 7.8

13.8g of N

was placed in a 1L reaction

vessel at 400K and allowed to attain

equilibrium

(g) 2NO

(g)

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208 CHEMISTRY

Let x mol per litre of PCl

be dissociated,

At equilibrium:

(3-x) x x

= [PCl

][Cl

]/[PCl

]

1.8 = x

/ (3 – x)

+ 1.8x – 5.4 = 0

x = [–1.8 ± √(1.8)

– 4(–5.4)]/2

x = [–1.8 ± √3.24 + 21.6]/2

x = [–1.8 ± 4.98]/2

x = [–1.8 + 4.98]/2 = 1.59

[PCl

] = 3.0 – x = 3 –1.59 = 1.41 M

[PCl

] = [Cl

] = x = 1.59 M

7.7 RELATIONSHIP BETWEEN

EQUILIBRIUM CONSTANT K,

REACTION QUOTIENT Q AND

GIBBS ENERGY G

The value of K

for a reaction does not depend

on the rate of the reaction. However, as you

have studied in Unit 6, it is directly related

to the thermodynamics of the reaction and

in particular, to the change in Gibbs energy,

∆G. If,

• ∆G is negative, then the reaction is

spontaneous and proceeds in the forward

direction.

• ∆

G is positive, then reaction is considered

non-spontaneous. Instead, as reverse

reaction would have a negative ∆G, the

products of the forward reaction shall be

converted to the reactants.

• ∆G is 0, reaction has achieved equilibrium;

at this point, there is no longer any free

energy left to drive the reaction.

A mathematical expression of this

thermodynamic view of equilibrium can be

described by the following equation:

∆G = ∆G

+ RT lnQ (7.21)

where, G

is standard Gibbs energy.

At equilibrium, when ∆G = 0 and Q = K

the equation (7.21) becomes,

∆G = ∆G

+ RT ln K = 0

∆G

= – RT lnK (7.22)

lnK = – ∆G

/ RT

Taking antilog of both sides, we get,

K = e

–∆G

/RT

(7.23)

Hence, using the equation (7.23), the

reaction spontaneity can be interpreted in

terms of the value of ∆G

• If ∆G

< 0, then –∆G

/RT is positive, and

>1, making K >1, which implies

a spontaneous reaction or the reaction

which proceeds in the forward direction to

such an extent that the products are

present predominantly.

• If ∆

> 0, then –∆G

/RT is negative, and

< 1, that is , K < 1, which implies

a non-spontaneous reaction or a reaction

which proceeds in the forward direction to

such a small degree that only a very minute

quantity of product is formed.

Problem 7.10

The value of ∆G

for the phosphorylation

of glucose in glycolysis is 13.8 kJ/mol.

Find the value of K

at 298 K.

Solution

∆G

= 13.8 kJ/mol = 13.8 × 10

J/mol

Also, ∆G

= – RT lnK

Hence, ln K

= –13.8 × 10

J/mol

(8.314 J mol

–1

× 298 K)

ln K

= – 5.569

= e

–5.569

= 3.81 × 10

–3

Problem 7.11

Hydrolysis of sucrose gives,

Sucrose + H

O Glucose + Fructose

Equilibrium constant K

for the reaction

is 2 ×10

at 300K. Calculate ∆G

300K.

Solution

∆G

= – RT lnK

∆G

= – 8.314J mol

–1

300K × ln(2×10

)

∆

= – 7.64 ×10

J mol

–1

7.8 FACTORS AFFECTING EQUILIBRIA

One of the principal goals of chemical synthesis

is to maximise the conversion of the reactants

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209EQUILIBRIUM

Fig. 7.8 Effect of addition of H

on change of

concentration for the reactants and

products in the reaction,

(g) + I

(g) 2HI(g)

to products while minimizing the expenditure

of energy. This implies maximum yield of

products at mild temperature and pressure

conditions. If it does not happen, then the

experimental conditions need to be adjusted.

For example, in the Haber process for the

synthesis of ammonia from N

and H

, the

choice of experimental conditions is of real

economic importance. Annual world

production of ammonia is about hundred

million tones, primarily for use as fertilizers.

Equilibrium constant, K

is independent of

initial concentrations. But if a system at

equilibrium is subjected to a change in the

concentration of one or more of the reacting

substances, then the system is no longer at

equilibrium; and net reaction takes place in

some direction until the system returns to

equilibrium once again. Similarly, a change in

temperature or pressure of the system may

also alter the equilibrium. In order to decide

what course the reaction adopts and make a

qualitative prediction about the effect of a

change in conditions on equilibrium we use

Le Chatelier’s principle. It states that a

change in any of the factors that

determine the equilibrium conditions of a

system will cause the system to change

in such a manner so as to reduce or to

counteract the effect of the change. This

is applicable to all physical and chemical

equilibria.

We shall now be discussing factors which

can influence the equilibrium.

7.8.1 Effect of Concentration Change

In general, when equilibrium is disturbed by

the addition/removal of any reactant/

products, Le Chatelier’s principle predicts that:

• The concentration stress of an added

reactant/product is relieved by net reaction

in the direction that consumes the added

substance.

• The concentration stress of a removed

reactant/product is relieved by net reaction

in the direction that replenishes the

removed substance.

or in other words,

“When the concentration of any of the

reactants or products in a reaction at

equilibrium is changed, the composition

of the equilibrium mixture changes so as

to minimize the effect of concentration

changes”.

Let us take the reaction,

(g) + I

(g)

2HI(g)

If H

is added to the reaction mixture at

equilibrium, then the equilibrium of the

reaction is disturbed. In order to restore it, the

reaction proceeds in a direction wherein H

consumed, i.e., more of H

and I

react to form

HI and finally the equilibrium shifts in right

(forward) direction (Fig.7.8). This is in

accordance with the Le Chatelier’s principle

which implies that in case of addition of a

reactant/product, a new equilibrium will be

set up in which the concentration of the

reactant/product should be less than what it

was after the addition but more than what it

was in the original mixture.

The same point can be explained in terms

of the reaction quotient, Q

= [HI]

/ [H

][I

]

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210 CHEMISTRY

Addition of hydrogen at equilibrium results

in value of Q

being less than K

. Thus, in order

to attain equilibrium again reaction moves in

the forward direction. Similarly, we can say

that removal of a product also boosts the

forward reaction and increases the

concentration of the products and this has

great commercial application in cases of

reactions, where the product is a gas or a

volatile substance. In case of manufacture of

ammonia, ammonia is liquified and removed

from the reaction mixture so that reaction

keeps moving in forward direction. Similarly,

in the large scale production of CaO (used as

important building material) from CaCO

constant removal of CO

from the kiln drives

the reaction to completion. It should be

remembered that continuous removal of a

product maintains Q

at a value less than K

and reaction continues to move in the forward

direction.

Effect of Concentration – An experiment

This can be demonstrated by the following

reaction:

(aq)+ SCN

–

(aq)

[Fe(SCN)]

(aq) (7.24)

yellow colourless deep red

(7.25)

A reddish colour appears on adding two

drops of 0.002 M potassium thiocynate

solution to 1 mL of 0.2 M iron(III) nitrate

solution due to the formation of [Fe(SCN)]

The intensity of the red colour becomes

constant on attaining equilibrium. This

equilibrium can be shifted in either forward

or reverse directions depending on our choice

of adding a reactant or a product. The

equilibrium can be shifted in the opposite

direction by adding reagents that remove Fe

or SCN

–

ions. For example, oxalic acid

), reacts with Fe

ions to form the

stable complex ion [Fe(C

)

]

3–

, thus

decreasing the concentration of free Fe

(aq).

In accordance with the Le Chatelier’s principle,

the concentration stress of removed Fe

relieved by dissociation of [Fe(SCN)]

replenish the Fe

ions. Because the

concentration of [Fe(SCN)]

decreases, the

intensity of red colour decreases.

Addition of aq. HgCl

also decreases red

colour because Hg

reacts with SCN

–

ions to

form stable complex ion [Hg(SCN)

]

2–

. Removal

of free SCN

–

(aq) shifts the equilibrium in

equation (7.24) from right to left to replenish

SCN

–

ions. Addition of potassium thiocyanate

on the other hand increases the colour

intensity of the solution as it shift the

equilibrium to right.

7.8.2 Effect of Pressure Change

A pressure change obtained by changing the

volume can affect the yield of products in case

of a gaseous reaction where the total number

of moles of gaseous reactants and total

number of moles of gaseous products are

different. In applying Le Chatelier’s principle

to a heterogeneous equilibrium the effect of

pressure changes on solids and liquids can

be ignored because the volume (and

concentration) of a solution/liquid is nearly

independent of pressure.

Consider the reaction,

CO(g) + 3H

(g) CH

(g) + H

O(g)

Here, 4 mol of gaseous reactants (CO + 3H

)

become 2 mol of gaseous products (CH

O). Suppose equilibrium mixture (for above

reaction) kept in a cylinder fitted with a piston

at constant temperature is compressed to one

half of its original volume. Then, total pressure

will be doubled (according to

pV = constant). The partial pressure and

therefore, concentration of reactants and

products have changed and the mixture is no

longer at equilibrium. The direction in which

the reaction goes to re-establish equilibrium

can be predicted by applying the Le Chatelier’s

principle. Since pressure has doubled, the

equilibrium now shifts in the forward

direction, a direction in which the number of

moles of the gas or pressure decreases (we

know pressure is proportional to moles of the

gas). This can also be understood by using

reaction quotient, Q

. Let [CO], [H

], [CH

] and

O] be the molar concentrations at

equilibrium for methanation reaction. When

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211EQUILIBRIUM

Fig. 7.9 Effect of temperature on equilibrium for

the reaction, 2NO

(g) N

(g)

volume of the reaction mixture is halved, the

partial pressure and the concentration are

doubled. We obtain the reaction quotient by

replacing each equilibrium concentration by

double its value.

( )









( )









( )









( )









CH g H O g

CO g H g

4 2

As Q

< K

, the reaction proceeds in the

forward direction.

In reaction C(s) + CO

(g)



2CO(g), when

pressure is increased, the reaction goes in the

reverse direction because the number of moles

of gas increases in the forward direction.

7.8.3 Effect of Inert Gas Addition

If the volume is kept constant and an inert gas

such as argon is added which does not take

part in the reaction, the equilibrium remains

undisturbed. It is because the addition of an

inert gas at constant volume does not change

the partial pressures or the molar

concentrations of the substance involved in the

reaction. The reaction quotient changes only

if the added gas is a reactant or product

involved in the reaction.

7.8.4 Effect of Temperature Change

Whenever an equilibrium is disturbed by a

change in the concentration, pressure or

volume, the composition of the equilibrium

mixture changes because the reaction

quotient, Q

no longer equals the equilibrium

constant, K

. However, when a change in

temperature occurs, the value of equilibrium

constant, K

is changed.

In general, the temperature dependence of

the equilibrium constant depends on the sign

of ∆H for the reaction.

• The equilibrium constant for an exothermic

reaction (negative ∆H) decreases as the

temperature increases.

• The equilibrium constant for an

endothermic reaction (positive ∆H)

increases as the temperature increases.

Temperature changes affect the

equilibrium constant and rates of reactions.

Production of ammonia according to the

reaction,

(g) + 3H

(g) 2NH

(g) ;

∆

H= – 92.38 kJ mol

–

is an exothermic process. According to

Le Chatelier’s principle, raising the

temperature shifts the equilibrium to left and

decreases the equilibrium concentration of

ammonia. In other words, low temperature is

favourable for high yield of ammonia, but

practically very low temperatures slow down

the reaction and thus a catalyst is used.

Effect of Temperature – An experiment

Effect of temperature on equilibrium can be

demonstrated by taking NO

gas (brown in

colour) which dimerises into N

gas

(colourless).

2NO

(g) N

(g); ∆H = –57.2 kJ mol

–1

gas prepared by addition of Cu

turnings to conc. HNO

is collected in two

5 mL test tubes (ensuring same intensity of

colour of gas in each tube) and stopper sealed

with araldite. Three 250 mL beakers 1, 2 and

3 containing freezing mixture, water at room

temperature and hot water (363K),

respectively, are taken (Fig. 7.9). Both the test

tubes are placed in beaker 2 for 8-10 minutes.

After this one is placed in beaker 1 and the

other in beaker 3. The effect of temperature

on direction of reaction is depicted very well

in this experiment. At low temperatures in

beaker 1, the forward reaction of formation of

is preferred, as reaction is exothermic, and

thus, intensity of brown colour due to NO

decreases. While in beaker 3, high

temperature favours the reverse reaction of

2020-21

212 CHEMISTRY

formation of NO

and thus, the brown colour

intensifies.

Effect of temperature can also be seen in

an endothermic reaction,

[Co(H

]

(aq) + 4Cl

–

(aq) [CoCl

]

2–

(aq) +

O(l)

pink colourless blue

At room temperature, the equilibrium

mixture is blue due to [CoCl

]

2–

. When cooled

in a freezing mixture, the colour of the mixture

turns pink due to [Co(H

]

7.8.5 Effect of a Catalyst

A catalyst increases the rate of the chemical

reaction by making available a new low energy

pathway for the conversion of reactants to

products. It increases the rate of forward and

reverse reactions that pass through the same

transition state and does not affect

equilibrium. Catalyst lowers the activation

energy for the forward and reverse reactions

by exactly the same amount. Catalyst does not

affect the equilibrium composition of a

reaction mixture. It does not appear in the

balanced chemical equation or in the

equilibrium constant expression.

Let us consider the formation of NH

from

dinitrogen and dihydrogen which is highly

exothermic reaction and proceeds with

decrease in total number of moles formed as

compared to the reactants. Equilibrium

constant decreases with increase in

temperature. At low temperature rate

decreases and it takes long time to reach at

equilibrium, whereas high temperatures give

satisfactory rates but poor yields.

German chemist, Fritz Haber discovered

that a catalyst consisting of iron catalyse the

reaction to occur at a satisfactory rate at

temperatures, where the equilibrium

concentration of NH

is reasonably favourable.

Since the number of moles formed in the

reaction is less than those of reactants, the

yield of NH

can be improved by increasing

the pressure.

Optimum conditions of temperature and

pressure for the synthesis of NH

using

catalyst are around 500°C and 200 atm.

Similarly, in manufacture of sulphuric

acid by contact process,

2SO

(g) + O

(g)

2SO

(g); K

= 1.7 × 10

though the value of K is suggestive of reaction

going to completion, but practically the oxidation

of SO

to SO

is very slow. Thus, platinum or

divanadium penta-oxide (V

) is used as

catalyst to increase the rate of the reaction.

Note: If a reaction has an exceedingly small

K, a catalyst would be of little help.

7.9 IONIC EQUILIBRIUM IN SOLUTION

Under the effect of change of concentration on

the direction of equilibrium, you have

incidently come across with the following

equilibrium which involves ions:

(aq) + SCN

–

(aq) [Fe(SCN)]

(aq)

There are numerous equilibria that involve

ions only. In the following sections we will

study the equilibria involving ions. It is well

known that the aqueous solution of sugar

does not conduct electricity. However, when

common salt (sodium chloride) is added to

water it conducts electricity. Also, the

conductance of electricity increases with an

increase in concentration of common salt.

Michael Faraday classified the substances into

two categories based on their ability to conduct

electricity. One category of substances

conduct electricity in their aqueous solutions

and are called electrolytes while the other do

not and are thus, referred to as non-

electrolytes. Faraday further classified

electrolytes into strong and weak electrolytes.

Strong electrolytes on dissolution in water are

ionized almost completely, while the weak

electrolytes are only partially dissociated.

For example, an aqueous solution of

sodium chloride is comprised entirely of

sodium ions and chloride ions, while that

of acetic acid mainly contains unionized

acetic acid molecules and only some acetate

ions and hydronium ions. This is because

there is almost 100% ionization in case of

sodium chloride as compared to less

than 5% ionization of acetic acid which is

a weak electrolyte. It should be noted

that in weak electrolytes, equilibrium is

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213EQUILIBRIUM

Fig.7.10 Dissolution of sodium chloride in water.

and Cl

–

ions are stablised by their

hydration with polar water molecules.

established between ions and the unionized

molecules. This type of equilibrium involving

ions in aqueous solution is called ionic

equilibrium. Acids, bases and salts come

under the category of electrolytes and may act

as either strong or weak electrolytes.

7.10 ACIDS, BASES AND SALTS

Acids, bases and salts find widespread

occurrence in nature. Hydrochloric acid

present in the gastric juice is secreted by the

lining of our stomach in a significant amount

of 1.2-1.5 L/day and is essential for digestive

processes. Acetic acid is known to be the main

constituent of vinegar. Lemon and orange

juices contain citric and ascorbic acids, and

tartaric acid is found in tamarind paste. As

most of the acids taste sour, the word “acid”

has been derived from a latin word “acidus”

meaning sour. Acids are known to turn blue

litmus paper into red and liberate dihydrogen

on reacting with some metals. Similarly, bases

are known to turn red litmus paper blue, taste

bitter and feel soapy. A common example of a

base is washing soda used for washing

purposes. When acids and bases are mixed in

the right proportion they react with each other

to give salts. Some commonly known

examples of salts are sodium chloride, barium

sulphate, sodium nitrate. Sodium chloride

(common salt ) is an important component of

our diet and is formed by reaction between

hydrochloric acid and sodium hydroxide. It

exists in solid state as a cluster of positively

charged sodium ions and negatively charged

chloride ions which are held together due to

electrostatic interactions between oppositely

charged species (Fig.7.10). The electrostatic

forces between two charges are inversely

proportional to dielectric constant of the

medium. Water, a universal solvent, possesses

a very high dielectric constant of 80. Thus,

when sodium chloride is dissolved in water,

the electrostatic interactions are reduced by a

factor of 80 and this facilitates the ions to move

freely in the solution. Also, they are well-

separated due to hydration with water

molecules.

Faraday was born near London into a family of very limited means. At the age of 14 he

was an apprentice to a kind bookbinder who allowed Faraday to read the books he

was binding. Through a fortunate chance he became laboratory assistant to Davy, and

during 1813-4, Faraday accompanied him to the Continent. During this trip he gained

much from the experience of coming into contact with many of the leading scientists of

the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories,

and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first

important work was on analytical chemistry. After 1821 much of his work was on

electricity and magnetism and different electromagnetic phenomena. His ideas have led to the establishment

of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest

and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to

work alone and never had any assistant. He disseminated science in a variety of ways including his

Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his

Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers.

Michael Faraday

(1791–1867)

Comparing, the ionization of hydrochloric

acid with that of acetic acid in water we find

that though both of them are polar covalent

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214 CHEMISTRY

molecules, former is completely ionized into

its constituent ions, while the latter is only

partially ionized (< 5%). The extent to which

ionization occurs depends upon the strength

of the bond and the extent of solvation of ions

produced. The terms dissociation and

ionization have earlier been used with different

meaning. Dissociation refers to the process of

separation of ions in water already existing as

such in the solid state of the solute, as in

sodium chloride. On the other hand, ionization

corresponds to a process in which a neutral

molecule splits into charged ions in the

solution. Here, we shall not distinguish

between the two and use the two terms

interchangeably.

7.10.1 Arrhenius Concept of Acids and

Bases

According to Arrhenius theory, acids are

substances that dissociates in water to give

hydrogen ions H

(aq) and bases are

substances that produce hydroxyl ions

–

(aq). The ionization of an acid HX (aq) can

be represented by the following equations:

HX (aq) → H

(aq) + X

–

(aq)

HX(aq) + H

O(l) → H

(aq) + X

–

(aq)

A bare proton, H

is very reactive and

cannot exist freely in aqueous solutions. Thus,

it bonds to the oxygen atom of a solvent water

molecule to give trigonal pyramidal

hydronium ion, H

{[H (H

O)]

} (see box).

In this chapter we shall use H

(aq) and H

(aq)

interchangeably to mean the same i.e., a

hydrated proton.

Similarly, a base molecule like MOH

ionizes in aqueous solution according to the

equation:

MOH(aq) → M

(aq) + OH

–

(aq)

The hydroxyl ion also exists in the hydrated

form in the aqueous solution. Arrhenius

concept of acid and base, however, suffers

from the limitation of being applicable only to

aqueous solutions and also, does not account

for the basicity of substances like, ammonia

which do not possess a hydroxyl group.

7.10.2 The Brönsted-Lowry Acids and

Bases

The Danish chemist, Johannes Brönsted and

the English chemist, Thomas M. Lowry gave a

more general definition of acids and bases.

According to Brönsted-Lowry theory, acid is

a substance that is capable of donating a

hydrogen ion H

and bases are substances

capable of accepting a hydrogen ion, H

. In

short, acids are proton donors and bases are

proton acceptors.

Consider the example of dissolution of NH

in H

O represented by the following equation:

Hydronium and Hydroxyl Ions

Hydrogen ion by itself is a bare proton with

very small size (~10

–15

m radius) and

intense electric field, binds itself with the

water molecule at one of the two available

lone pairs on it giving H

. This species

has been detected in many compounds

(e.g., H

–

) in the solid state. In aqueous

solution the hydronium ion is further

hydrated to give species like H

, H

and

. Similarly the hydroxyl ion is hydrated

to give several ionic species like H

–

, H

–

and H

–

etc.

The basic solution is formed due to the

presence of hydroxyl ions. In this reaction,

water molecule acts as proton donor and

ammonia molecule acts as proton acceptor

and are thus, called Lowry-Brönsted acid and

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215EQUILIBRIUM

Svante Arrhenius

(1859-1927)

Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities

of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he

travelled extensively and visited a number of research centers in Europe. In 1895 he was

appointed professor of physics at the newly formed University of Stockholm, serving its

rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry

at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic

solutions. In 1899 he discussed the temperature dependence of reaction rates on the

basis of an equation, now usually known as Arrhenius equation.

He worked in a variety of fields, and made important contributions to

immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the

first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in

Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development

of chemistry.

base, respectively. In the reverse reaction, H

is transferred from NH

to OH

–

. In this case,

acts as a Bronsted acid while OH

–

acted

as a Brönsted base. The acid-base pair that

differs only by one proton is called a conjugate

acid-base pair. Therefore, OH

–

is called the

conjugate base of an acid H

O and NH

called conjugate acid of the base NH

. If

Brönsted acid is a strong acid then its

conjugate base is a weak base and vice-

versa. It may be noted that conjugate acid

has one extra proton and each conjugate base

has one less proton.

Consider the example of ionization of

hydrochloric acid in water. HCl(aq) acts as an

acid by donating a proton to H

O molecule

which acts as a base.

ammonia it acts as an acid by donating a

proton.

Problem 7.12

What will be the conjugate bases for the

following Brönsted acids: HF, H

and

HCO

–

Solution

The conjugate bases should have one

proton less in each case and therefore the

corresponding conjugate bases are: F

–

HSO

–

and CO

2–

respectively.

Problem 7.13

Write the conjugate acids for the following

Brönsted bases: NH

–

, NH

and HCOO

–

Solution

The conjugate acid should have one extra

proton in each case and therefore the

corresponding conjugate acids are: NH

and HCOOH respectively.

Problem 7.14

The species: H

O, HCO

–

, HSO

–

and NH

can act both as Bronsted acids and bases.

For each case give the corresponding

conjugate acid and conjugate base.

Solution

The answer is given in the following Table:

Species Conjugate

Conjugate

acid base

O H

–

HCO

–

2–

HSO

–

2–

–

It can be seen in the above equation, that

water acts as a base because it accepts the

proton. The species H

is produced when

water accepts a proton from HCl. Therefore,

–

is a conjugate base of HCl and HCl is the

conjugate acid of base Cl

–

. Similarly, H

O is a

conjugate base of an acid H

and H

is a

conjugate acid of base H

It is interesting to observe the dual role of

water as an acid and a base. In case of reaction

with HCl water acts as a base while in case of

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216 CHEMISTRY

7.10.3 Lewis Acids and Bases

G.N. Lewis in 1923 defined an acid as a

species which accepts electron pair and base

which donates an electron pair. As far as

bases are concerned, there is not much

difference between Brönsted-Lowry and Lewis

concepts, as the base provides a lone pair in

both the cases. However, in Lewis concept

many acids do not have proton. A typical

example is reaction of electron deficient species

with NH

does not have a proton but still acts

as an acid and reacts with NH

by accepting

its lone pair of electrons. The reaction can be

represented by,

+ :NH

→ BF

:NH

Electron deficient species like AlCl

, Co

, etc. can act as Lewis acids while species

like H

O, NH

, OH

–

etc. which can donate a pair

of electrons, can act as Lewis bases.

Problem 7.15

Classify the following species into Lewis

acids and Lewis bases and show how

these act as such:

(a) HO

–

(b)F

–

(d) BCl

Solution

(a) Hydroxyl ion is a Lewis base as it can

donate an electron lone pair (:OH

–

(b) Flouride ion acts as a Lewis base as

it can donate any one of its four

electron lone pairs.

accept a lone pair of electrons from

bases like hydroxyl ion and fluoride

ion.

(d) BCl

acts as a Lewis acid as it can

accept a lone pair of electrons from

species like ammonia or amine

molecules.

7.11 IONIZATION OF ACIDS AND BASES

Arrhenius concept of acids and bases becomes

useful in case of ionization of acids and bases

as mostly ionizations in chemical and

biological systems occur in aqueous medium.

Strong acids like perchloric acid (HClO

hydrochloric acid (HCl), hydrobromic acid

(HBr), hyrdoiodic acid (HI), nitric acid (HNO

)

and sulphuric acid (H

) are termed strong

because they are almost completely

dissociated into their constituent ions in an

aqueous medium, thereby acting as proton

) donors. Similarly, strong bases like

lithium hydroxide (LiOH), sodium hydroxide

(NaOH), potassium hydroxide (KOH), caesium

hydroxide (CsOH) and barium hydroxide

Ba(OH)

are almost completely dissociated into

ions in an aqueous medium giving hydroxyl

ions, OH

–

. According to Arrhenius concept

they are strong acids and bases as they are

able to completely dissociate and produce H

and OH

–

ions respectively in the medium.

Alternatively, the strength of an acid or base

may also be gauged in terms of Brönsted-

Lowry concept of acids and bases, wherein a

strong acid means a good proton donor and a

strong base implies a good proton acceptor.

Consider, the acid-base dissociation

equilibrium of a weak acid HA,

HA(aq) + H

O(l)

(aq) + A

–

(aq)

conjugate conjugate

acid base acid base

In section 7.10.2 we saw that acid (or base)

dissociation equilibrium is dynamic involving

a transfer of proton in forward and reverse

directions. Now, the question arises that if the

equilibrium is dynamic then with passage of

time which direction is favoured? What is the

driving force behind it? In order to answer

these questions we shall deal into the issue of

comparing the strengths of the two acids (or

bases) involved in the dissociation equilibrium.

Consider the two acids HA and H

present

in the above mentioned acid-dissociation

equilibrium. We have to see which amongst

them is a stronger proton donor. Whichever

exceeds in its tendency of donating a proton

over the other shall be termed as the stronger

acid and the equilibrium will shift in the

direction of weaker acid. Say, if HA is a

stronger acid than H

, then HA will donate

protons and not H

, and the solution will

mainly contain A

–

and H

ions. The

equilibrium moves in the direction of

formation of weaker acid and weaker base

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217EQUILIBRIUM

because the stronger acid donates a proton

to the stronger base.

It follows that as a strong acid dissociates

completely in water, the resulting base formed

would be very weak i.e., strong acids have

very weak conjugate bases. Strong acids like

perchloric acid (HClO

), hydrochloric acid

(HCl), hydrobromic acid (HBr), hydroiodic acid

(HI), nitric acid (HNO

) and sulphuric acid

) will give conjugate base ions ClO

–

, Cl,

–

, I

–

, NO

–

and HSO

–

, which are much weaker

bases than H

O. Similarly a very strong base

would give a very weak conjugate acid. On the

other hand, a weak acid say HA is only partially

dissociated in aqueous medium and thus, the

solution mainly contains undissociated HA

molecules. Typical weak acids are nitrous acid

(HNO

), hydrofluoric acid (HF) and acetic acid

(CH

COOH). It should be noted that the weak

acids have very strong conjugate bases. For

example, NH

–

, O

2–

and H

–

are very good proton

acceptors and thus, much stronger bases than

Certain water soluble organic compounds

like phenolphthalein and bromothymol blue

behave as weak acids and exhibit different

colours in their acid (HIn) and conjugate base

(In

–

) forms.

HIn(aq) + H

O(l)

(aq) + In

–

(aq)

acid conjugate conjugate

indicator acid base

colour A colourB

Such compounds are useful as indicators

in acid-base titrations, and finding out H

ion

concentration.

7.11.1 The Ionization Constant of Water

and its Ionic Product

Some substances like water are unique in their

ability of acting both as an acid and a base.

We have seen this in case of water in section

7.10.2. In presence of an acid, HA it accepts a

proton and acts as the base while in the

presence of a base, B

–

it acts as an acid by

donating a proton. In pure water, one H

molecule donates proton and acts as an acid

and another water molecules accepts a proton

and acts as a base at the same time. The

following equilibrium exists:

O(l) + H

O(l)

(aq) + OH

–

(aq)

acid base conjugate conjugate

acid base

The dissociation constant is represented by,

K = [H

] [OH

–

] / [H

O] (7.26)

The concentration of water is omitted from

the denominator as water is a pure liquid and

its concentration remains constant. [H

O] is

incorporated within the equilibrium constant

to give a new constant, K

, which is called the

ionic product of water.

= [H

][OH

–

] (7.27)

The concentration of H

has been found

out experimentally as 1.0 × 10

–7

M at 298 K.

And, as dissociation of water produces equal

number of H

and OH

–

ions, the concentration

of hydroxyl ions, [OH

–

] = [H

] = 1.0 × 10

–7

Thus, the value of K

at 298K,

= [H

][OH

–

] = (1 × 10

–7

)

= 1 × 10

–14

(7.28)

The value of K

is temperature dependent

as it is an equilibrium constant.

The density of pure water is 1000 g / L

and its molar mass is 18.0 g /mol. From this

the molarity of pure water can be given as,

O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.

Therefore, the ratio of dissociated water to that

of undissociated water can be given as:

–7

/ (55.55) = 1.8 × 10

–9

~ 2 in 10

–9

(thus,

equilibrium lies mainly towards undissociated

water)

We can distinguish acidic, neutral and

basic aqueous solutions by the relative values

of the H

and OH

–

concentrations:

Acidic: [H

] > [OH

–

]

Neutral: [H

] = [OH

–

]

Basic : [H

] < [OH

–

]

7.11.2 The pH Scale

Hydronium ion concentration in molarity is

more conveniently expressed on a logarithmic

scale known as the pH scale. The pH of a

solution is defined as the negative logarithm

to base 10 of the activity

( )

of hydrogen

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218 CHEMISTRY

ion. In dilute solutions (< 0.01 M), activity of

hydrogen ion (H

) is equal in magnitude to

molarity represented by [H

]. It should

be noted that activity has no units and is

defined as:

= [H

] / mol L

–1

From the definition of pH, the following

can be written,

pH = – log a

H+

= – log {[H

] / mol L

–1

}

Thus, an acidic solution of HCl (10

–2

will have a pH = 2. Similarly, a basic solution

of NaOH having [OH

–

] =10

–4

M and [H

] =

–10

M will have a pH = 10. At 25 °C, pure

water has a concentration of hydrogen ions,

] = 10

–7

M. Hence, the pH of pure water is

given as:

pH = –log(10

–7

) = 7

Acidic solutions possess a concentration

of hydrogen ions, [H

] > 10

–7

M, while basic

solutions possess a concentration of hydrogen

ions, [H

] < 10

–7

M. thus, we can summarise

that

Acidic solution has pH < 7

Basic solution has pH > 7

Neutral solution has pH = 7

Now again, consider the equation (7.28) at

298 K

= [H

] [OH

–

] = 10

–14

Taking negative logarithm on both sides

of equation, we obtain

–log K

= – log {[H

] [OH

–

]}

= – log [H

] – log [OH

–

]

= – log 10

–14

= pH + pOH = 14 (7.29)

Note that although K

may change with

temperature the variations in pH with

temperature are so small that we often

ignore it.

is a very important quantity for

aqueous solutions and controls the relative

concentrations of hydrogen and hydroxyl ions

as their product is a constant. It should be

noted that as the pH scale is logarithmic, a

change in pH by just one unit also means

change in [H

] by a factor of 10. Similarly, when

the hydrogen ion concentration, [H

] changes

by a factor of 100, the value of pH changes by

2 units. Now you can realise why the change

in pH with temperature is often ignored.

Measurement of pH of a solution is very

essential as its value should be known when

dealing with biological and cosmetic

applications. The pH of a solution can be found

roughly with the help of pH paper that has

different colour in solutions of different pH.

Now-a-days pH paper is available with four

strips on it. The different strips have different

colours (Fig. 7.11) at the same pH. The pH in

the range of 1-14 can be determined with an

accuracy of ~0.5 using pH paper.

Fig.7.11 pH-paper with four strips that may

have different colours at the same pH

For greater accuracy pH meters are used.

pH meter is a device that measures the

pH-dependent electrical potential of the test

solution within 0.001 precision. pH meters of

the size of a writing pen are now available in

the market. The pH of some very common

substances are given in Table 7.5 (page 212).

Problem 7.16

The concentration of hydrogen ion in a

sample of soft drink is 3.8 × 10

–3

M. what

is its pH ?

Solution

pH = – log[3.8 × 10

–3

]

= – {log[3.8] + log[10

–3

]}

= – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42

Therefore, the pH of the soft drink is 2.42

and it can be inferred that it is acidic.

Problem 7.17

Calculate pH of a 1.0 × 10

–8

M solution

of HCl.

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219EQUILIBRIUM

Table 7.6 The Ionization Constants of Some

Selected Weak Acids (at 298K)

Acid Ionization Constant,

Hydrofluoric Acid (HF) 3.5 × 10

–4

Nitrous Acid (HNO

) 4.5 × 10

–4

Formic Acid (HCOOH) 1.8 × 10

–4

Niacin (C

NCOOH) 1.5 × 10

–5

Acetic Acid (CH

COOH) 1.74 × 10

–5

Benzoic Acid (C

COOH) 6.5 × 10

–5

Hypochlorous Acid (HCIO) 3.0 × 10

–8

Hydrocyanic Acid (HCN) 4.9 × 10

–10

Phenol (C

OH) 1.3 × 10

–10

Solution

O (l) H

(aq) + OH

–

(aq)

= [OH

–

][H

]

= 10

–14

Let, x = [OH

–

] = [H

] from H

O. The H

concentration is generated (i) from

the ionization of HCl dissolved i.e.,

HCl(aq) + H

O(l) H

(aq) + Cl

–

(aq),

and (ii) from ionization of H

O. In these

very dilute solutions, both sources of

must be considered:

] = 10

–8

+ x

= (10

–8

+ x)(x) = 10

–14

or x

+ 10

–8

x – 10

–14

= 0

[OH

–

] = x = 9.5 × 10

–8

So, pOH = 7.02 and pH = 6.98

7.11.3 Ionization Constants of Weak Acids

Consider a weak acid HX that is partially

ionized in the aqueous solution. The

equilibrium can be expressed by:

HX(aq) + H

O(l) H

(aq) + X

–

(aq)

Initial

concentration (M)

c 0 0

Let α be the extent of ionization

Change (M)

-cα +cα +cα

Equilibrium concentration (M)

c-cα cα cα

Here, c = initial concentration of the

undissociated acid, HX at time, t = 0. α = extent

up to which HX is ionized into ions. Using

these notations, we can derive the equilibrium

constant for the above discussed acid-

dissociation equilibrium:

= c

/ c(1-α) = cα

/ 1-α

is called the dissociation or ionization

constant of acid HX. It can be represented

alternatively in terms of molar concentration

as follows,

= [H

][X

–

] / [HX] (7.30)

At a given temperature T, K

is a

measure of the strength of the acid HX

i.e., larger the value of K

, the stronger is

the acid. K

is a dimensionless quantity

with the understanding that the standard

state concentration of all species is 1M.

The values of the ionization constants of

some selected weak acids are given in

Table 7.6.

Table 7.5 The pH of Some Common Substances

Name of the Fluid pH Name of the Fluid pH

Saturated solution of NaOH ~15 Black Coffee 5.0

0.1 M NaOH solution 13 Tomato juice ~4.2

Lime water 10.5 Soft drinks and vinegar ~3.0

Milk of magnesia 10 Lemon juice ~2.2

Egg white, sea water 7.8 Gastric juice ~1.2

Human blood 7.4 1M HCl solution ~0

Milk 6.8 Concentrated HCl ~–1.0

Human Saliva 6.4

The pH scale for the hydrogen ion

concentration has been so useful that besides

, it has been extended to other species and

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220 CHEMISTRY

quantities. Thus, we have:

= –log (K

) (7.31)

Knowing the ionization constant, K

of an

acid and its initial concentration, c, it is

possible to calculate the equilibrium

concentration of all species and also the degree

of ionization of the acid and the pH of the

solution.

A general step-wise approach can be

adopted to evaluate the pH of the weak

electrolyte as follows:

Step 1. The species present before dissociation

are identified as Brönsted-Lowry

acids / bases.

Step 2. Balanced equations for all possible

reactions i.e., with a species acting both as

acid as well as base are written.

Step 3. The reaction with the higher K

identified as the primary reaction whilst the

other is a subsidiary reaction.

Step 4. Enlist in a tabular form the following

values for each of the species in the primary

reaction

(a) Initial concentration, c.

(b) Change in concentration on proceeding to

equilibrium in terms of α, degree of

ionization.

Step 5. Substitute equilibrium concentrations

into equilibrium constant equation for

principal reaction and solve for α.

Step 6. Calculate the concentration of species

in principal reaction.

Step 7. Calculate pH = – log[H

]

The above mentioned methodology has

been elucidated in the following examples.

Problem 7.18

The ionization constant of HF is

3.2 × 10

–4

. Calculate the degree of

dissociation of HF in its 0.02 M solution.

Calculate the concentration of all species

present (H

, F

–

and HF) in the solution

and its pH.

Solution

The following proton transfer reactions are

possible:

1) HF + H

O H

+ F

–

= 3.2 × 10

–4

2) H

O + H

+ OH

–

= 1.0 × 10

–14

As K

>> K

, [1] is the principle reaction.

HF + H

O H

+ F

–

Initial

concentration (M)

0.02 0 0

Change (M)

–0.02α +0.02α +0.02α

Equilibrium

concentration (M)

0.02 – 0.02 α 0.02 α 0.02α

Substituting equilibrium concentrations

in the equilibrium reaction for principal

reaction gives:

= (0.02α)

/ (0.02 – 0.02α)

= 0.02 α

/ (1 –α) = 3.2 × 10

–4

We obtain the following quadratic

equation:

+ 1.6 × 10

–2

α – 1.6 × 10

–2

= 0

The quadratic equation in α can be solved

and the two values of the roots are:

α = + 0.12 and – 0.12

The negative root is not acceptable and

hence,

α = 0.12

This means that the degree of ionization,

α = 0.12, then equilibrium concentrations

of other species viz., HF, F

–

and H

are

given by:

] = [F

–

] = cα = 0.02 × 0.12

= 2.4 × 10

–3

[HF] = c(1 – α) = 0.02 (1 – 0.12)

= 17.6 × 10

-3

pH = – log[H

] = –log(2.4 × 10

–3

) = 2.62

Problem 7.19

The pH of 0.1M monobasic acid is 4.50.

Calculate the concentration of species H

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221EQUILIBRIUM

Table 7.7 The Values of the Ionization

Constant of Some Weak Bases at

298 K

Base K

Dimethylamine, (CH

)

NH 5.4 × 10

–4

Triethylamine, (C

)

N 6.45 × 10

–5

Ammonia, NH

or NH

OH 1.77 × 10

–5

Quinine, (A plant product) 1.10 × 10

–6

Pyridine, C

N 1.77 × 10

–9

Aniline, C

4.27 × 10

–10

Urea, CO (NH

)

1.3 × 10

–14

–

and HA at equilibrium. Also, determine

the value of K

and pK

of the monobasic

acid.

Solution

pH = – log [H

]

Therefore, [H

] = 10

–pH

= 10

–4.50

= 3.16 × 10

–5

] = [A

–

] = 3.16 × 10

–5

Thus, K

= [H

][A

] / [HA]

[HA]

eqlbm

= 0.1 – (3.16 × 10

-5

) ∫ 0.1

= (3.16 × 10

–5

)

/ 0.1 = 1.0 × 10

–8

= – log(10

–8

) = 8

Alternatively, “Percent dissociation” is

another useful method for measure of

strength of a weak acid and is given as:

Percent dissociation

= [HA]

dissociated

/[HA]

initial

× 100% (7.32)

Problem 7.20

Calculate the pH of 0.08M solution of

hypochlorous acid, HOCl. The ionization

constant of the acid is 2.5 × 10

–5

Determine the percent dissociation of

HOCl.

Solution

HOCl(aq) + H

O (l)

(aq) + ClO

–

(aq)

Initial concentration (M)

0.08 0 0

Change to reach

equilibrium concentration

(M)

– x + x +x

equilibrium concentartion (M)

0.08 – x x x

= {[H

][ClO

–

] / [HOCl]}

= x

/ (0.08 –x)

As x << 0.08, therefore 0.08 – x ∫ 0.08

/ 0.08 = 2.5 × 10

–5

= 2.0 × 10

–6

, thus, x = 1.41 × 10

–3

] = 1.41 × 10

–3

Therefore,

Percent dissociation

= {[HOCl]

dissociated

/ [HOCl]

initial

}× 100

= 1.41 × 10

–3

× 10

/ 0.08 = 1.76 %.

pH = –log(1.41 × 10

–3

) = 2.85.

7.11.4 Ionization of Weak Bases

The ionization of base MOH can be represented

by equation:

MOH(aq) M

(aq) + OH

–

(aq)

In a weak base there is partial ionization

of MOH into M

and OH

–

, the case is similar to

that of acid-dissociation equilibrium. The

equilibrium constant for base ionization is

called base ionization constant and is

represented by K

. It can be expressed in terms

of concentration in molarity of various species

in equilibrium by the following equation:

= [M

][OH

–

] / [MOH] (7.33)

Alternatively, if c = initial concentration of

base and α = degree of ionization of base i.e.

the extent to which the base ionizes. When

equilibrium is reached, the equilibrium

constant can be written as:

= (cα)

/ c (1-α) = cα

/ (1-α)

The values of the ionization constants of

some selected weak bases, K

are given in

Table 7.7.

Many organic compounds like amines are

weak bases. Amines are derivatives of

ammonia in which one or more hydrogen

atoms are replaced by another group. For

example, methylamine, codeine, quinine and

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222 CHEMISTRY

nicotine all behave as very weak bases due to

their very small K

. Ammonia produces OH

–

in aqueous solution:

(aq) + H

O(l)

(aq) + OH

–

(aq)

The pH scale for the hydrogen ion

concentration has been extended to get:

= –log (K

) (7.34)

Problem 7.21

The pH of 0.004M hydrazine solution is

9.7. Calculate its ionization constant K

and pK

Solution

+ H

O NH

+ OH

–

From the pH we can calculate the

hydrogen ion concentration. Knowing

hydrogen ion concentration and the ionic

product of water we can calculate the

concentration of hydroxyl ions. Thus we

have:

] = antilog (–pH)

= antilog (–9.7) = 1.67 ×10

–10

[OH

–

] = K

/ [H

] = 1 × 10

–14

/ 1.67 × 10

–10

= 5.98 × 10

–5

The concentration of the corresponding

hydrazinium ion is also the same as that

of hydroxyl ion. The concentration of both

these ions is very small so the

concentration of the undissociated base

can be taken equal to 0.004M.

Thus,

= [NH

][OH

–

] / [NH

]

= (5.98 × 10

–5

)

/ 0.004 = 8.96 × 10

–7

= –logK

= –log(8.96 × 10

–7

) = 6.04.

Problem 7.22

Calculate the pH of the solution in which

0.2M NH

Cl and 0.1M NH

are present. The

of ammonia solution is 4.75.

Solution

+ H

O NH

+ OH

–

The ionization constant of NH

= antilog (–pK

) i.e.

= 10

–4.75

= 1.77 × 10

–5

+ H

+ OH

–

Initial concentration (M)

0.10 0.20 0

Change to reach

equilibrium (M)

–x +x +x

At equilibrium (M)

0.10 – x 0.20 + x x

= [NH

][OH

–

] / [NH

]

= (0.20 + x)(x) / (0.1 – x) = 1.77 × 10

–5

As K

is small, we can neglect x in

comparison to 0.1M and 0.2M. Thus,

[OH

–

] = x = 0.88 × 10

–5

Therefore, [H

] = 1.12 × 10

–9

pH = – log[H

] = 8.95.

7.11.5 Relation between

and K

As seen earlier in this chapter, K

and K

represent the strength of an acid and a base,

respectively. In case of a conjugate acid-base

pair, they are related in a simple manner so

that if one is known, the other can be deduced.

Considering the example of NH

and NH

we see,

(aq) + H

O(l) H

(aq) + NH

(aq)

= [H

][ NH

] / [NH

] = 5.6 × 10

–10

(aq) + H

O(l) NH

(aq) + OH

–

(aq)

=[ NH

][ OH

–

] / NH

= 1.8 × 10

–5

Net: 2 H

O(l) H

(aq) + OH

–

(aq)

= [H

][ OH

–

] = 1.0 × 10

–14

Where, K

represents the strength of NH

an acid and K

represents the strength of NH

as a base.

It can be seen from the net reaction that

the equilibrium constant is equal to the

product of equilibrium constants K

and K

for the reactions added. Thus,

× K

= {[H

][ NH

] / [NH

]} × {[NH

]

[ OH

–

] / [NH

]}

= [H

][ OH

–

] = K

= (5.6x10

–10

) × (1.8 × 10

–5

) = 1.0 × 10

–14

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223EQUILIBRIUM

This can be extended to make a

generalisation. The equilibrium constant for

a net reaction obtained after adding two

(or more) reactions equals the product of

the equilibrium constants for individual

reactions:

NET

= K

× K

× …… (3.35)

Similarly, in case of a conjugate acid-base

pair,

× K

= K

(7.36)

Knowing one, the other can be obtained. It

should be noted that a strong acid will have

a weak conjugate base and vice-versa.

Alternatively, the above expression

= K

× K

, can also be obtained by

considering the base-dissociation equilibrium

reaction:

B(aq) + H

O(l) BH

(aq) + OH

–

(aq)

= [BH

][OH

–

] / [B]

As the concentration of water remains

constant it has been omitted from the

denominator and incorporated within the

dissociation constant. Then multiplying and

dividing the above expression by [H

], we get:

= [BH

][OH

–

][H

] / [B][H

]

={[ OH

–

][H

]}{[BH

] / [B][H

]}

= K

/ K

or K

× K

= K

It may be noted that if we take negative

logarithm of both sides of the equation, then

pK values of the conjugate acid and base are

related to each other by the equation:

+ pK

= pK

= 14 (at 298K)

Problem 7.23

Determine the degree of ionization and pH

of a 0.05M of ammonia solution. The

ionization constant of ammonia can be

taken from Table 7.7. Also, calculate the

ionization constant of the conjugate acid

of ammonia.

Solution

The ionization of NH

in water is

represented by equation:

+ H

O NH

+ OH

–

We use equation (7.33) to calculate

hydroxyl ion concentration,

[OH

–

] = c α = 0.05 α

= 0.05 α

/ (1 – α)

The value of α is small, therefore the

quadratic equation can be simplified by

neglecting α in comparison to 1 in the

denominator on right hand side of the

equation,

Thus,

= c α

or α = √ (1.77 × 10

–5

/ 0.05)

= 0.018.

[OH

–

] = c α = 0.05 × 0.018 = 9.4 × 10

–4

] = K

/ [OH

–

] = 10

–14

/ (9.4 × 10

–4

)

= 1.06 × 10

–11

pH = –log(1.06 × 10

–11

) = 10.97.

Now, using the relation for conjugate

acid-base pair,

× K

= K

using the value of K

of NH

from

Table 7.7.

We can determine the concentration of

conjugate acid NH

= K

/ K

= 10

–14

/ 1.77 × 10

–5

= 5.64 × 10

–10

7.11.6 Di- and Polybasic Acids and Di- and

Polyacidic Bases

Some of the acids like oxalic acid, sulphuric

acid and phosphoric acids have more than one

ionizable proton per molecule of the acid.

Such acids are known as polybasic or

polyprotic acids.

The ionization reactions for example for a

dibasic acid H

X are represented by the

equations:

X(aq)

(aq) + HX

–

(aq)

–

(aq) H

(aq) + X

2–

(aq)

And the corresponding equilibrium

constants are given below:

= {[H

][HX

–

]} / [H

X] and

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224 CHEMISTRY

= {[H

][X

]} / [HX

]

Here, K

and K

are called the first and second

ionization constants respectively of the acid H

X. Similarly, for tribasic acids like H

have three ionization constants. The values of

the ionization constants for some common

polyprotic acids are given in Table 7.8.

In general, when strength of H-A bond

decreases, that is, the energy required to break

the bond decreases, HA becomes a stronger

acid. Also, when the H-A bond becomes more

polar i.e., the electronegativity difference

between the atoms H and A increases and

there is marked charge separation, cleavage

of the bond becomes easier thereby increasing

the acidity.

But it should be noted that while

comparing elements in the same group of the

periodic table, H-A bond strength is a more

important factor in determining acidity than

its polar nature. As the size of A increases

down the group, H-A bond strength decreases

and so the acid strength increases. For

example,

Size increases

HF << HCl << HBr << HI

Acid strength increases

Similarly, H

S is stronger acid than H

But, when we discuss elements in the same

row of the periodic table, H-A bond polarity

becomes the deciding factor for determining

the acid strength. As the electronegativity of A

increases, the strength of the acid also

increases. For example,

Electronegativity of A increases

< NH

< H

O < HF

Acid strength increases

7.11.8 Common Ion Effect in the

Ionization of Acids and Bases

Consider an example of acetic acid dissociation

equilibrium represented as:

COOH(aq)

(aq) + CH

COO

–

(aq)

or HAc(aq) H

(aq) + Ac

–

(aq)

= [H

][Ac

–

] / [HAc]

Addition of acetate ions to an acetic acid

solution results in decreasing the

concentration of hydrogen ions, [H

]. Also, if

ions are added from an external source then

the equilibrium moves in the direction of

undissociated acetic acid i.e., in a direction of

reducing the concentration of hydrogen ions,

]. This phenomenon is an example of

Table 7.8 The Ionization Constants of Some

Common Polyprotic Acids (298K)

It can be seen that higher order ionization

constants (K

, K

) are smaller than the lower

order ionization constant (K

) of a polyprotic

acid. The reason for this is that it is more

difficult to remove a positively charged proton

from a negative ion due to electrostatic forces.

This can be seen in the case of removing a

proton from the uncharged H

compared from a negatively charged HCO

–

Similarly, it is more difficult to remove a proton

from a doubly charged HPO

2–

anion as

compared to H

–

Polyprotic acid solutions contain a mixture

of acids like H

A, HA

–

and A

2–

in case of a

diprotic acid. H

A being a strong acid, the

primary reaction involves the dissociation of

A, and H

in the solution comes mainly

from the first dissociation step.

7.11.7 Factors Affecting Acid Strength

Having discussed quantitatively the strengths

of acids and bases, we come to a stage where

we can calculate the pH of a given acid

solution. But, the curiosity rises about why

should some acids be stronger than others?

What factors are responsible for making them

stronger? The answer lies in its being a

complex phenomenon. But, broadly speaking

we can say that the extent of dissociation of

an acid depends on the strength and polarity

of the H-A bond.

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225EQUILIBRIUM

common ion effect. It can be defined as a

shift in equilibrium on adding a substance

that provides more of an ionic species already

present in the dissociation equilibrium. Thus,

we can say that common ion effect is a

phenomenon based on the Le Chatelier’s

principle discussed in section 7.8.

In order to evaluate the pH of the solution

resulting on addition of 0.05M acetate ion to

0.05M acetic acid solution, we shall consider

the acetic acid dissociation equilibrium once

again,

HAc(aq)

(aq) + Ac

–

(aq)

Initial concentration (M)

0.05 0 0.05

Let x be the extent of ionization of acetic

acid.

Change in concentration (M)

–x +x +x

Equilibrium concentration (M)

0.05-x x 0.05+x

Therefore,

= [H

][Ac

–

]/[H Ac] = {(0.05+x)(x)}/(0.05-x)

As K

is small for a very weak acid, x<<0.05.

Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05

Thus,

1.8 × 10

–5

= (x) (0.05 + x) / (0.05 – x)

= x(0.05) / (0.05) = x = [H

] = 1.8 × 10

–5

pH = – log(1.8 × 10

–5

) = 4.74

Problem 7.24

Calculate the pH of a 0.10M ammonia

solution. Calculate the pH after 50.0 mL

of this solution is treated with 25.0 mL of

0.10M HCl. The dissociation constant of

ammonia, K

= 1.77 × 10

–5

Solution

+ H

O → NH

+ OH

–

= [NH

][OH

–

] / [NH

] = 1.77 × 10

–5

Before neutralization,

[NH

] = [OH

–

] = x

[NH

] = 0.10 – x ∫ 0.10

/ 0.10 = 1.77 × 10

–5

Thus, x = 1.33 × 10

–3

= [OH

–

]

Therefore,[H

] = K

/ [OH

–

] = 10

–14

(1.33 × 10

–3

) = 7.51 × 10

–12

pH = –log(7.5 × 10

–12

) = 11.12

On addition of 25 mL of 0.1M HCl

solution (i.e., 2.5 mmol of HCl) to 50 mL

of 0.1M ammonia solution (i.e., 5 mmol

of NH

), 2.5 mmol of ammonia molecules

are neutralized. The resulting 75 mL

solution contains the remaining

unneutralized 2.5 mmol of NH

molecules

and 2.5 mmol of NH

+ HCl → NH

+ Cl

–

2.5 2.5 0 0

At equilibrium

0 0 2.5 2.5

The resulting 75 mL of solution contains

2.5 mmol of NH

ions (i.e., 0.033 M) and

2.5 mmol (i.e., 0.033 M ) of uneutralised

molecules. This NH

exists in the

following equilibrium:

OH NH

+ OH

–

0.033M – y y y

where, y = [OH

–

] = [NH

]

The final 75 mL solution after

neutralisation already contains

2.5 m mol NH

ions (i.e. 0.033M), thus

total concentration of NH

ions is given as:

[NH

] = 0.033 + y

As y is small, [NH

OH] ∫ 0.033 M and

[NH

] ∫ 0.033M.

We know,

= [NH

][OH

–

] / [NH

OH]

= y(0.033)/(0.033) = 1.77 × 10

–5

Thus, y = 1.77 × 10

–5

= [OH

–

]

] = 10

–14

/ 1.77 × 10

–5

= 0.56 × 10

–9

Hence, pH = 9.24

7.11.9 Hydrolysis of Salts and the pH of

their Solutions

Salts formed by the reactions between acids

and bases in definite proportions, undergo

ionization in water. The cations/anions formed

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226 CHEMISTRY

on ionization of salts either exist as hydrated

ions in aqueous solutions or interact with

water to reform corresponding acids/bases

depending upon the nature of salts. The later

process of interaction between water and

cations/anions or both of salts is called

hydrolysis. The pH of the solution gets affected

by this interaction. The cations (e.g., Na

, K

, Ba

, etc.) of strong bases and anions

(e.g., Cl

–

, Br

–

, NO

–

, ClO

–

etc.) of strong acids

simply get hydrated but do not hydrolyse, and

therefore the solutions of salts formed from

strong acids and bases are neutral i.e., their

pH is 7. However, the other category of salts

do undergo hydrolysis.

We now consider the hydrolysis of the salts

of the following types :

(i) salts of weak acid and strong base e.g.,

COONa.

(ii) salts of strong acid and weak base e.g.,

Cl, and

(iii) salts of weak acid and weak base, e.g.,

COONH

In the first case, CH

COONa being a salt of

weak acid, CH

COOH and strong base, NaOH

gets completely ionised in aqueous solution.

COONa(aq) → CH

COO

–

(aq)+ Na

(aq)

Acetate ion thus formed undergoes

hydrolysis in water to give acetic acid and OH

–

ions

COO

–

(aq)+H

O(l)

COOH(aq)+OH

–

(aq)

Acetic acid being a weak acid

= 1.8 × 10

–5

) remains mainly unionised in

solution. This results in increase of OH

–

ion

concentration in solution making it alkaline.

The pH of such a solution is more than 7.

Similarly, NH

Cl formed from weak base,

OH and strong acid, HCl, in water

dissociates completely.

Cl(aq) → NH

(aq) +Cl

–

(aq)

Ammonium ions undergo hydrolysis with

water to form NH

OH and H

ions

(aq) + H

O (1) NH

OH(aq) + H

(aq)

Ammonium hydroxide is a weak base

= 1.77 × 10

–5

) and therefore remains almost

unionised in solution. This results in

increased of H

ion concentration in solution

making the solution acidic. Thus, the pH of

Cl solution in water is less than 7.

Consider the hydrolysis of CH

COONH

salt

formed from weak acid and weak base. The

ions formed undergo hydrolysis as follow:

COO

–

+ NH

+ H

O CH

COOH +

COOH and NH

OH, also remain into

partially dissociated form:

COOH CH

COO

–

+ H

OH NH

+ OH

–

+ OH

–

Without going into detailed calculation, it

can be said that degree of hydrolysis is

independent of concentration of solution, and

pH of such solutions is determined by their pK

values:

pH = 7 + ½ (pK

– pK

) (7.38)

The pH of solution can be greater than 7,

if the difference is positive and it will be less

than 7, if the difference is negative.

Problem 7.25

The pK

of acetic acid and pK

ammonium hydroxide are 4.76 and 4.75

respectively. Calculate the pH of

ammonium acetate solution.

Solution

pH = 7 + ½ [pK

– pK

]

= 7 + ½ [4.76 – 4.75]

= 7 + ½ [0.01] = 7 + 0.005 = 7.005

7.12 BUFFER SOLUTIONS

Many body fluids e.g., blood or urine have

definite pH and any deviation in their pH

indicates malfunctioning of the body. The

control of pH is also very important in many

chemical and biochemical processes. Many

medical and cosmetic formulations require

that these be kept and administered at a

particular pH. The solutions which resist

change in pH on dilution or with the

addition of small amounts of acid or alkali

are called Buffer Solutions. Buffer solutions

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227EQUILIBRIUM

of known pH can be prepared from the

knowledge of pK

of the acid or pK

of base

and by controlling the ratio of the salt and acid

or salt and base. A mixture of acetic acid and

sodium acetate acts as buffer solution around

pH 4.75 and a mixture of ammonium chloride

and ammonium hydroxide acts as a buffer

around pH 9.25. You will learn more about

buffer solutions in higher classes.

7.12.1 Designing Buffer Solution

Knowledge of pK

, pK

and equilibrium

constant help us to prepare the buffer solution

of known pH. Let us see how we can do this.

Preparation of Acidic Buffer

To prepare a buffer of acidic pH we use weak

acid and its salt formed with strong base. We

develop the equation relating the pH, the

equilibrium constant, K

of weak acid and ratio

of concentration of weak acid and its conjugate

base. For the general case where the weak acid

HA ionises in water,

HA + H

O Ç H

+ A

–

For which we can write the expression

Rearranging the expression we have,

Taking logarithm on both the sides and

rearranging the terms we get —

(7.39)

(7.40)

The expression (7.40) is known as

Henderson–Hasselbalch equation. The

quantity is the ratio of concentration of

conjugate base (anion) of the acid and the acid

present in the mixture. Since acid is a weak

acid, it ionises to a very little extent and

concentration of [HA] is negligibly different from

concentration of acid taken to form buffer. Also,

most of the conjugate base, [A

—

], comes from

the ionisation of salt of the acid. Therefore, the

concentration of conjugate base will be

negligibly different from the concentration of

salt. Thus, equation (7.40) takes the form:

p pK

[Salt]

[Acid]

+ log

In the equation (7.39), if the concentration

of [A

—

] is equal to the concentration of [HA],

then pH = pK

because value of log 1 is zero.

Thus if we take molar concentration of acid and

salt (conjugate base) same, the pH of the buffer

solution will be equal to the pK

of the acid. So

for preparing the buffer solution of the required

pH we select that acid whose pK

is close to the

required pH. For acetic acid pK

value is 4.76,

therefore pH of the buffer solution formed by

acetic acid and sodium acetate taken in equal

molar concentration will be around 4.76.

A similar analysis of a buffer made with a

weak base and its conjugate acid leads to the

result,

pOH= p +log

[Conjugate acid,BH

]

[Base,B]

(7.41)

pH of the buffer solution can be calculated

by using the equation pH + pOH =14.

We know that pH + pOH = pK

and

+ pK

= pK

. On putting these values in

equation (7.41) it takes the form as follows:

p - pH = p p

[Conjugate acid,BH ]

[Base,B]

w w

K K K

− +

log

pH = p

[Conjugate acid,BH ]

[Base,B]

log

(7.42)

If molar concentration of base and its

conjugate acid (cation) is same then pH of the

buffer solution will be same as pK

for the base.

value for ammonia is 9.25; therefore a

buffer of pH close to 9.25 can be obtained by

taking ammonia solution and ammonium

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228 CHEMISTRY

chloride solution of same molar concentration.

For a buffer solution formed by ammonium

chloride and ammonium hydroxide, equation

(7.42) becomes:

pH =

[Conjugate acid,BH ]

[Base,B]

9 25. log+

pH of the buffer solution is not affected by

dilution because ratio under the logarithmic

term remains unchanged.

7.13 SOLUBILITY EQUILIBRIA OF

SPARINGLY SOLUBLE SALTS

We have already known that the solubility of

ionic solids in water varies a great deal. Some

of these (like calcium chloride) are so soluble

that they are hygroscopic in nature and even

absorb water vapour from atmosphere. Others

(such as lithium fluoride) have so little solubility

that they are commonly termed as insoluble.

The solubility depends on a number of factors

important amongst which are the lattice

enthalpy of the salt and the solvation enthalpy

of the ions in a solution. For a salt to dissolve

in a solvent the strong forces of attraction

between its ions (lattice enthalpy) must be

overcome by the ion-solvent interactions. The

solvation enthalpy of ions is referred to in terms

of solvation which is always negative i.e. energy

is released in the process of solvation. The

amount of solvation enthalpy depends on the

nature of the solvent. In case of a non-polar

(covalent) solvent, solvation enthalpy is small

and hence, not sufficient to overcome lattice

enthalpy of the salt. Consequently, the salt does

not dissolve in non-polar solvent. As a general

rule , for a salt to be able to dissolve in a

particular solvent its solvation enthalpy must

be greater than its lattice enthalpy so that the

latter may be overcome by former. Each salt has

its characteristic solubility which depends on

temperature. We classify salts on the basis of

their solubility in the following three categories.

Category I Soluble Solubility > 0.1M

Category II Slightly 0.01M<Solubility< 0.1M

Soluble

Category III Sparingly Solubility < 0.01M

Soluble

We shall now consider the equilibrium

between the sparingly soluble ionic salt and

its saturated aqueous solution.

7.13.1 Solubility Product Constant

Let us now have a solid like barium sulphate

in contact with its saturated aqueous solution.

The equilibrium between the undisolved solid

and the ions in a saturated solution can be

represented by the equation:

BaSO

(s)

(aq) + SO

2–

(aq),

The equilibrium constant is given by the

equation:

K = {[Ba

][SO

2–

]} / [BaSO

]

For a pure solid substance the

concentration remains constant and we can

write

= K[BaSO

] = [Ba

][SO

2–

] (7.43)

We call K

the solubility product constant

or simply solubility product. The experimental

value of K

in above equation at 298K is

1.1 × 10

–10

. This means that for solid barium

sulphate in equilibrium with its saturated

solution, the product of the concentrations of

barium and sulphate ions is equal to its

solubility product constant. The

concentrations of the two ions will be equal to

the molar solubility of the barium sulphate. If

molar solubility is S, then

1.1 × 10

–10

= (S)(S) = S

or S = 1.05 × 10

–5

Thus, molar solubility of barium sulphate

will be equal to 1.05 × 10

–5

mol L

–1

A salt may give on dissociation two or more

than two anions and cations carrying different

charges. For example, consider a salt like

zirconium phosphate of molecular formula

(Zr

)

(PO

3–

)

. It dissociates into 3 zirconium

cations of charge +4 and 4 phosphate anions

of charge –3. If the molar solubility of

zirconium phosphate is S, then it can be seen

from the stoichiometry of the compound that

[Zr

] = 3S and [PO

3–

] = 4S

and K

= (3S)

(4S)

= 6912 (S)

or S = {K

/ (3

× 4

)}

1/7

= (K

/ 6912)

1/7

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229EQUILIBRIUM

A solid salt of the general formula

M X

q+ −

with molar solubility S in equilibrium

with its saturated solution may be represented

by the equation:

(s)

(aq) + yX

q–

(aq)

(where x × p

= y × q

–

)

And its solubility product constant is given

by:

= [M

]

q–

]

= (xS)

(yS)

(7.44)

= x

. y

. S

(x + y)

= K

/ x

. y

S = (K

/ x

. y

)

1 / x + y

(7.45)

The term K

in equation is given by Q

(section 7.6.2) when the concentration of one

or more species is not the concentration under

equilibrium. Obviously under equilibrium

conditions K

= Q

but otherwise it gives the

direction of the processes of precipitation or

dissolution. The solubility product constants

of a number of common salts at 298K are given

in Table 7.9.

Table 7.9 The Solubility Product Constants,

of Some Common Ionic Salts at

298K.

Problem 7.26

Calculate the solubility of A

in pure

water, assuming that neither kind of ion

reacts with water. The solubility product

of A

, K

= 1.1 × 10

–23

Solution

→ 2A

+ 3X

2–

= [A

]

2–

]

= 1.1 × 10

–23

If S = solubility of A

, then

] = 2S; [X

2–

] = 3S

therefore, K

= (2S)

(3S)

= 108S

= 1.1 × 10

–23

thus, S

= 1 × 10

–25

S = 1.0 × 10

–5

mol/L.

Problem 7.27

The values of K

of two sparingly soluble

salts Ni(OH)

and AgCN are 2.0 × 10

–15

and 6 × 0

–17

respectively. Which salt is

more soluble? Explain.

Solution

AgCN Ag

+ CN

–

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230 CHEMISTRY

= [Ag

][CN

–

] = 6 × 10

–17

Ni(OH)

+ 2OH

–

= [Ni

][OH

–

]

= 2 × 10

–15

Let [Ag

] = S

, then [CN

] = S

Let [Ni

] = S

, then [OH

–

] = 2S

= 6 × 10

–17

, S

= 7.8 × 10

–9

)(2S

)

= 2 × 10

–15

, S

= 0.58 × 10

–4

Ni(OH)

is more soluble than AgCN.

7.13.2 Common Ion Effect on Solubility

of Ionic Salts

It is expected from Le Chatelier’s principle that

if we increase the concentration of any one of

the ions, it should combine with the ion of its

opposite charge and some of the salt will be

precipitated till once again K

= Q

. Similarly,

if the concentration of one of the ions is

decreased, more salt will dissolve to increase

the concentration of both the ions till once

again K

= Q

. This is applicable even to

soluble salts like sodium chloride except that

due to higher concentrations of the ions, we

use their activities instead of their molarities

in the expression for Q

. Thus if we take a

saturated solution of sodium chloride and

pass HCl gas through it, then sodium chloride

is precipitated due to increased concentration

(activity) of chloride ion available from the

dissociation of HCl. Sodium chloride thus

obtained is of very high purity and we can get

rid of impurities like sodium and magnesium

sulphates. The common ion effect is also used

for almost complete precipitation of a particular

ion as its sparingly soluble salt, with very low

value of solubility product for gravimetric

estimation. Thus we can precipitate silver ion

as silver chloride, ferric ion as its hydroxide

(or hydrated ferric oxide) and barium ion as

its sulphate for quantitative estimations.

Dissolution of S mol/L of Ni(OH)

provides

S mol/L of Ni

and 2S mol/L of OH

–

, but

the total concentration of OH

–

= (0.10 +

2S) mol/L because the solution already

contains 0.10 mol/L of OH

–

from NaOH.

= 2.0 × 10

–15

= [Ni

] [OH

–

]

= (S) (0.10 + 2S)

As K

is small, 2S << 0.10,

thus, (0.10 + 2S) ≈

0.10

Hence,

2.0 × 10

–15

= S (0.10)

S = 2.0 × 10

–13

M = [Ni

]

The solubility of salts of weak acids like

phosphates increases at lower pH. This is

because at lower pH the concentration of the

anion decreases due to its protonation. This

in turn increase the solubility of the salt so

that K

= Q

. We have to satisfy two equilibria

simultaneously i.e.,

= [M

] [X

–

] / [HX] = K

/ [H

]

Taking inverse of both side and adding 1

we get

HX H

[ ]









+ =









[ ]

























−

1 1

Now, again taking inverse, we get

–

] / {[X

–

] + [HX]} = f = K

/ (K

+ [H

]) and it

can be seen that ‘f’ decreases as pH decreases. If

S is the solubility of the salt at a given pH then

= [S] [f S] = S

/ (K

+ [H

])} and

S = {K

([H

] + K

) / K

}

1/2

(7.46)

Thus solubility S increases with increase

in [H

] or decrease in pH.

Problem 7.28

Calculate the molar solubility of Ni(OH)

in 0.10 M NaOH. The ionic product of

Ni(OH)

is 2.0 × 10

–15

Solution

Let the solubility of Ni(OH)

be equal to S.

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231EQUILIBRIUM

SUMMARY

When the number of molecules leaving the liquid to vapour equals the number of

molecules returning to the liquid from vapour, equilibrium is said to be attained and is

dynamic in nature. Equilibrium can be established for both physical and chemical

processes and at this stage rate of forward and reverse reactions are equal. Equilibrium

constant, K

is expressed as the concentration of products divided by reactants, each

term raised to the stoichiometric coefficient.

For reaction, a A + b B c C +d D

= [C]

[D]

/[A]

[B]

Equilibrium constant has constant value at a fixed temperature and at this stage

all the macroscopic properties such as concentration, pressure, etc. become constant.

For a gaseous reaction equilibrium constant is expressed as K

and is written by replacing

concentration terms by partial pressures in K

expression. The direction of reaction can

be predicted by reaction quotient Q

which is equal to K

at equilibrium. Le Chatelier’s

principle states that the change in any factor such as temperature, pressure,

concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce

or counteract the effect of the change. It can be used to study the effect of various

factors such as temperature, concentration, pressure, catalyst and inert gases on the

direction of equilibrium and to control the yield of products by controlling these factors.

Catalyst does not effect the equilibrium composition of a reaction mixture but increases

the rate of chemical reaction by making available a new lower energy pathway for

conversion of reactants to products and vice-versa.

All substances that conduct electricity in aqueous solutions are called electrolytes.

Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous

solutions is due to anions and cations produced by the dissociation or ionization of

electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In

weak electrolytes there is equilibrium between the ions and the unionized electrolyte

molecules. According to Arrhenius, acids give hydrogen ions while bases produce

hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined

an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry

acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding

to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one

proton. Lewis further generalised the definition of an acid as an electron pair acceptor

and a base as an electron pair donor. The expressions for ionization (equilibrium)

constants of weak acids (K

) and weak bases (K

) are developed using Arrhenius definition.

The degree of ionization and its dependence on concentration and common ion are

discussed. The pH scale (pH = -log[H

]) for the hydrogen ion concentration (activity) has

been introduced and extended to other quantities (pOH = – log[OH

–

]) ; pK

= –log[K

] ;

= –log[K

]; and pK

= –log[K

] etc.). The ionization of water has been considered and

we note that the equation: pH + pOH = pK

is always satisfied. The salts of strong acid

and weak base, weak acid and strong base, and weak acid and weak base undergo

hydrolysis in aqueous solution.The definition of buffer solutions, and their importance

are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed

and the equilibrium constant is introduced as solubility product constant (K

). Its

relationship with solubility of the salt is established. The conditions of precipitation of

the salt from their solutions or their dissolution in water are worked out. The role of

common ion and the solubility of sparingly soluble salts is also discussed.

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232 CHEMISTRY

SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT

(a) The student may use pH paper in determining the pH of fresh juices of various

vegetables and fruits, soft drinks, body fluids and also that of water samples available.

(b) The pH paper may also be used to determine the pH of different salt solutions and

from that he/she may determine if these are formed from strong/weak acids and

bases.

and acetic acid and determine their pH using pH paper.

(d) They may be provided with different indicators to observe their colours in solutions

of varying pH.

(e) They may perform some acid-base titrations using indicators.

(f) They may observe common ion effect on the solubility of sparingly soluble salts.

(g) If pH meter is available in their school, they may measure the pH with it and compare

the results obtained with that of the pH paper.

EXERCISES

7.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed

temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final

vapour pressure?

7.2 What is K

for the following equilibrium when the equilibrium concentration of

each substance is: [SO

]= 0.60M, [O

] = 0.82M and [SO

] = 1.90M ?

2SO

(g) + O

(g)

2SO

(g)

7.3 At a certain temperature and total pressure of 10

Pa, iodine vapour contains 40%

by volume of I atoms

(g) 2I (g)

Calculate K

for the equilibrium.

7.4 Write the expression for the equilibrium constant, K

for each of the following

reactions:

(i) 2NOCl (g) 2NO (g) + Cl

(g)

(ii) 2Cu(NO

)

(s) 2CuO (s) + 4NO

(g) + O

(g)

(iii) CH

COOC

(aq) + H

O(l) CH

COOH (aq) + C

OH (aq)

(iv) Fe

(aq) + 3OH

–

(aq) Fe(OH)

(s)

(v) I

(s) + 5F

2IF

7.5 Find out the value of K

for each of the following equilibria from the value of K

(i) 2NOCl (g) 2NO (g) + Cl

(g); K

= 1.8 × 10

–2

at 500 K

(ii) CaCO

(s) CaO(s) + CO

(g); K

= 167 at 1073 K

7.6 For the following equilibrium, K

= 6.3 × 10

at 1000 K

NO (g) + O

(g) NO

(g) + O

(g)

Both the forward and reverse reactions in the equilibrium are elementary

bimolecular reactions. What is K

, for the reverse reaction?

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233EQUILIBRIUM

7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium

constant expression?

7.8 Reaction between N

and O

2–

takes place as follows:

(g) + O

(g)

O (g)

If a mixture of 0.482 mol N

and 0.933 mol of O

is placed in a 10 L reaction

vessel and allowed to form N

O at a temperature for which K

= 2.0 × 10

–37

determine the composition of equilibrium mixture.

7.9 Nitric oxide reacts with Br

and gives nitrosyl bromide as per reaction given

below:

2NO (g) + Br

(g)

2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br

are mixed in a closed container at

constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate

equilibrium amount of NO and Br

7.10 At 450K, K

= 2.0 × 10

/bar for the given reaction at equilibrium.

2SO

(g) + O

(g)

2SO

(g)

What is K

at this temperature ?

7.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the

partial pressure of HI(g) is 0.04 atm. What is K

for the given equilibrium ?

2HI (g) H

(g) + I

(g)

7.12 A mixture of 1.57 mol of N

, 1.92 mol of H

and 8.13 mol of NH

is introduced into

a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant,

for the reaction N

(g) + 3H

(g) 2NH

(g) is 1.7 × 10

. Is the reaction mixture

at equilibrium? If not, what is the direction of the net reaction?

7.13 The equilibrium constant expression for a gas reaction is,

[ ]

NH O

NO H O

Write the balanced chemical equation corresponding to this expression.

7.14 One mole of H

O and one mole of CO are taken in 10 L vessel and heated to

725 K. At equilibrium 40% of water (by mass) reacts with CO according to the

equation,

O (g) + CO (g) H

(g) + CO

(g)

Calculate the equilibrium constant for the reaction.

7.15 At 700 K, equilibrium constant for the reaction:

(g) + I

(g) 2HI (g)

is 54.8. If 0.5 mol L

–1

of HI(g) is present at equilibrium at 700 K, what are the

concentration of H

(g) and I

(g) assuming that we initially started with HI(g) and

allowed it to reach equilibrium at 700K?

7.16 What is the equilibrium concentration of each of the substances in the

equilibrium when the initial concentration of ICl was 0.78 M ?

2ICl (g) I

(g) + Cl

(g); K

= 0.14

7.17 K

= 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium

concentration of C

when it is placed in a flask at 4.0 atm pressure and allowed

to come to equilibrium?

(g) C

(g) + H

(g)

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234 CHEMISTRY

7.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the

equilibrium is represented as:

COOH (l) + C

OH (l)

COOC

(l) + H

O (l)

(i) Write the concentration ratio (reaction quotient), Q

, for this reaction (note:

water is not in excess and is not a solvent in this reaction)

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol,

there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate

the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining

it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium

been reached?

7.19 A sample of pure PCl

was introduced into an evacuated vessel at 473 K. After

equilibrium was attained, concentration of PCl

was found to be

0.5 × 10

–1

mol L

–1

. If value of K

is 8.3 × 10

–3

, what are the concentrations of PCl

and Cl

at equilibrium?

PCl

(g) PCl

(g) + Cl

(g)

7.20 One of the reaction that takes place in producing steel from iron ore is the

reduction of iron(II) oxide by carbon monoxide to give iron metal and CO

FeO (s) + CO (g) Fe (s) + CO

(g); K

= 0.265 atm at 1050K

What are the equilibrium partial pressures of CO and CO

at 1050 K if the

initial partial pressures are: p

= 1.4 atm and = 0.80 atm?

7.21 Equilibrium constant, K

for the reaction

(g) + 3H

(g) 2NH

(g) at 500 K is 0.061

At a particular time, the analysis shows that composition of the reaction mixture

is 3.0 mol L

–1

, 2.0 mol L

–1

and 0.5 mol L

–1

. Is the reaction at equilibrium?

If not in which direction does the reaction tend to proceed to reach equilibrium?

7.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches

the equilibrium:

2BrCl (g) Br

(g) + Cl

(g)

for which K

= 32 at 500 K. If initially pure BrCl is present at a concentration of

3.3 × 10

–3

mol L

–1

, what is its molar concentration in the mixture at equilibrium?

7.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO

in equilibrium

with soild carbon has 90.55% CO by mass

C (s) + CO

(g) 2CO (g)

Calculate K

for this reaction at the above temperature.

7.24 Calculate a) ∆G

and b) the equilibrium constant for the formation of NO

from

NO and O

at 298K

NO (g) + ½ O

(g) NO

(g)

where

∆

(NO

) = 52.0 kJ/mol

∆

(NO) = 87.0 kJ/mol

∆

) = 0 kJ/mol

7.25 Does the number of moles of reaction products increase, decrease or remain

same when each of the following equilibria is subjected to a decrease in pressure

by increasing the volume?

(a) PCl

(g) PCl

(g) + Cl

(g)

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235EQUILIBRIUM

(b) CaO (s) + CO

(g) CaCO

(s)

O (g) Fe

(s) + 4H

(g)

7.26 Which of the following reactions will get affected by increasing the pressure?

Also, mention whether change will cause the reaction to go into forward or

backward direction.

(i) COCl

(g) CO (g) + Cl

(g)

(ii) CH

(g) + 2S

(g) CS

(g) + 2H

S (g)

(iii) CO

(g) + C (s) 2CO (g)

(iv) 2H

(g) + CO (g) CH

OH (g)

(v) CaCO

(s)

CaO (s) + CO

(g)

(vi) 4 NH

(g) + 5O

(g) 4NO (g) + 6H

O(g)

7.27 The equilibrium constant for the following reaction is 1.6 ×10

at 1024K

(g) + Br

(g) 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a

sealed container at 1024K.

7.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as

per following endothermic reaction:

(g) + H

O (g) CO (g) + 3H

(g)

(a) Write as expression for K

for the above reaction.

(b) How will the values of K

and composition of equilibrium mixture be affected

(i) increasing the pressure

(ii) increasing the temperature

(iii) using a catalyst ?

7.29 Describe the effect of :

a) addition of H

b) addition of CH

c) removal of CO

d) removal of CH

on the equilibrium of the reaction:

(g) + CO (g) CH

OH (g)

7.30 At 473 K, equilibrium constant K

for decomposition of phosphorus pentachloride,

PCl

is 8.3 ×10

-3

. If decomposition is depicted as,

PCl

(g) PCl

(g) + Cl

(g) ∆

= 124.0 kJ mol

–1

a) write an expression for K

for the reaction.

b) what is the value of K

for the reverse reaction at the same temperature ?

c) what would be the effect on K

if (i) more PCl

is added (ii) pressure is increased

(iii) the temperature is increased ?

7.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from

natural gas with high temperature steam. The first stage of two stage reaction

involves the formation of CO and H

. In second stage, CO formed in first stage is

reacted with more steam in water gas shift reaction,

CO (g) + H

O (g) CO

(g) + H

(g)

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236 CHEMISTRY

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and

steam such that p

= p

= 4.0 bar, what will be the partial pressure of H

equilibrium? K

= 10.1 at 400°C

7.32 Predict which of the following reaction will have appreciable concentration of

reactants and products:

a) Cl

(g) 2Cl (g) K

= 5 ×10

–39

b) Cl

(g) + 2NO (g) 2NOCl (g) K

= 3.7 × 10

c) Cl

(g) + 2NO

(g) 2NO

Cl (g) K

= 1.8

7.33 The value of K

for the reaction 3O

(g) 2O

(g) is 2.0 ×10

–50

at 25°C. If the

equilibrium concentration of O

in air at 25°C is 1.6 ×10

–2

, what is the

concentration of O

7.34 The reaction, CO(g) + 3H

(g)

(g) + H

O(g)

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol

of H

and 0.02 mol of H

O and an unknown amount of CH

in the flask. Determine

the concentration of CH

in the mixture. The equilibrium constant, K

for the

reaction at the given temperature is 3.90.

7.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base

for the following species:

HNO

, CN

–

, HClO

–

, OH

–

, CO

2–

, and S

2–

7.36 Which of the followings are Lewis acids? H

O, BF

, H

, and NH

7.37 What will be the conjugate bases for the Brönsted acids: HF, H

and HCO

–

7.38 Write the conjugate acids for the following Brönsted bases: NH

–

, NH

and HCOO

–

7.39 The species: H

O, HCO

–

, HSO

–

and NH

can act both as Brönsted acids and

bases. For each case give the corresponding conjugate acid and base.

7.40 Classify the following species into Lewis acids and Lewis bases and show how

these act as Lewis acid/base: (a) OH

–

(b) F

–

(d) BCl

7.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10

–3

M. what

is its pH?

7.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen

ion in it.

7.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10

–4

1.8 × 10

–4

and 4.8 × 10

–9

respectively. Calculate the ionization constants of the

corresponding conjugate base.

7.44 The ionization constant of phenol is 1.0 × 10

–10

. What is the concentration of

phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization

if the solution is also 0.01M in sodium phenolate?

7.45 The first ionization constant of H

S is 9.1 × 10

–8

. Calculate the concentration of

–

ion in its 0.1M solution. How will this concentration be affected if the solution

is 0.1M in HCl also ? If the second dissociation constant of H

S is

1.2 × 10

–13

, calculate the concentration of S

2–

under both conditions.

7.46 The ionization constant of acetic acid is 1.74 × 10

–5

. Calculate the degree of

dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of

acetate ion in the solution and its pH.

7.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15.

Calculate the concentration of the anion, the ionization constant of the acid

and its pK

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237EQUILIBRIUM

7.48 Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

7.49 Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution.

b) 0.3 g of Ca(OH)

dissolved in water to give 500 mL of solution.

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

7.50 The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate

the pH of the solution and the pK

of bromoacetic acid.

7.51 The pH of 0.005M codeine (C

) solution is 9.95. Calculate its ionization

constant and pK

7.52 What is the pH of 0.001M aniline solution ? The ionization constant of aniline

can be taken from Table 7.7. Calculate the degree of ionization of aniline in the

solution. Also calculate the ionization constant of the conjugate acid of aniline.

7.53 Calculate the degree of ionization of 0.05M acetic acid if its pK

value is 4.74.

How is the degree of dissociation affected when its solution also contains

(a) 0.01M (b) 0.1M in HCl ?

7.54 The ionization constant of dimethylamine is 5.4 × 10

–4

. Calculate its degree of

ionization in its 0.02M solution. What percentage of dimethylamine is ionized if

the solution is also 0.1M in NaOH?

7.55 Calculate the hydrogen ion concentration in the following biological fluids whose

pH are given below:

(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2

7.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,

5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion

concentration in each.

7.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.

Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What

is its pH?

7.58 The solubility of Sr(OH)

at 298 K is 19.23 g/L of solution. Calculate the

concentrations of strontium and hydroxyl ions and the pH of the solution.

7.59 The ionization constant of propanoic acid is 1.32 × 10

–5

. Calculate the degree of

ionization of the acid in its 0.05M solution and also its pH. What will be its

degree of ionization if the solution is 0.01M in HCl also?

7.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization

constant of the acid and its degree of ionization in the solution.

7.61 The ionization constant of nitrous acid is 4.5 × 10

–4

. Calculate the pH of 0.04 M

sodium nitrite solution and also its degree of hydrolysis.

7.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the

ionization constant of pyridine.

7.63 Predict if the solutions of the following salts are neutral, acidic or basic:

NaCl, KBr, NaCN, NH

, NaNO

and KF

7.64 The ionization constant of chloroacetic acid is 1.35 × 10

–3

. What will be the pH of

0.1M acid and its 0.1M sodium salt solution?

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238 CHEMISTRY

7.65 Ionic product of water at 310 K is 2.7 × 10

–14

. What is the pH of neutral water at

this temperature?

7.66 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)

+ 25 mL of 0.1M HCl

b) 10 mL of 0.01M H

+ 10 mL of 0.01M Ca(OH)

c) 10 mL of 0.1M H

+ 10 mL of 0.1M KOH

7.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,

lead chloride and mercurous iodide at 298K from their solubility product

constants given in Table 7.9. Determine also the molarities of individual ions.

7.68 The solubility product constant of Ag

CrO

and AgBr are 1.1 × 10

–12

and

5.0 × 10

–13

respectively. Calculate the ratio of the molarities of their saturated

solutions.

7.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are

mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate

= 7.4 × 10

–8

7.70 The ionization constant of benzoic acid is 6.46 × 10

–5

and K

for silver benzoate

is 2.5 × 10

–13

. How many times is silver benzoate more soluble in a buffer of pH

3.19 compared to its solubility in pure water?

7.71 What is the maximum concentration of equimolar solutions of ferrous sulphate

and sodium sulphide so that when mixed in equal volumes, there is no

precipitation of iron sulphide? (For iron sulphide, K

= 6.3 × 10

–18

7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate

at 298 K? (For calcium sulphate, K

is 9.1 × 10

–6

7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen

sulphide is 1.0 × 10

–19

M. If 10 mL of this is added to 5 mL of 0.04 M solution of

the following: FeSO

, MnCl

, ZnCl

and CdCl

. in which of these solutions

precipitation will take place?

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