L1 Ch-07 Sol Set-1 Equilibrium

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Level - I

SECTION - A

School/Board Exam. Type Questions

Very Short Answer Type Questions :

1. For an exothermic reaction, what happens to the equilibrium constant if the temperature is raised?

Sol. For an exothermic reaction Kb will increase with rise in temperature much more than K

f. As a result f

b

K

K will

decrease.

2. Write expression for Kp and K

c for the decomposition reaction of calcium carbonate.

Sol. The reaction is

CaCO3(s) CaO(s) + CO

2(g)

2c

3

[CaO(s)][CO (g)]K

[CaCO (s)]

Active mass of solids to be taken as unity.

Kc = [CO

2(g)]

Kp =

2COp

3. At 500 K, Kc for the reaction

H2(g) + D

2(g) 2HD(g) is 3.6.

What is the value of Kc for the following reaction?

2HD(g) H2(g) + D

2(g)

Sol. Kc will be 1

0.283.6

.

Chapter 7

Equilibrium

Solutions (Set-1)

2 Equilibrium Solutions of Assignment (Set-1) (Level-I)


4. Write the expression for equilibrium constant Kp for the reaction

3Fe(s) + 4H2O(g) Fe

3O4(s) + 4H

2(g)

Sol.2

2

4H

p 4H O

(p )K

(p )

5. Under what condition, a reversible process becomes irreversible?

Sol. If one of the products (gaseous) is allowed to escape out (i.e., in open vessel) or the reaction results in the

formation of precipitate.

6. What is the conjugate base of Al(H2O)

6]+3 ion?

Sol. [Al(H2O)

5OH]+2

7. What will be the pH of 1 M Na2SO4 solution?

Sol. Na2SO

4 is a salt of strong acid and strong base and its aqueous solution will be neutral. Therefore, its pH

will be 7.

8. What happens to ionic product of water if some acid is added to it?

Sol. Ionic product will remain unchanged.

9. Why is ammonia termed as a base though it does not contain OH– ions?

Sol. The basic nature of Ammonia is due to its tendency to donate electron pair. Therefore, it is a Lewis base.

10. Calculate the pH of 0.001 M HCl solution.

Sol. 0.001 M HCl = 10–3 M HCl

HCl H+ + Cl–

10–3 M 10–3 M

pH = –log[H+] = –log10–3

pH = 3

Short Answer Type Questions :

11. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

Sol. Kw = [H

3O+] [OH–] = 2.7 × 10–14

[H3O+] = [OH–]

[H3O+]2 = 2.7 × 10–14 = K

w

[H3O+] = 1.643 × 10–7

pH = –log[H3O+]

= –log(1.643 × 10–7)

= –0.2156 – 7

= 6.75

3Solutions of Assignment (Set-1) (Level-I) Equilibrium


12. At 450 K, Kp = 2.0 × 1010 bar for the given reaction at equilibrium

2SO2(g) + O

2(g) 2SO

3(g)

What is the Kc at this temperature?

Sol. Kp = K

c(RT)n ; n = 2 – 3 = –1

p

c n

KK

(RT)

10

c 1

(2 10 )K

(0.0831 450)

= 2.0 × 1010 × 0.0831 × 450

= 7.47 × 1011 M

13. What is the effect of temperature on the ionic product of water? How will it change the pH value of a neutral

solution?

Sol. Ionic product of water is

2H2O H

3O+ + OH–

Ionic product Kw = [H3O+][OH–]

Ionic product increases with increase of temperature because the dissociation of water increases with increase

of temperature.

Now, pH = –log[H3O+]

With increase in concentration of H3O+ ions, the pH of the neutral solution will decrease.

14. Does the number of moles of product increase, decrease or remains same when each of the following

equilibrium is subjected to a decrease in pressure?

(a) PCl5(g) PCl

3(g) + Cl

2(g)

(b) CaO(s) + CO2(g) CaCO

3(s)

(c) 3Fe(s) + 4H2O(g) Fe

3O4(s) + 4H

2(g)

Sol. (a) Increase.

(b) Decrease.

(c) Remain same.

15. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl

5 is 8.3 × 10–3. If

decomposition is depicted as

PCl5(g) PCl

3(g) + Cl

2(g)

H° = 124.0 kJ mol–1

(a) Write an expression for Kc of the reaction.

(b) What is the value of Kc for reverse reaction at the same temperature?

(c) What would be the effect on Kc if

(i) More PCl5 is added?

(ii) Pressure is increased?

(iii) Temperature is increased?



Sol. (a) 3 2

c

5

[PCl ][Cl ]K

[PCl ]

(b)c 3

c

1 1K 120.48

K 8.3 10

(c) (i) No effect, because Kc is constant at a given temperature.

(ii) No effect.

(iii) Increases. Since reaction is endothermic on increasing temperature, Kf will increase so that

f

c

b

KK

K will also increase.

16. For the reaction, H2 + I

2 2HI, if initially 25 ml of H

2 and 20 ml of I

2 are present in a container and at

equilibrium, 30 ml of HI is formed, then calculate equilibrium constant.

Sol. H2 + I

2 2HI

Initial 25 ml 20 ml 0

At equilibrium 25 – x 20 – x 2x

But 2x = 30 ml

x = 15 ml

At equilibrium, H2 = 25 – 15 = 10 ml

I2 = 20 – 15 = 5 ml

HI = 30 ml

As equal volumes contain equal number of moles, therefore, volumes can be used in place of moles.

Hence,

2 2

2 2

[HI] (30) 30 30K 18

[H ][I ] 10 5 10 5

17. 2N2O(g) + O

2(g) 4NO(g); H > 0

What will be the effect on equilibrium when

(i) Volume of the vessel increases?

(ii) Temperature decreases?

Sol. (i) For the given reaction, 4

2

2 2

[NO]K

[N O] [O ] .

When volume of the vessel increases number of moles per unit volume (i.e., molar concentration) of each

reactant and product decreases. As there are 4 concentration terms in the numerator but 3 concentration

terms in the denominator, to keep K constant, the decrease in [N2O] and [O

2] should be more i.e.,

equilibrium will shift in the forward direction.

(ii) As H is positive i.e., reaction is endothermic decrease of temperature will favour the direction in which

heat is absorbed i.e., backward direction.



18. The following reaction has attained equilibrium

CO(g) + 2H2(g) CH

3OH(g), H° = –92.0 kJ mol–1

What will happen if

(i) Volume of the reaction vessel is suddenly reduced to half?

(ii) The partial pressure of hydrogen is suddenly doubled?

(iii) An inert gas is added to the system at constant volume?

Sol. 3

2

CH OH3c p2 2

2 CO H

(p )[CH OH]K , K

[CO][H ] (p )(p )

(i) When volume of the vessel is reduced to half, the concentration of each reactant or product becomes

double. Thus, 3

c c2

2

2[CH OH] 1Q K

42[CO] (2[H ])

.

As Qc < K

c, equilibrium will shift in the forward direction, producing more of CH

3OH to make Q

c = K

c.

(ii) Again Qp < K

p, equilibrium will shift in the forward direction to make Q

p = K

p.

(iii) As volume remain constant, molar concentrations will not change. Hence, there is no effect on the state

of equilibrium.

19. What qualitative information can be obtained from the magnitude of equilibrium constant?

Sol. (i) Large value of equilibrium constant (> 103) shows that forward reaction is favoured i.e., concentration of

products is much larger than that of the reactants at equilibrium.

(ii) Intermediate value of K(10–3 to 103) shows that the concentration of the reactants and products are

comparable.

(iii) Low value of K(< 10–3) shows that backward reaction is favoured i.e., concentration of reactants is much

larger than that of the products.

20. Two processes are given below. What happens to the process if it is subjected to a change given in the

brackets?

(i) Ice Melting point

��⇀↽�� Water (Pressure is increased)

(ii) N2(g) + O

2(g) 2NO(g) – 180.7 kJ (Pressure is increased and temperature is decreased)

Sol. (i) Equilibrium will shift in the forward direction i.e., more of ice will melt.

(ii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.

21. Ionization constant (Ka) for formic acid (HCOOH) and acetic acid (CH

3COOH) are 17.7 × 10–5 and 1.77 × 10–5.

Which acid is stronger and how many times the other the other gets ionised if equimolar concentrations of

the two are taken?

Sol. Ka for HCOOH > K

a for CH

3COOH

Hence, formic acid is stronger.

3

HCOOH

3 CH COOH

KStrength of HCOOH10 3.16 times

Strength of CH COOH K



22. If pH = 7.40 find [H3O+].

Sol. pH = 7.4

–log[H3O+] = 7.4

log[H3O+] = –7.4

log[H3O+] = 8.6

[H3O+] = 4 × 10–8 M

23. The values of Ksp

of two sparingly soluble salts. Ni(OH)2 and AgCN are 2.0 × 10–15 and 6.0 × 10–17 respectively.

Which salt is more soluble? Explain.

Sol. Ni(OH)2 Ni+2 + 2OH–

1 0 0

1 – s s 2s

Ksp

= s × (2s)2 = 4s3 = 2 × 10–15

s = 5.8 × 10–5 mol L–1

AgCN Ag+ + CN–

1 0 0

1 – s s s

Ksp

= [Ag+][CN–] = s × s = s2

17 9 1s 6.0 10 7.8 10 mol L

Ni(OH)

2 is more soluble.

24. Derive the equation of solubility for A2X3 and AX

2 salt w.r.t. solubility product (K

sp), if solubility is ‘s’.

Sol. A2X3 2A+3 + 3X–2

1 0 0

1 – s 2s 3s

Ksp

= [A+3]2 [X–2]3

= (2s)2 × (3s)3 = 4s2 × 27s3

Ksp

= 108s5

1/5

spKs

108

⎛ ⎞ ⎜ ⎟⎝ ⎠

AX2 A2+ + 2X–

1 0 0

1 – s s 2s

Ksp

= [A2+] [X–]2

= s × (2s)2 = s × 4s2 = 4s3

1/3

spKs

4

⎛ ⎞ ⎜ ⎟⎝ ⎠



25. The solubility product of AgCl in water is 1.5 × 10–10. Calculate its solubility in 0.01 M NaCl aqueous

solution.

Sol. As NaCl dissociates completely, therefore, in 0.01 M NaCl solution, [Cl–] = 0.01 M.

If solubility of AgCl in 0.01 M NaCl solution is s mol L–1, then from AgCl that dissolves,

[Ag+] = [Cl–] = s mol L–1

Total [Cl–] = 0.01 + s � 0.01 M

Ksp for AgCl = [Ag+] [Cl–] = s × 0.01 = 0.01 × s

0.01 × s = 1.5 × 10–10

s = 1.5 × 10–8 M

26. The pH of a solution obtained by dissolving 0.1 mole of an acid HA in 100 ml of the aqueous solution was

found to be 3.0. Calculate the dissociation constant of the acid.

Sol. pH = –log[H3O+] = 3

[H3O+] = Antilog(–3) = 10–3 mole/L = 0.001 mole/L

HA + H2O H

3O+ + A–

1 M 0 0

1 – 0.001 0.001 M 0.001 M

Dissociation constant (K) = 3[H O ][A ]

[HA]

6

0.001 0.001 10

1 0.001 1

= 10–6

27. How many grams of NaOH must be dissolved in one litre of the solution to give it a pH value of 12?

Sol. pH = –log[H3O+]

log[H3O+] = –pH = –12

[H+] = 10–12

Kw = [H+] [OH–] = 10–14

142w

12

K 10[OH ] 10

[H ] 10

Since, NaOH is a strong electrolyte, it undergoes complete ionization as

NaOH Na+ + OH–

[NaOH] = [OH–] = 10–2 M

Molecular mass of NaOH = 40

Amount of NaOH dissolve per litre = 10–2 × 40

= 0.4 g



28. Calculate the degree of dissociation and concentration of H3O+ ions in 0.01 M solution of formic acid,

Ka = 2.1 × 10–6 at 298 K.

Sol. Formic acid is weak electrolyte and ionizes in water to give H3O+ ions according to the equation.

HCOOH + H2O H

3O+ + HCOO–

0.01 0 0

0.01(1 – ) � 0.01 0.01 0.01

2a

0.01 0.01K 0.01

0.01

6

2 2.1 10

0.01

= 0.0145

[H3O+] = × c = 0.0145 × 0.01 = 1.45 × 10–4 mol L–1

29. For the reaction, N2(g) + 3H

2(g) 2NH

3(g), at 400 K, K

p = 41. Find the value of K

p for each of the

following reactions at the same temperature.

(i) 2NH3(g) N

2(g) + 3H

2(g)

(ii)1

2N2(g) +

3

2H2(g) NH

3(g)

(iii) 2N2(g) + 6H

2(g) 4NH

3(g)

Sol. (i) It is the reverse of the given reaction. Hence, Kp =

1

41.

(ii) It is obtained by dividing the given equation by 2. Hence, Kp = 41 .

(iii) It is obtained by multiplying the given equation by 2. Hence, Kp = (41)2.

30. AB2 dissociates as

AB2(g) AB(g) + B(g)

If the initial pressure is 500 mm of Hg and the total pressure at equilibrium is 700 mm of Hg. Calculate Kp

for the reaction.

Sol. After dissociation, suppose the decrease in the pressure of AB2 at equilibrium is p mm. Then

AB2 AB + B

500 mm 0 0

500 – p p p

Total pressure at equilibrium = 500 – p + p + p

= 500 + p mm = 700

p = 200 mm



Hence, at equilibrium pAB2

= 500 – 200 = 300 mm

pAB

= 200 mm

pB = 200 mm

2

AB Bp

AB

p p 200 200K 133.3 mm

p 300

Long Answer Type Questions :

31. Classify each of the following substances into an acid or base and mention the concept on the basis of which

you can do so.

(i) HCl (ii) NH3

(iii) CO32– (iv) CH

3COOH (v) CO

2

(vi) BF3

(vii) AlCl3

(viii) CN– (ix) H2O (x) SiF

4

Sol. (i) HCl Acid (Arrhenius concept and Bronsted-Lowry concept)

(ii) NH3 Base (Bronsted concept and Lewis concept)

(iii) CO32– Base (Bronsted concept)

(iv) CH3COOH Acid (Arrhenius and Bronsted concept)

(v) CO2 Acid (Bronsted and Lewis concept)

(vi) BF3 Acid Lewis concept

(vii) AlCl3 Acid

(viii) CN– Base (Lewis concept and Bronsted concept)

(ix) H2O Amphoteric (Bronsted concept)

(x) SiF4 Acid (Lewis concept)

32. What is Kc and K

p? Derive a relation between them.

Sol. aA + bB cC + dD

c d c d (c d)C D

p a b a b (a b)A B

(p )(p ) [C] [D] (RT)K

(p )(p ) [A] [B] (RT)

c d(c d) (a b)

a b

[C] [D](RT)

[A] [B]

c d

n n

ca b

[C] [D](RT) K (RT)

[A] [B]

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 Kp, pressure should be expressed in bar)

1 pascal, Pa = 1 Nm–2, and 1 bar = 105 Pa



33. Briefly explain the dynamic nature of chemical equilibrium.

Sol. We have studied the equilibria in the different types of physical systems. In a similar way, equilibrium can also

be achieved in the chemical systems involving reversible chemical reactions carried in closed containers.

For example :

CaCO3(s) CaO(s) + CO2(g)

H2(g) + I

2(g) 2HI(g)

N2O4(g) 2NO

2(g)

PCl5(g) PCl3(g) + Cl2(g)

N2(g) + 3H

2(g) 2NH

3(g)

In above chemical reaction, the sign ( ) is for a reversible reaction. It may be noted that equilibrium can

be obtained only in the reversible reactions but not in the irreversible reactions in which the reactions cannot

proceed in the backward direction even if they are carried in closed containers.

For example :

2KClO3(s) 2KCl(s) + 3O

2(g)

Now, let us consider a general case of a reversible reaction carried in a closed container and how equilibrium

is attained.

A + B C + D

Products

Reactants

Time Equilibrium

Concentration

Change in concentration with time

State ofequilibrium

With passage of time there is accumulation of the products C and D, and depletion of the reactants A and

B. This leads to a decrease in the rate of forward reaction and an increase in the rate of the reverse reaction.

A stage will be ultimately reached when their concentration becomes constant i.e., there will be no further

change in concentration of either of the reactants A and B or of the products C and D. This represents a state

of equilibrium. At this point, the forward and the backward reactions will proceed at the same rate.

We have studied that an equilibrium is achieved in a reversible chemical reaction carried in a closed container

when the speeds of the forward and backward reactions become equal. Now, a question strikes our minds

as to what is the state of these reactions at the equilibrium. Have they come to a stop or they are still

continuing. Actually, both the forward and the backward reactions are still taking place even after the equilibrium

is achieved but with the same speed. In other words, the equilibrium is of dynamic nature.



The dynamic nature of the chemical equilibrium can be demonstrated with the help of the reversible reaction

involving the formation of ammonia in Haber’s process.

N2(g) + 3H

2(g) 2NH

3(g)

In this reaction in a state of equilibrium it is possible to determine the amount of N2 and H

2 left at some definite

intervals and also the amount of NH3 which is formed. The molar concentrations can also be determined. It

is found that they remain the same. This means that both the forward and backward reactions are still taking

place even at equilibrium or the equilibrium is of dynamic nature.

This can also be shown graphically as follows

Hydrogen (Reactant)

Nitrogen (Reactant)

Ammonia (Product)

Time

Mola

r concen

tration

Equilibrium in the Haber's process

The dynamic nature of chemical equilibrium can be illustrated by another reversible reaction.

H2(g) + I

2(g) 2HI(g)

We have stated earlier that the equilibrium is achieved in reaction when carried in a closed container at

773 K. To demonstrate the dynamic nature of equilibrium, add a small amount of radioactive iodine (I2*) to

the equilibrium mixture. After some time, HI is also found to be radioactive (HI*). This clearly shows that both

the forward and the reverse reactions are taking place even at the equilibrium point or the chemical equilibrium

is of dynamic nature.

34. Briefly explain the important characteristics of chemical equilibrium.

Sol. Characteristics of chemical equilibrium

A few important characteristics of chemical equilibrium are:

(i) At equilibrium both the forward and the backward reactions proceed at the same rate and hence, the

equilibrium is dynamic in nature.

(ii) At equilibrium all macroscopic properties such as pressure, concentration, density and colour etc. of the

system become constant and remain unchanged thereafter.

(iii) A chemical equilibrium can be established only when none of the products is allowed to escape out.

(iv) Chemical equilibrium can be approached from either direction.

Consider the equilibrium

2 4 2(Colourless) (Reddishbrown)

N O 2NO��⇀↽��

(v) A catalyst does not affect the position of chemical equilibrium, as it affects the forward and the backward

reactions equally. So, a catalyst only helps in attaining the equilibrium earlier.



35. What are Homogeneous and Heterogeneous equilibria? Give three examples of each.

Sol. Homogeneous equilibria :

When in an equilibrium reaction, all the reactants and the products are present in the same phase, it is called

a homogeneous equilibrium.

Example :

H2(g) + I

2(g) 2HI(g)

CO(g) + H2O(g) CO

2(g) + H

2(g)

PCl5(g) PCl

3(g) + Cl

2(g)

Heterogeneous equilibria :

When in an equilibrium reaction, the reactants and the products are presents in two or more than two phases,

it is called a heterogeneous equilibrium.

Example :

CaCO3(s) CaO(s) + CO2(g)

3Fe(s) + 4H2O(g) Fe

3O4(s) + 4H

2(g)

C(s) + H2O(g) CO(g) + 2

Water gas

H (g)

36. Prove that the pressure necessary to obtain 50% dissociation of PCl5 at 500 K is numerically equal to three

times the value of the equilibrium constant Kp.

Sol. PCl5 PCl

3 + Cl

2

1 0 0

1 – 0.5 = 0.5 0.5 0.5

Total number of moles = 1 – 0.5 + 0.5 + 0.5 = 1.5 mole.

If p is the total required pressure, then

5 3(PCl ) (PCl )

0.5 p 0.5 pp p , p p

1.5 3 1.5 3

2(Cl )

0.5 pp p

1.5 3

3 2

5

PCl Cl

pPCl

p pp p p3 3K

pp 3

3

pp 3 K



37. What are strong and weak electrolytes and derive how the degree of ionization is related to the concentration

of the solution of the electrolyte?

Sol. As strong electrolytes are completely ionised in the aqueous solution, therefore, their ionisation is represented

by putting a single arrow pointing towards right. e.g.,

HCl + H2O H

3O+ + Cl–

NaOH(aq) Na+(aq) + OH–(aq)

In case of weak electrolytes, as they are partly ionized, an equilibrium is set up between the ions and the

unionized electrolyte. Hence, their ionization is represented by putting double arrows ( ) in between e.g.,

CH3COOH + H

2O CH

3COO– + H

3O+

NH4OH + aq NH+

4(aq) + OH–(aq)

In general, the ionisation of a weak electrolyte, AB, is represented as follows :

AB(s) + aq A+(aq) + B–(aq)

Such an equilibrium is called ionic equilibrium between the ions and the undissociated electrolyte.

Applying the law of chemical equilibrium to the above equilibrium, we get

[A ][B ]

[AB]

+ −

= Ki called ionisation constant

Ionization constants of Weak Acids and Weak Bases

The dissociation of weak acids or weak bases in water can be represented as an equilibrium process. For

example

CH3COOH + H

2O CH

3COO–(aq) + H

3O+(aq)

NH3 + H

2O NH

4

+(aq) + OH–(aq) etc.

In general, if a weak acid is represented by HA, its dissociation in water can be represented by the equilibrium

HA + H2O A–(aq) + H

3O+(aq)

Applying the law of chemical equilibrium, the expression for equilibrium constant will be

3

2

[A ][H O ]K

[HA][H O]

As H2O is solvent, its concentration is large and remains almost constant. We put k[H

2O] = K

a called

dissociation constant of the acid. Thus,

3

a

[A ][H O ]K

[HA]

Knowing the value of the dissociation constant of the acid, Ka and the concentration (c) of the weak acid HA

taken, the concentration of H3O+ or H+(aq) in the solution can be calculated as follows



HA + H2O H

3O+ + A–

Initial concentration c 0 0

Concentration at equilibrium (c – c) c c

c(1 – )

2 2

3

a

[H O ][A ] c c cK

[HA] c(1 ) c(1 )

( is degree of dissociation)

As usually x is very small as compared to the initial concentration c, x can be neglected in comparison to

c. Hence, the equation is simplified to the form

Ka = 2c

1/2

aK

c

⎛ ⎞ ⎜ ⎟⎝ ⎠

1/2

a(K V)

Here V is the volume of the solution in litres containing 1 mole of the electrolyte, c = 1

V.

In case of a weak electrolyte at a given temperature, the degree of ionisation is inversely proportional to the

square root of the molar concentration or directly proportional to square root of the volume of the solution which

contains one mole electrolyte.

38. What are acids and bases according to Bronsted-Lowry concept? Explain with example.

Sol. The Brönsted-Lowry Acids and Bases :

Bronsted (a Danish chemist) and Lowry (an English chemist) proposed this concept simultaneously and

independently in 1923. According to it, “An acid is a substance (molecule or ion) which has a tendency to

donate proton and base is a substance (molecule or ion) which has a tendency to accept proton.”

Thus, an acid is a proton donor species and a base is a proton acceptor species.

H+

Acid(Proton donor)

Base(Proton acceptor)

Acid

HCl H Cl

3 4

Base

NH H NH

HClAcid

NH3

Base

H+

��⇀↽��3 4

Acid Base

HCl NH NH Cl



Here HCl is an acid as it donates a H+ while NH3 is a base since it accepts a proton. An acid on losing a

proton produce a species which have the tendency of a base which is called conjugate base of the acid.

HAcid Conjugate base

H3 3Acid Conjugate base

CH COOH CH COO

These acid-base pair is known as conjugate pair. Similarly

HBase Conjugate acid

A conjugate pair of acid and a base differs by a proton only i.e.,

H

H

Conjugate acid Conjugate base��⇀↽��

If Bronsted acid is a strong acid then its conjugate base is a weak base and vice-versa.

CH COOH + H O3 2

Acid Base

CH COO + H O3

–

3

+

Conjugatebase

Conjugateacid

Conjugate pair

Conjugate pair

39. Derive expression for the calculation of pH of a salt of strong acid and weak base.

Sol. Relation between hydrolysis constant and dissociation constant and pH :

Salts of strong acid and weak base : It is known as cationic hydrolysis and proceeds as follows

B+ + H2O ��⇀↽�� BOH + H+

Now,h

[BOH][H ]k

[B ]

For the dissociation of weak base,

BOH ��⇀↽�� B+ + OH–

b

[B ][OH ]k

[BOH]

Ionic product (kw) for water is

kw = [H+][OH–]



Now, h b

w

k k [BOH][H ] [B ][OH ] 11

k [BOH][B ] [H ][OH ]

w

h

b

kk

k

B+ + H2O ��⇀↽�� BOH + H+

Initial concentration c 0 0

Concentration at equilibrium c(1 – h) ch ch

Now, 2

h

H ][BOH] ch ch chk

c(1 h) 1 h[B ]

kh = ch2 (1 – h = 1 for weak electrolyte)

hk

hc

w

b

kh

k c

1/2

w w

b b

k k c[H ] ch c

k c k

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

pH = –log[H+] = –log

1/2

w

b

k c

k

⎡ ⎤⎢ ⎥⎣ ⎦

pH = 1

2 [logk

w – logk

b + logc]

b

1pH 7 [pk logc] at 298 K

2

40. What is a Buffer solution? Discuss the buffer action.

Sol. 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 solution which resists a change in the pH value on the addition of a small amount of acid or base to it”.

Buffer Action : Let us consider the buffer action of the acidic buffer containing CH3COOH and CH

3COONa.

Acetic acid dissociates to a small extent where as sodium acetate is almost completely dissociated in the

aqueous solution as follows

CH COOH + H O3 2

CH COO + H O3

–

3

+

CH COONa3

CH COO + Na3

– +

By ‘common ion effect’, the ionisation of CH3COOH is further suppressed. Thus, in the solution, there are

excess of acetate (CH3COO–) ions and a small amount of H

3O+ ions.

When a few drops of an acid are added to the above mixture solution, the H3O+ ions given by the acid combine

with the CH3COO– ions to form weakly ionized molecules of CH

3COOH.

CH3COO– + H

3O+ CH

3COOH + H

2O–

Thus, the H3O+ ion concentration and hence the pH of the solution remains almost constant.

Similarly, when a few drops of a base are added, the OH– ions given by the base combine with the H3O+ ions

already present to form weakly ionized molecules of H2O.

H3O+ + OH– 2H

2O

(Given by

the base)

As the H3O+ ions are consumed, the equilibrium of acetic acid shifts towards right (according to Le Chatelier’s

principle). Thus, more of CH3COOH dissociates to make up the loss of H

3O+ ions. Hence, the H

3O+ ion

concentration or the pH of the solution does not change.

41. Write a relation between G and Q and define the meaning of each term and answer the following

(i) Why a reaction proceeds forward when Q < K?

(ii) And no net reaction occurs when Q = K.

Sol. G = G° + RT lnQ

Where, G = change in free energy as the reaction proceeds.

G° = Standard free energy change

Q = Reaction quotient

R = Gas constant

T = Temperature in °K

As, G° = –RT lnk

G = –RT lnK + RT lnQ = RT lnQ

K

If Q < K, G will be negative, reaction will proceed in the forward direction.

If Q = K, G = 0, reaction is in equilibrium and there is no net reaction.



42. Explain the characteristics of chemical equilibrium constant.

Sol. Characteristics of equilibrium constant

(i) The equilibrium constant does not depend upon initial concentration and has a definite value for every

chemical reaction at a given temperature and pressure.

(ii) If the reaction is reversed, the value of equilibrium constant is inverted.

Example :

For reaction, H2(g) + I

2(g) ��⇀↽�� 2HI(g)

Kc = 50

If reaction is reversed

2HI(g) ��⇀↽�� H2(g) + I

2(g)

The value of the equilibrium constant will be

2 2c

c

[H ][I ]1 1K 0.02

K [HI] 50

(iii) If the reaction is multiplied by a factor then the value of equilibrium constant becomes Kc = (K

c)n

Example :

��⇀↽��A B 2C

��⇀↽��2A 2B 4C

Kc = (K

c)2

(iv) The pure solids and liquids are not considered for calculation of Kc because they have constant

concentration.

43. What are law of chemical equilibrium? Explain it and derive the equation of equilibrium.

Sol. The law of chemical equilibrium is simply a mathematical expression which can be derived by the application

of the law of mass action to a reversible reaction.

Consider the following reversible reaction, taking place at constant temperature

A + B C + D

According to law of mass action

Rate of the forward reaction [A][B]

= K1[A][B]

where [A] and [B] are the active masses of A and B respectively and K1 is the proportionality constant, known

as the rate constant for the forward reaction.



Similarly,

Rate of the backward reaction [C][D]

= K2[C][D]

Where [C] and [D] are the active masses of C and D respectively and K2 is the rate constant for the backward

reaction.

At equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction.

K1[A][B] = K

2[C][D]

or 1c

2

K [C][D]K

K [A][B]

Where Kc is called the equilibrium constant for a general reaction.

aA + bB cC + dD

The equilibrium constant may be given by

c d

1

c a b2

K [C] [D]K

K [A] [B]

The above relationship is the mathematical expression for the law of chemical equilibrium. Thus, the law of

the chemical equilibrium may be defined as

‘At equilibrium in a reversible reaction, the product of the molar concentration of products, each raised to the

power equal to its coefficient, divided by the product of the molar concentration of the reactant each raised

to the power equal to its coefficient, is constant at a constant temperature and is called equilibrium constant’.

44. Derive the relation between equilibrium constant (K), reaction quotient (Q) and Gibb’s free energy (G).

Sol. Relationship between Equilibrium constant (K), Reaction Quotient (Q) and Gibbs Energy (G) :

The value of the equilibrium constant (Kc) for a reaction does not depend upon the rate of the reaction at

equilibrium point. However, it is related to G known as Gibbs free energy change.

If G = 0; the reaction is in a state of equilibrium.

If G is negative, the reaction is spontaneous in the forward direction.

If G is positive, the reaction is non-spontaneous in forward direction or the reaction is spontaneous in

the reverse direction.

The relation between free energy change and reaction quotient (Q) may be given as

G G RT lnQ

If the reaction has attained the equilibrium, under the reaction conditions, then G = 0



Thus,

0 = G° + RT lnK

G° = –RT lnK

G 2.303 RT logK

The equation may also be expressed as

G /RTK e

Hence, using the above equation, the reaction spontaneity can be interpreted in terms of the value of G°.

If G° < 0, then –G°/RT is positive and e–G°/RT > 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 G° > 0, then –G°/RT is negative and e–G°/RT < 1, i.e., 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.

45. The ionisation constant of formic acid (HCOOH) is 1.8 × 10–5. Calculate the ratio of sodium formate and formic

acid in a buffer of pH = 4.91.

Sol. pH = pKa + log

[Salt]

[Acid] i.e., 4.91 = 4.76 + log

[Salt]

[Acid]

or[Salt]

log 0.15[Acid]

[Salt]Antilog 0.15 = 1.41

[Acid]

SECTION - B

Model Test Paper

Very Short Answer Type Questions :

1. The following system is in equilibrium

SO2Cl2 + Heat SO2 + Cl2

What will happen to the system if some Cl2 is added into it? Give reason.

Sol. With increase in the concentration of Cl2 or product the equilibrium shift towards left/reactant/backward side

to minimize the change.

2. What is meant by buffer solution? Give one example of an acidic buffer.

Sol. The solution which resist the pH change by the addition of small amount of acid/base is called buffer solution.

Acid buffer : CH3COOH + CH

3COONa

Weak acid Salt of

strong base




3. Out of CH3COO– and OH– which is a stronger base and why?

Sol. The conjugate acid of CH3COO– and OH– is CH

3COOH and H

2O. As CH

3COOH is stronger acid than H

2O,

so OH– is stronger base.

4. Consider the following equilibrium

CO2(g) + C(graphite) 2CO(g)

Write the equilibrium expression for Kc and calculate its units.

Sol.2

c

2

[CO]K

[CO ]

units = mol L–1

5. Write the conjugate bases for the following acids.

(i) HF

(ii) HCO3–

(iii) CH NH33

(iv) NH3

Sol. (i) F–

(ii) CO3–2

(iii) CH3NH

2

(iv) NH2–


6. Calculate the pH of a solution obtained by mixing 300 ml of 0.2 M HCl with 400 ml of 0.1 M NaOH solution.

Sol. HCl

0.2 × 0.3

= 0.06 moles

+ NaOH 0.1 × 0.4

= 0.04 moles

NaCl +H O2

0.04 moles– 0.04 moles 0

0 0

At (t = 0),

At (t ),

Left HCl = 0.02 moles the V = 700 molTotal

0.04 moles

Final Molarity of (H+) = 0.02

0.03 M0.7

pH = –log 0.03

= 2 – log3

= 1.5229

7. PCl5, PCl

3 and Cl

2 are at equilibrium at 500 K and having concentration 1.59 M PCl

3, 1.59 M Cl

2 and

1.41 M PCl5. Calculate Kc for the reaction

PCl5 PCl

3 + Cl

2



Sol. 3 2

c

5

[PCl ][Cl ]K

[PCl ]

2(1.59)1.79

1.41

8. What is the expression for Ksp

of Ag2CrO

4?

Sol. Ag2CrO

4 2Ag+ + CrO

4–

1 0 0

1 – S 2S S

Ksp

= [Ag+]2 [CrO4–]

(2S)2 (S)

Ksp

= 4S3

9. A B; K1 = 1

B C; K2 = 2

C D; K3 = 3

D E; K4 = 4

What is the value of K for A E?

Sol. A B K1 = 1

B C K2 = 2

C D K3 = 3

D E K4 = 4

Value for A K

��⇀↽�� E

Here K = K1 × K

2 × K

3 × K

4

= 1 × 2 × 3 × 4

= 24

Long Answer Type Questions :

10. If Kw = 49 × 10–14, what will be the neutral pH value of H

2O?

Sol. Kw = [H+] [OH–]

Kw = [H+]2 for neutral solution [H+] = [OH–]

Kw = 49 × 10–14 = [H+]2

[H+] = 7 × 10–7

pH = –log[H+]

= 7 – log 7

= 7 – 0.84

= 6.16



11. State Le-Chatelier’s principle. Discuss the effect of increase in temperature and pressure on the following

equation.

2SO(g) + O2(g) 2SO

2(g) + Heat

Sol. Le Chatelier’s principle : If a system in equilibrium is subjected to a change of concentration, temperature

or pressure, the equilibrium shifts in a direction that tends to undo the effect of the change imposed.

Effect of pressure :

2SO + O2 2SO

2 + Heat

with increasing pressure, equilibrium shifts towards forward or right.

Effect of temperature :

for exothermic reaction

with increasing temperature equilibrium shifts towards reactant or left or backward.

12. (i) How does Lewis theory explain the acidic character of CO2?

(ii) Select the Lewis acid and Lewis base in

H2PO

4–, SO

3–2, ClO–, Fe+3, BCl

3, NH

4+

Sol. (i) According to Lewis concept, CO2 is Lewis acid because carbon atom attached to highly electronegative

element oxygen and form double bonds. It can easily accept an electron pair from the Lewis base such

as OH– ion.

OH + O–

C O

Lewis

base

Lewis

acid

C O–

O

OH

or

HCO3

–

Bicarbonate ion

(ii) Lewis acid = Fe+3, BCl3

Lewis base = H2PO

4–, SO

3–2, ClO–

13. The solubility product constants of Ag2CrO

4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate

the ratio of the molarities of their saturated solutions.

Sol. Ag2CrO4(s) 2Ag+(aq) + CrO4–2(aq)

0 0

2S S

Ksp

= [Ag+]2 [CrO4–2] = (2S)2 × S = 4S3



12sp 4 133

K 1.1 10S 0.65 10 mol L

4 4

AgBr(s) Ag+(aq) + Br–(aq)

0 0

S S

Ksp

= [Ag+] [Br –] = S2

spS K = (5.0 × 10–13)1/2 = 0.707 × 10–6 mol L–1

2 4

4(Ag CrO )

6(AgBr )

S 0.65 1091.9

S 0.707 10

� � �



Objective Type Questions

(Physical and Chemical Equilibrium, Expression for Kc, K

p and applications of Equilibrium constant)

1. The Kc for given reaction will be

2A (g) 2B(g) C(g) 2D(s) ��⇀

↽��

(1)2

2

2

[C][D]K

[A ][B]

(2)2

2

[C]K

[A ][B]

(3)2

2

2

[A ][B]K

[C][D]

(4)2

2[A ][B]

K[C]

Sol. Answer (2)

2

c 22

[C] [D]K

[A ] [B]

∵ D is solid, its concentration is taken unity

c 2

2

[C]K

[A ] [B]

2. For which of the following reaction, the degree of dissociation () and equilibrium constant (Kp) are related as

2

p 2

4 PK

(1 )

?

(1) N2O4(g) 2NO2(g)

(2) H2(g) + I2(g) 2HI(g)

(3) N2(g) + 3H2(g) 2NH3(g)

(4) PCl3(g) + Cl2(g) PCl5(g)

Sol. Answer (1)

The first reaction is a decomposition reaction for which

2

p 2

4 PK

1

; Where P is total pressure at equilibrium and is degree of dissociation of N

2O4.

Solutions (Set-2)



3. In which of the following does the reaction go almost to completion?

(1) Kc = 103 (2) Kc = 102 (3) Kc = 10–2 (4) Kc = 10–3

Sol. Answer (1)

Highest value of Kc means reactions maximum approaches completion.

4. In a chemical equilibrium the rate constant of the backward reaction is 7.5 × 10–4 and the equilibrium constant

is 1.5. So the rate constant of the forward reaction is

(1) 2 × 10–3 (2) 15 × 10–4 (3) 1.125 × 10–3 (4) 9.0 × 10–4

Sol. Answer (3)

We know,

f

b

KK

K

or, Kf = K × K

b

= 1.5 × 7.5 × 10–4

= 1.125 × 10–3

5. Kp is how many times equal to Kc for the given reaction?

N2(g) + 3H2(g) 2NH3(g)

(1)22

TR

1(2) R2T2 (3)

T

R(4) RT

Sol. Answer (1)

We know,

n

p cK K RT

In the reaction,

n = 2 – (1 + 3) = – 2

Kp = (RT)–2 × K

c

= c2 2

1K

R T

6. 4 g H2, 32 g O2, 14 g N2 and 11g CO2 are taken in a bulb of 500 ml. Which one of these has maximum active

mass?

(1) H2 (2) O2 (3) N2 (4) CO2

Sol. Answer (1)

Active mass of H2 =

4

24

500

1000

Active mass of O2 =

32

322

500

1000



Active mass of N2 =

14

281

500

1000

Active mass of CO2 =

11

44 0.5500

1000

7. For reaction, 2A + B 2C, K = x. Equilibrium constant for C A + 1/2B

will be

(1) x (2)2

x(3)

x

1

(4) x

Sol. Answer (3)

Given reaction :

2A B 2C ; K x ��⇀↽��

1

A B C ; K x2

��⇀↽��

1 1

C A B ; K2 x

��⇀↽��

8. XY2 dissociates as,

XY2(g) XY(g) + Y(g)

Initial pressure of XY2 is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of

system to remain constant, the value of Kp is

(1) 50 (2) 100 (3) 20 (4) 400

Sol. Answer (2)

2XY g XY g Y g

t 0 600 mmHg 0 0

At equilibrium 600 P P P

��⇀↽��

Ptotal

= 600 – P + P + P = 600 + P

600 + P = 800

P = 200 mm Hg

2

pP 200 200

K 100600 P 600 200



9. The initial pressure of COCl2 is 1000 torr. The total pressure of the system becomes 1500 torr, when the

equilibrium COCl2(g) CO(g)+Cl2(g) is attained at constant temperature. The value of Kp of a reaction

(1) 1500 (2) 1000 (3) 2500 (4) 500

Sol. Answer (4)

2 2COCl CO Cl

t 0 1000 torr 0 0

At equilibrium 1000 P P P

��⇀↽��

Ptotal

= 1000 + P = 1500

P = 500 torr

2

pP 500 500

K 5001000 P 500

10. Hydrogen (a moles) and iodine (b moles) react to give 2x moles of the HI at equilibrium. The total number of

moles at equilibrium is

(1) a + b + 2x (2) (a – b) + (b – 2x)

(3) (a + b) (4) a + b – x

Sol. Answer (3)

2 2H l 2 Hl

t 0 a b 0

Equilibrium a x b x 2x

��⇀↽��

Total moles = a + b

11. When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when

two–third of the acid and alcohol are consumed. The equilibrium constant of the reaction will be

(1) 0.4 (2) 4 (3) 40 (4) 0.04

Sol. Answer (2)

2 5 3 3 2 5 2C H OH CH COOH CH COOC H H O

t 0 1 1 0 0

2 2 2 2Equilibrium 1 1

3 3 3 3

��⇀↽��

c

2 2

3 3K 4 [Considering 1L container]1 1

3 3

12. Two moles of N2 and two moles of H2 are taken in a closed vessel of 5 litres capacity and suitable conditions

are provided for the reaction. When the equilibrium is reached, it is found that a half mole of N2 is used up.

The equilibrium concentration of NH3 is

(1) 0.3 (2) 0.4 (3) 0.2 (4) 0.1



Sol. Answer (3)

2 2 3N 3H 2NH

t 0 2 2 0

Equilibrium 2 x 2 3x 2x

��⇀↽��

Given: x = 0.5

Number of moles of NH3 at equilibrium = 1

1 13

1NH mol L 0.2 mol L

5

⎡ ⎤⎣ ⎦

13. 1 mole of NO2 and 2 moles of CO are enclosed in a one litre vessel to attain the following equilibrium

NO2 + CO NO + CO2. It was estimated that at the equilibrium, 25% of initial amount of CO is consumed.

The equilibrium constant Kp is

(1) 1 (2) 1/2 (3) 1/4 (4) 1/3

Sol. Answer (4)

2 2NO CO NO CO

t 0 1 2 0 0

Equilibrium 1 x 2 x x x

��⇀↽��

Given : 25

x 2 0.5100

∵ Kp = K

c [n = 0]

2

2p c

2

[NO] [CO ] xK K

[NO ] [CO] (1 x) (2 x)

0.5 0.5 1

0.5 0.5 3

14. Two moles of NH3 gas are introduced into a previously evacuated one litre vessel in which it partially

dissociates at high temperature as 2NH3 (g) . N2 (g) + 3H2 (g). At equilibrium, one mole of NH3(g) remain.

The value of Kc is

(1) 3 (2) 27/16 (3) 3/2 (4) 27/64

Sol. Answer (2)

3 2 22NH N 3H

t 0 2 0 0

Equilibrium 2 2x x 3x

��⇀↽��

2 – 2x = 1

x = 0.5

3 3

c 2 2

x 3x 0.5 1.5 75 27K 0.75

100 162 2x 1



15. 4.0 moles of PCl5 dissociate at 760 K in a 2 litre flask, PCl5 (g) PCl3(g) + Cl2(g) at equilibrium. 0.8 mole

of Cl2 was present in the flask. The equilibrium constant would be

(1) 1.0 × 10–1 (2) 1.0 × 10–4 (3) 1.0 × 10–2 (4) 1.0 × 10–3

Sol. Answer (1)

5 3 2PCl PCl Cl

t 0 4 0 0

Equilibrium 4 x x x

4 x x xActive mass

2 2 2

��⇀↽��

2

c

x x

x2 2K

4 x 2 4 x

2

Given : x = 0.8 mole

1c

0.8 0.8K 10

8 1.6

16. When 3.00 mole of A and 1.00 mole of B are mixed in a 1.00 litre vessel, the following reaction takes place

A(g) + B(g) 2C(g)

The equilibrium mixture contains 0.5 mole of C. What is the value of equilibrium constant for the reaction?

(1) 0.12 (2) 6 (3) 1.5 (4) 3

Sol. Answer (1)

A B 2C

t 0 3 1 0

Equilibrium 3 x 1 x 2x

��⇀↽��

Given : 2x = 0.5

x = 0.25

2 2

c

2x 0.5K 0.12

3 x 1 x 2.75 0.75

17. At 700 K, the equilibrium constant, Kp, for the reaction 2SO3(g) 2SO2(g) + O2 (g) is 1.8 × 10–3 atm. The value

of Kc for the above reaction at the same temperature in moles per litre would be

(1) 1.1 × 10–7 (2) 3.1 × 10–5 (3) 6.2 × 10–7 (4) 9.3 × 10–7

Sol. Answer (2)

For the reaction,

n = 1

Kp = K

c (RT)n = K

c RT

3

p 5c 1 1

K 1.8 10 atmK 3.13 10

RT 0.082 L atmK mol 700 K



(Factors Affecting Equilibria: Le Chatelier's Principles)

18. Which one of the following equilibrium moves backward when pressure is applied?

(1) N2 (g) + 3H2 (g) 2NH3 (g) (2) N2 (g) + O2 (g) 2NO (g)

(3) Water Ice (4) I2 (g) I2 (s)

Sol. Answer (3)

When pressure is applied, boiling point is increased and freezing point is depressed.

Equilibrium shifts backward.

19. In melting of ice, which one of the conditions will be more favourable?

(1) High temperature and high pressure (2) Low temperature and low pressure

(3) Low temperature and high pressure (4) High temperature and low pressure

Sol. Answer (1)

For melting of ice, high temperature and high pressure would be favourable.

20. Given the reaction,

2X(g) + Y(g) Z(g) + 80 kcal

Which combination of pressure and temperature gives the highest yield of Z at equilibrium?

(1) 1000 atm and 500°C (2) 500 atm and 500°C

(3) 1000 atm and 100°C (4) 500 atm and 100°C

Sol. Answer (3)

2X g Y(g) Z g 80 kcal ��⇀↽��

Endothermic reaction

Low temperature high yield of Z.

Also,

high pressure increase in concentration

increase in yield of Z

1000 atm and 100°C

(Ionic Equilibrium in solution, Acids, Bases and Salts, pH calculations)

21. Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl Ka of acetic acid is 1.8 × 10–5

(1) 0.18% (2) 0.018% (3) 1.8% (4) 18%

Sol. Answer (2)

3 3CH COOH CH COO H

0.01 x x x 0.1

HCl H Cl

0.1 0.1

○

��⇀↽��

5a

x 0.1 xK 1.8 10

0.01 x

x is very low,



So,

5

x 0.1 x 0.1x 1.8 10

0.01 x 0.01

x = 1.8 × 10–6

Number of moles ionizedNow % ionization 100

Number of moles taken

= x

100 0.018%0.01

22. A 0.2 molar solution of formic acid is 3.2% ionised, its ionisation constant is

(1) 9.6 × 10–3 (2) 2.1 × 10–4 (3) 1.25 × 10–6 (4) 2.1 × 10–8

Sol. Answer (2)

–

HCOOH HCOO H

At t 0 0.2 0 0

0.2 x x x

○��⇀↽��

Given, 3.2

x 0.2100

– 2

i

[HCOO ] [H ] xK

[HCOOH] 0.2 x

○

=

2

4

3.2 0.2

1002.1 10

3.2 0.20.2

100

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

23. At 100°C, Kw = 10–12. pH of pure water at 100°C will be

(1) 7.0 (2) 6.0 (3) 8.0 (4) 12.0

Sol. Answer (2)

Given Kw = 10–12

[H+] [OH–1] = 10–12

[H+] = 10–6 [∵ Water is pure]

pH = – log10

[H+] = 6

24. A monoprotic acid in a 0.1 M solution ionises to 0.001%. Its ionisation constant is

(1) 1.0 × 10–3 (2) 1.0 × 10–6 (3) 1.0 × 10–8 (4) 1.0 × 10–11

Sol. Answer (4)

HA H A

At t 0 0.1 0 0

0.001 0.001 0.0010.1 0.1 0.1 0.1

100 100 100

��⇀↽��

6 6

11i 6

[H ] [A ] 10 10K 10

[HA] 0.1 10



25. When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. The solution is found to have

a hydroxide ion concentration of 1.34 × 10–3. The dissociation constant of ammonia is

(1) 1.8 × 10–5 (2) 1.6 × 10–6 (3) 1.34 × 10–3 (4) 1.8 × 10–4

Sol. Answer (1)

3 2 4

4 4

NH H O NH OH

0.1 0.1

NH OH NH OH

0.1 x x x

��⇀↽��

24

b4

[NH ] [OH ] xK

[NH OH] 0.1 x

Given : x = 1.34 × 10–3

23

5b

1.34 10

K 1.79 100.1 x

26. A solution of NaOH contain 0.04 gm of NaOH per litre. Its pH is

(1) 10 (2) 9 (3) 11 (4) 12

Sol. Answer (3)

0.04 g of NaOH

3

1

0.04 g10 moles of NaOH

40 g mol

3

3 3

NaOH Na OH

10 0 0

0 10 10

[OH–]= 10–3

p[OH] = 3

pH = 14 – 3 = 11

27. 1 c.c of 0.1 N HCl is added to 1 litre solution of sodium chloride. The pH of the resulting solution will be

(1) 7 (2) 0 (3) 10 (4) 4

Sol. Answer (4)

0.1 N HCl 0.1 mole of HCl in 1 L solution

0.1 × 10–3 moles of HCl in 1 c.c. (1 mL) solution

pH = – log10

[0.1 × 10–3] = 4

28. 100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to

1 litre. The pH of the resulting solution will be

(1) 1 (2) 2 (3) 3 (4) 4



Sol. Answer (2)

We know, for an aicd base mixture,

N3(V

1 + V

2) = |N

1V

1 – N

2V

2|

Where, N3(V

1 + V

2) = Number of gram equivalents of HCl (∵ Acid is in excess)

Now,

g. equivalents of HCl = | 0.1 × 0.1 – 0.1 × 0.2 | = 0.01

Number of g equivalentsNumber of moles =

n factor of HCl

0.010.01

1

HCl H Cl

0.01 0.01 0.01

○

pH = – log10

[H+] = 2

29. The pH of a solution is zero. The solution is

(1) Neutral (2) Normal acid (3) Decinormal acid (4) Strongly alkaline

Sol. Answer (2)

pH of solution is 0

[H+] = 1 M

Normal acid

30. 100 ml of 0.1 N NaOH is mixed with 100 ml of 0.1 N H2SO4. The pH of the resultant solution is

(1) < 7 (2) > 7

(3) = 7 (4) Cannot be predicted

Sol. Answer (3)

∵ Number of g equivalents of acid

= Number of g equivalents of base

pH = 7 and solution is neutral

31. The pH of 0.016 M NaOH solution is

(1) 1.796 (2) 12.204 (3) 11 (4) None of these

Sol. Answer (2)

0.016 M NaOH

[OH–] = 0.016

pOH = – log10

(0.016)

= – log10

(16 × 10–3)

= 3 – log10

16

= 3 – 4 log 2

= 3 – 4 × 0.3

= 3 – 12 = 1.8

pH = 14 – 1.8 = 12.2



32. pH of 1 M HCl is

(1) Zero (2) –2 (3) 7 (4) 14

Sol. Answer (1)

1 M HCl

[H+] = 1 M

pH = – log10

1 = 0

33. For a acid 'A' pH = 2 and for acid 'B' pH is 4. Then

(1) A is more basic than B (2) B is more acidic than A

(3) A is more acidic than B (4) B is more basic than A

Sol. Answer (3)

Lower pH stronger acid (more [H+] ion concentration)

34. The following reactions are known to occur in the body

CO2 + H

2O H

2CO

3 H+ + HCO

3–

If CO2 escapes from the system

(1) pH will decrease (2) Hydrogen ion concentration will diminish

(3) H2CO

3 concentration will be promoted (4) The forward reaction will be promoted

Sol. Answer (2)

If CO2 decreases

H2CO

3 concentration decreases

H+ concentration decreases

(Buffer solution)

35. A buffer solution can be prepared from a mixture of

I. Sodium acetate and acetic acid in water

II. Sodium chloride and HCl in water

III. Ammonia and NH4Cl in water

IV. Ammonia and sodium hydroxide in water

(1) I, III, IV (2) II, III, IV (3) I, II, IV (4) I, IV

Sol. Answer (4)

Buffer solution can be prepared by

3 3CH COO Na and CH COOH

Weak acid + Strong base Weak acid

AND



4 4NH OH and NH Cl

Weak base Weak base + Strong acid

Where NH4OH is formed as follows

3 2 4NH H O NH OH

(Hydrolysis of Salts)

36. The addition of solid sodium carbonate to pure water causes

(1) An increase in the hydronium ion concentration

(2) An increase in pH

(3) No change in pH

(4) A decrease in the hydroxide ion concentration

Sol. Answer (2)

2 3 2 2 3Na CO H O NaOH H CO ��⇀↽��

Now, NaOH is a strong electrolyte and dissociates completely

NaOH Na OH

On the other hand, H2CO3 is weak and dissociates partially

22 3 3H CO 2H CO

��⇀↽��

[OH–] > [H+]

Basic solution (increase in pH)

37. A salt of strong acid and weak base is dissolved in water. Its hydrolysis in solution is

(1) Unaffected on heating (2) Increased by adding strong acid

(3) Suppressed by diluting (4) Suppressed by adding strong acid

Sol. Answer (4)

Strong acid + weak base Acidic salt

Acidic salt upon hydrolysis gives out H+. Hence adding more H+ results in increase in [H+] and hence shiftof equilibrium in backward direction.

38. Which of the following salts undergoes hydrolysis?

(1) CH3COONa (2) KNO3 (3) NaCl (4) K2SO4

Sol. Answer (1)

Salt containing at least one conjugate acid or base of a weak acid or base undergoes hydrolysis to give

corresponding acids or base. (e.g., 3CH COO Na

)

Strong electrolytes do not get hydrolysed to form acids and bases. (e.g., NaCl, KNO3)



39. Which will undergo cationic hydrolysis?

(1) NaCl (2) CH3COONa (3) (NH4)2SO4 (4) H2CO3

Sol. Answer (3)

Acidic salt undergoes cationic hydrolysis 4 42NH SO

40. pH of a salt of a strong base with weak acid

(1) Clog2

1pK

2

1pK

2

1pH

aw (2) Clog

2

1pK

2

1pK

2

1pH

aw

(3) Clog2

1pK

2

1pK

2

1pH

aw (4) None of these

Sol. Answer (1)

For salt of a strong base with weak acid,

w a

1 1 1pH pK pK log C

2 2 2

a

1 17 pK log C

2 2

41. An example of a salt dissolved in water to give acidic solution is

(1) Ammonium chloride (2) Sodium acetate (3) Potassium nitrate (4) Barium bromide

Sol. Answer (1)

An acidic salt gives acidic solution NH4Cl is made up of weak base and strong acid, hence gives acidic

solution.

(Solubility and Solubility Product)

42. Solubility product principle can be applied when

(1) A solid is insoluble in a liquid

(2) A liquid is insoluble in another liquid

(3) Any ionic compound is sparingly soluble in a liquid

(4) Substance is ionic

Sol. Answer (3)

Solubility product principle can only be applied for saturated solutions of sparingly soluble salts.

43. The solubility product of AgCl is Ksp. Then the solubility of AgCl in xM KCl is

(1) Ksp × x2 (2)spK

x(3) 2

sp

x

K

(4)x

Ksp

Sol. Answer (4)

AgCl(s) Ag (aq) Cl (aq)

0 0

S (S x) [x due to KCl]

��⇀↽��

S(S + x) = Ksp

or, S2 + xS = Ksp

∵ S2 is very small, it is neglected

xS = Ksp

spKS

x



44. The correct representation for the Ksp of SnS2 is

(1) [Sn2+][S2–]2 (2) [Sn4+][S–2]2 (3) [Sn2+][2S–2] (4) [Sn4+][2S2–]2

Sol. Answer (2)

4 2–2SnS (s) Sn (aq) 2S (aq) ��⇀

↽��

Ksp

= [Sn4+] [S2–]2

45. The Ksp for a sparingly soluble Ag2CrO4 is 4 × 10–12. The molar solubility of the salt is

(1) 2.0 × 10–6 mol L–1 (2) 1.0 × 10–4 mol L–1

(3) 2.0 × 10–12 mol L–1 (4) 1.0 × 10–15 mol L–1

Sol. Answer (2)

2–

2 4 4Ag CrO (s) 2 Ag (aq) CrO (aq)

0 0

2 S S

��⇀↽��

Ksp

= (2S)2S = 4 S3

or, 4 S3 = Ksp

= 4 × 10–12

S3 = 10–12

S = 10–4 mol L–1

46. Precipitation occurs only if I.P (Ionic Product)

(1) Equals KSP (2) Exceeds KSP (3) Less than KSP (4) Is very small

Sol. Answer (2)

Precipitation occurs only when Ionic product > solubility product due to which the reaction shifts backward

and solid precipitate is obtained.

47. The precipitate of CaF2 (Ksp = 1.7 × 10–10) is obtained when equal volumes of the following are mixed

(1) 10–4 M Ca2+ + 10–4 M F– (2) 10–2 M Ca2+ + 10–3 M F–

(3) 10–4 M Ca2+ + 10–3 M F– (4) 10–3 M Ca2+ + 10–5 M F–

Sol. Answer (2)

On mixing equal volumes, concentrations become half.

2 –

2Ca (aq) 2F (aq) CaF (s) ��⇀

↽��

For 10–2 M Ca2+ and 10–3 M F–

Ionic product

2–2 –3

–1010 1012.5 10

2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Ionic product > Solubility product

Precipitation occurs

� � �

L1 Ch-07 Sol Set-1 Equilibrium

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