NCERT Solutions for Class 11 Chemistry Chapter 7: Equilibrium

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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?

Ans. (a) When the volume of the container is suddenly increased, the vapour pressure would decrease initially because the amount of vapour remains the same, but the volume increases suddenly. Hence, the same amount of vapour is distributed in a larger volume.

(b) As the temperature is fixed, the rate of evaporation also remains constant but the rate of condensation decreases initially. The reason for this is that as the volume of the container is increased, the density of the vapour phase decreases. This decreases the rate of collision of the vapour particles. Hence, the rate of condensation decreases initially.

(c) As equilibrium is restored, the rate of evaporation becomes equal to the rate of condensation. As the temperature remains constant, the vapour pressure which varies according to change in temperature (and not on volume), will be equal to the original vapour pressure of the system. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

7.2. What is Kc for the following equilibrium when the equilibrium concentration of each substance is:

[SO₂] = 0.60M, [O₂] = 0.82M and [SO₃] = 1.90M ?

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Ans. The equilibrium constant (K_c) for the given reaction is:

$$\text{k}_c=\frac{[\text{SO}_3]^{2}}{[\text{SO}_2]^{2}[\text{O}_2]}\\=\frac{(1.90)^{2}\text{M}^{2}}{(0.60)^{2}(0.82)\text{M}^{3}}$$

= 12.23 M^–1 (approx)

Therefore, K_c for the equilibrium is 12.23 M^–1

7.3. At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms.

I₂(g) ⇌ 2I(g)

Calculate K_p for the equilibrium.

Ans. Total pressure = P_total = 10⁵ P_a

Since out of total volume, 40% by volume are I atoms.

∴ 60% are I₂ molecules (gaseous)

Partial pressure of I(g) atoms,

$$\text{P}_1=\bigg(\frac{40}{100}\bigg)×\text{P}_{\text{Total}}\\=\bigg(\frac{40}{100}\bigg)×10^{5}$$

= 4 × 10⁴ Pa

Partial pressure of I₂ molecules,

$$\text{PI}_2=\bigg(\frac{60}{100}\bigg)×\text{P}_{\text{Total}}\\=\bigg(\frac{40}{100}\bigg)×10^{5}$$

= 6 × 10⁴ Pa

Now, for the given reaction,

$$\text{K}_p=\frac{(\text{PI})2}{\text{P}_\text{I2}}\\=\frac{(4×10^{4})^{2}}{6×10^{4}}$$

= 2.67 × 10⁴ Pa

7.4. Write the expression for the equilibrium constant, Kc for each of the following reactions:

(i) 2NOCl (g) ⇌ 2NO (g) + Cl₂ (g)

(ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)

(iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH
(aq) + C₂H5OH(aq)

(iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)

(v) I₂(s) + 5F₂ ⇌ 2IF₅

$$\textbf{Ans.}\space\text{(i)}\space \text{K}_c=\frac{[\text{NO(g)}]^{2}[\text{Cl}_2\text{(g)}]}{\text{[NOCL(g)]}^{2}}\\\text{(ii) K}_c=\frac{[\text{CuO(s)}]^{2}[\text{NO}_2\text{(g)}]^{4}[\text{O}_2\text{(g)}]}{[\text{Cu}(\text{NO}_3)_2\text{(s)}]^{2}}\\\text{(iii)}\space \text{K}_\text{c}=\frac{[\text{CH}_3\text{COOH}(\text{aq})][\text{C}_2\text{H}_5 \text{OH}(aq)]}{[\text{CH}_3\text{COOC}_2\text{H}]}\\\text{(iv)} \text{K}_c=\frac{\text{Fe(OH)}_3(s)}{[\text{Fe}^{3+}(\text{aq})][\text{OH}^{-}(\text{aq})]^{3}}\\=\frac{1}{[\text{Fe}^{3+}(\text{aq})][\text{OH}^{-}(\text{aq})]^{3}}\\\text(v) \text{K}_c=\frac{[\text{lF}_5]^{2}}{[\text{I}_2\text{(g)}][\text{F}_2\text{(g)}]^{5}}\\=\frac{[\text{lF}_5]^{2}}{[l_2\text{(g)}][\text{F}_2\text{(g)}]^{5}}$$

$$=\frac{[\text{lF}_5]^{2}}{[\text{F}_2]^{5}}$$

7.5. Find out the value of Kc for each of the following equilibria from the value of K_p:

(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 × 10^–2 at 500 K

(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K

Ans. (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)

The relation between K_p and K_c given as:

K_p = K_c (RT)^Δn

In this reaction,

Δn = 3 – 2 = 1

Given values:

R = 0.0831 bar L mol^–1 K^–1

T = 500 K

K_p = 1.8 × 10^–2

K_p = K_c(RT)^Δn

1.8 × 10^–2 = Kc (0.0831 × 500)¹

$$\text{K}_c=\frac{1.8×10^{\normalsize-2}}{0.0831×500}$$

= 4.33 × 10^–4 (approx)

(ii) In this reaction,

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Δn = 2 – 1 = 1

Given values:

R = 0.0831 bar L mol^–1 K^–1

T = 1073 K,

K_p = 167

K_p = K_c (RT)^Δn

⇒ 167 = K_c (0.0831 × 1073 )¹

$$\Rarr\space \text{K}_c=\frac{167}{0.0831×1073}\\= 1.87 (\text{approx})$$

7.6. For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is ′K_c for the reverse reaction?

Ans. It is given that ′Kc for the forward reaction is
6.3 × 10¹⁴ .

Then, K_c for the reverse reaction will be,

$$\text{K}'_\text{C}=\frac{1}{\text{K}_\text{c}}\\=\frac{1}{6.3×10^{14}}$$

= 1.59 × 10^–15

7.7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Ans. For a pure substance like solids and liquids,

$$\text{Pure substance =}\frac{\text{Number of moles}}{\text{Volume}}\\=\frac{\text{Mass/molecular mass}}{\text{Volume}}\\=\frac{\text{Mass}}{\text{Volume} × \text{Molecular}}\\=\frac{\text{Density}}{\text{Molecular mass}}$$

At a particular temperature, the molecular mass and density of a pure substance is always fixed and is accounted for in the equilibrium constant. This is why the values of pure substances are not mentioned in the equilibrium constant expression.

7.8. Reaction between N₂ and O₂ takes place as follows:

2N₂(g) + O₂(g) ⇌ 2N₂O(g)

If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which Kc = 2.0 × 10^–37, determine the composition of equilibrium mixture.

Ans. Let x moles of O₂ and 2x moles of N₂ react before the equilirbium is reached. The equilibrium concentrations can be calculated as:

2N₂(g) + O₂(g) ⇌ 2N₂O (g)

Initial moles	0.482	0.933	0
Eqm. moles	(0.482 – 2x)	(0.933 – x)	2x

Eqm. concentrations

$$\frac{(0.482-2x)}{10}\frac{(0.933-x)}{10}\frac{2x}{10}$$

Therefore, at equilibrium, in the 10 L vessel:

$$\text{[N}_2] = 0.482-\frac{x}{10},\\\text{[O}_2]= 0.933 − \frac{x}{10},\\\text{[N}_2\text{O}]=\frac{x}{10}$$

The value of equilibrium constant i.e. K_c = 2.0 × 10⁻³⁷ is very small. Therefore, the amount of N₂ and O₂ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N₂ and O₂.
Then,

$$\text{[N]}_2=\frac{0.482}{10}\\= 0.0482 \space\text{mol L}^{–1}\\\text{[\text{O}}_2]=\frac{0.933}{10}\\= 0.0933\space\text{mol L}^{–1}\\\text{Now,}\\\text{K}_\text{c}=\frac{[\text{N}_2\text{O(g)}]^{2}}{[\text{N}_2\text{(g)}]^{2}[\text{O(g)]}}\\2.0×10^{\normalsize-37}=\frac{\bigg(\frac{x}{10}\bigg)}{(0.0482)^{2}(0.0933)}\\\Rarr\space\bigg(\frac{x^{2}}{100}\bigg)=2.0×10^{\normalsize-37}×(0.0482)^{2}×(0.0933)$$

⇒ x² = 43.35 × 10^–40

⇒ x = 6.6 × 10^–20

$$[\text{N}_2\text{O}]=\frac{x}{10}=\frac{6.6×10^{\normalsize-20}}{10}=6.6×10^{\normalsize-21}$$

The composition of equilibrium mixture = 6.6 × 10^–21.

7.9. Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

2NO(g) + Br₂(g) ⇌ 2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.

Ans. The given reaction is:

$$\underset{\text{2 mol}}{2\text{NO (g)}}+\underset{\text{1 mol}}{2\text{Br}_2\text{(g)}}⇌ \underset{2 mol}2\text{NOBr}(g)$$

Now, 2 mol of NOBr are formed from 2 mol of NO.

Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from

$$\bigg(\frac{0.0518}{2}\bigg)$$

mol of Br, or 0.0259 mol of NO. The amount of NO and Br present initially are as follows:

[NO] = 0.087 mol

[Br₂] = 0.0437 mol

Therefore, the amount of NO present at equilibrium are:

[NO] = 0.087 – 0.0518

= 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br₂] = 0.0437 – 0.0259

= 0.0178 mol

7.10. At 450K, K_p = 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

What is K_c at this temperature?

Ans. For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L mol^–1K^–1

K_p = 2.0 × 10¹⁰ bar ^–1

We know that,

K_p = K_c(RT)^Δn

⇒ (2.0 × 10¹⁰) bar^–1 = K_c(0.0831 L bar K^–1mol^–1× 450 K)^–1

⇒ Kc = 2.0 × 10¹⁰ bar^–1 × 0.0831 L bar K_–1 mol_–1× 450 K

= 74.79 × 10¹⁰ L mol^–1

= 7.48 × 10¹¹ L mol^–1

= 7.48 × 10¹¹ M^–1

Thus, the value of Kc at 450 K is 7.48 × 10¹¹ M^–1.

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_p for the given equilibrium?

2HI(g) ⇌ H₂(g) + I₂(g)

Ans. The initial concentration of HI is 0.2 atm.

At equilibrium, it has a partial pressure of 0.04 atm.

Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16.

The given reaction is:

2HI(g) ⇌ H₂(g) I₂(g)


Initial conc.	0.2 atm	0	0
At equilibrium	0.04 atm	$$\frac{0.16}{2}$$	$$\frac{2.15}{2}$$
		=0.08 atm	=0.08 atm

$$\text{Therefore,}\space \text{K}_p = \text{PH}_2×{\text{P}}_{\text{H}_2}×\frac{\text{P}_{I_{2}}}{\text{P}^{2}_{\text{HI}}}\\=0.08×\frac{0.08}{(0.04)^{2}}\\=\frac{0.0064}{0.0016}\\=4.0 $$

Thus, the value of K_p for the given equilibrium is 4.0.

7.12. A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans. The given reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The given concentration of various species is

$$\text{[N}_2] =\frac{1.57}{20}\text{mol}\space\text{L}^{\normalsize-1}\\\text{[H}_2] =\frac{1.92}{20}\text{mol \space L}^{\normalsize-1}\\\text{[NH}_3] =\frac{8.13}{20}\text{mol\space L}^{\normalsize -1}$$

Now, reaction quotient Q_c is:

$$\text{Q}_c=\frac{[\text{NH}_3]^{2}}{[\text{N}_3][\text{H}_2]^{3}}\\=\frac{\bigg(\frac{8.13}{20}\bigg)^{2}}{\bigg(\frac{1.57}{20}\bigg)\bigg(\frac{1.92}{20}\bigg)^{3}}$$

= 2.4 × 10³

Since, Q_c ≠ K_c (1.7 × 10²) , the reaction mixture is not at equilibrium. Again, Q_c (2.4 × 10³) > K_c (1.7 × 10²).

Hence, the reaction will proceed in the reverse direction.

7.13. The equilibrium constant expression for a gas reaction is,

$$\text{K}_c=\frac{[\text{NH}_3]^{4}[\text{O}_2]^{5}}{[\text{NO}]^{4}[\text{H}_2\text{O}]^{6}}$$

Write the balanced chemical equation corresponding to this expression.

Ans. Since,

$$\text{K}_c=\frac{[\text{NH}_3]^{4}[\text{O}_2]^{5}}{[\text{NO}]^{4}[\text{H}_2\text{O}]^{6}}$$

The products are 4 mol NH₃ and 5 mol O₂ and the reactants are 4 mol NO and 6 mol H₂O.

The balanced chemical equation corresponding to the given expression can be written as:

4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

7.14. One mole of H₂O 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,

H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

Calculate the equilibrium constant for the reaction.

Ans. The given reaction is:

H₂O(g) + CO(g) ⇌ H₂(g) CO₂(g)


Initial conc.	$$\frac{1}{10}\text{M}$$	$$\frac{1}{10}\text{M}$$	0	0
At equilibrium	$$\frac{1-0.4}{10}\text{M}$$	$$\frac{1-0.4}{10}\text{M}$$	$$\frac{0.4}{10}\text{M}$$	$$\frac{0.4}{10}\text{M}$$
	= 0.06 M	= 0.06 M	=0.04 M	=0.04 M

Therefore, the equilibrium constant for the reaction,

$$\text{K}_c=\frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]}\\=\frac{0.04×0.04}{0.06×0.06}$$

= 0.444 (approx)

Thus, the value of equilibrium constant is 0.444.

7.15. At 700 K, equilibrium constant for the reaction:

H₂(g) + I₂(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 H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

Ans. Equilibrium constant K_c for the given reaction is 54.8.

H₂(g) + I₂(g) ⇌ 2HI(g)

Therefore, at equilibrium, the equilibrium constant K_c for the reaction

$$\text{2HI(g)} ⇌ \text{H}_2\text{(g) + I}_2 \text{(g) will be}\space\frac{1}{54.8}$$

[HI] = 0.5 mol L^–1

Let the concentrations of hydrogen and iodine at equilibrium be x mol L^–1

[H₂] = [I₂] = x mol L^–1

$$\text{Therefore,}\frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]}=\text{K}_c\\\Rarr\frac{x^{2}}{(0.5)^{2}}=\frac{1}{54.8}\\\Rarr x^{2}=\frac{0.25}{54.8}$$

⇒ x = 0.06754

x = 0.068 mol L^–1 (approx)

Hence, at equilibrium, [H₂] = [I₂] = 0.068 mol L^–1.

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) ⇌ I₂(g) + Cl₂(g); K_c = 0.14

Ans. The given reaction is:

2ICl(g) ⇌ I₂(g) + Cl₂(g)


Initial concen.	0.78 M	0	0
At equilibrium	(0.78 – 2x)M	xM	xM

Now, we can write,

$$\text{K}_c=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^{2}}\\\Rarr\space 0.14 =\frac{x^{2}}{(0.78-2x)^{2}}\\0.374 =\frac{x^{2}}{(0.78-2x)^{2}}$$

⇒ x = 0.292 – 0.748x

⇒ 1.748 x = 0.292

⇒ x = 0.167

Hence, at equilibrium,

[I₂] = [Cl₂] = 0.167 M

[ICl] = (0.78 – 2 × 0.167) M

= 0.446 M

7.17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C₂H₆(g) ⇌ C₂H₄(g) + H2(g)

Ans. Let p be the pressure exerted by ethane and hydrogen gas (each) at equilibrium. Now, according to the reaction,

C₂H₆(g) ⇌ C₂H₄(g) + H2(g)


Initial concen.	4 atm	0	0
At equilibrium	(4 – p)M	pM	pM

Now, we can write,

$$\text{K}_p=\text{P}_{\text{c}_2\text{H}_4}×\frac{\text{P}_{\text{H}_2}}{\text{P}_{\text{C}_2\text{H}_4}}\\=0.04 =p×\frac{\text{P}}{\text{P}(4-p)}$$

⇒ p² = 0.16 × 0.04 p

⇒ p² + 0.04p – 0.16 = 0

$$\text{Now},p = -0.04\space\pm\space\frac{0.80}{2}\\=\frac{0.76}{2}\text{(Taking only positive values)}$$

= 0.38

Hence, at equilibrium,

[C₂H₆] = 4 – p

= 4 – 0.38

= 3.62 atm

7.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

(i) Write the concentration ratio (reaction quotient), Q_c, 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?

Ans. (i) Reaction quotient,

$$\text{Q}_c=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$$

(ii) Let the volume of the reaction mixture be V.

Also, here we will consider that water is a solvent and present in excess.

The given reaction is:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)


Initial conc.	$$\frac{1}{\text{V}}\text{M}$$	$$\frac{0.18}{\text{V}}\text{M}$$	0	0
At eq.	$$\frac{(1-0.171)}{\text{V}}$$	$$\frac{(0.18-0.171)}{\text{V}}$$	$$\frac{0.171}{\text{V}}$$	$$\frac{0.171}{\text{V}}$$
	$$=\frac{0.829}{\text{V}}$$	$$=\frac{0.009}{\text{V}}$$

Therefore, equilibrium constant for the given reaction is:

$$\text{K}_c=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}\\=\frac{\bigg[\frac{0.171}{\text{V}}×\frac{0.171}{\text{V}}\bigg]}{\bigg[\frac{0.829}{\text{V}}×\frac{0.009}{\text{V}}\bigg]}$$

= 3.919

= 3.92 (approx)

(iii) Let the volume of the reaction mixture be V.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)


Initial conc.	$$\frac{1}{\text{V}}\text{M}$$	$$\frac{0.5}{\text{V}}\text{M}$$	0	0
At eq.	$$\frac{(1-0.214)}{\text{V}}$$	$$\frac{(0.5-0.214)}{\text{V}}$$	$$\frac{0.214}{\text{V}}$$	$$\frac{0.214}{\text{V}}$$
	$$=\frac{0.786}{\text{V}}$$	$$=\frac{0.286}{\text{V}}$$

Therefore, the reaction quotient is,

$$\text{Q}_c=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}\\=\frac{\bigg[\frac{0.214}{\text{V}}×\frac{0.214}{\text{V}}\bigg]}{\bigg[\frac{0.286}{\text{V}}×\frac{0.786}{\text{V}}\bigg]}$$

= 0.2037

= 0.204 (approx)

Since Q_c< K_c, equilibrium has not been reached.

7.19. A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅was found to be 0.5 × 10^–1 mol L^–1. If value of Kc is 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Ans. Let the concentration of both PCl₃ and Cl₂ at equilibrium be × mol L^–1. The given reaction is:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)


At equilibrium	0.5 × 10^–1	x mol	x mol
	mol L^–1	L^–1	L^–1

Given that the value of equilibrium constant, Kc is 8.3 × 10^–3.

Now we can write the expression for equilibrium as:

$$\text{K}_c=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\\\Rarr\space 8.3 × 10^{\normalsize–3} =\frac{x×x}{0.5×10^{\normalsize-1}}$$

⇒ x² = 4.15 × 10^–4

⇒ x = 2.04 × 10^–2

= 0.0204

= 0.02 (approx)

Therefore, at equilibrium, [PCl₃] = [Cl₂] = 0.02 mol L^–1.

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 CO₂.

FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g);

K_p = 0.265 atm at 1050K

What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: P_CO = 1.4 atm and P_CO₂ = 0.80 atm.

Ans. For the given reaction,

FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g)


Initially,	1.4 atm	0.80 atm
At eq.	1.4 + p	0.8 – p

$$\text{Q}p =\frac{\text{P}_{\text{CO}_2}}{\text{P}_{\text{CO}}}\\=\frac{0.80}{1.4}$$

= 0.571

It is given that K_p = 0.265

Since Q_p > K_p, the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO₂will decrease.

Now, let the increase in pressure of CO
= decrease in pressure of CO₂ be p.

So,

$$\text{K}_p=\frac{\text{P}_{\text{CO}_2}}{\text{P}_\text{CO}}\\\Rarr 0.265 =\frac{0.80-p}{1.4+p}$$

⇒ 0.371 + 0.265 p = 0.80 – p

⇒ 1.265 p = 0.429

⇒ p = 0.339 atm

Therefore, equilibrium partial pressure of CO₂, PCO₂ is 0.80 – 0.339 = 0.461 atm.

And, equilibrium partial pressure of CO , PCO is 1.4 + 0.339 = 1.739 atm.

7.21. Equilibrium constant, Kc for the reaction

N₂(g) + 3H₂(g) ⇌ 2NH₃(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^–1N₂, 2.0 mol L^–1 H₂ and 0.5 mol L^–1 NH₃.

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

Ans. The given reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

At particular time 3.0 mol L^–12.0 mol L^–10.5 mol L^–1time

Now,

$$\text{Q}_c=\frac{[\text{NH}_3]^{2}}{[\text{N}_2][\text{H}_2]}\\=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}$$

= 0.0104

It is given that K_c = 0.061.

Since Q_c ≠ K_c, the reaction is not at equilibrium.

Since Q_c < K_c, the reaction will proceed in the forward direction to reach equilibrium.

7.22. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇌ Br₂(g) + Cl₂(g) for which K_c = 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?

Ans. Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

2BrCl(g) ⇌ Br₂(g) + Cl₂(g)


Initial con.	3.3 × 10^–3	0	0
At equilibrium	(3.3 × 10^–3 – 2x)	x	x

Now,

$$\frac{[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^{2}}=\text{K}_c\\\Rarr\frac{x^{2}}{(3.3×10^{\normalsize-3}-2x)^{2}}=32\\\Rarr\frac{x}{(3.3×10^{\normalsize-3}-2x)}=5.66$$

⇒ x = 18.678 × 10^–3 – 11.32x

⇒ 12.32x = 18.67 × 10^–3

⇒ x = 1.5 × 10^–3

Therefore, at equilibrium,

[BrCl] = (3.3×10^–3 – 2x)

= 3.3 × 10^–3 – (2 × 1.5 × 10^–3)

= 3.3 × 10^–3– 3.0 × 10^–3

= 0.33 × 10^–3

= 3.3 × 10–4 mol L^–1

7.23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass

C(s) + CO₂(g) ⇌ 2CO(g)

Calculate K_c for this reaction at the above temperature.

Ans. Let the total mass of the gaseous mixture of CO and CO₂ be 100 g.

Then, mass of CO = 90.55 g

And, mass of CO₂ = (100 – 90.55) = 9.45 g

$$\text{Now, number of moles of CO, n}_{CO} =\frac{90.55}{28}=3.234\space\text{mol}\\\text{Number of moles of CO}_2, \text{nCO}_2 =\frac{9.45}{44}= 0.215 \space\text{mol}\\\text{P}_\text{CO =}\bigg[\frac{n_{\text{co}_2}}{n_{co}+n_{co_{2}}}\bigg]×\text{P}_{\text{Total}}\\=\bigg[\frac{3.234}{3.234+0.215}\bigg]×1$$

= 0.938 atm

Partial pressure of CO₂,

$$\text{P}_{\text{co}_2}=\bigg[\frac{n_{co_2}}{n_{co}+n_{co_2}}\bigg]×\text{P}_{\text{Total}}\\=\bigg[\frac{0.215}{3.234+0.215}\bigg]×1$$

= 0.062 atm

K_p for the reaction C(s) + CO₂(g) ⇌ 2CO(g)

$$\text{K}_p=\frac{P^{2}\text{co}}{P\text{co}_2}=\frac{(0.938)^{2}}{0.062}=14.19$$

Now ∆n_g = 2 – 1 = 1, K_p = K_c(RT)^∆n_g

$$\text{or K}_c=\frac{\text{K}_P}{\text{RT}}=\frac{14.19}{0.0821×1127}=0.153$$

7.24. Calculate

(A) ∆G° and

(B) the equilibrium constant for the formation of NO2 from NO and O2 at 298K

NO(g) + ½O₂(g) ⇌ NO₂(g)

where ∆_fG° (NO₂) = 52.0 kJ/mol

∆_fG° (NO) = 87.0 kJ/mol

∆_fG° (O₂) = 0 kJ/mol
Ans. (A) For the given reaction,

∆_fG° = ∆G°(Products) – ∆G°(Reactants)

∆_rG° = 52.0 – [87 + 0]

= – 35.0 kJ mol^–1

(B) We know that,

∆G° = – RT ln K_c

∆G° = – 2.303 RT log K_c

$$\text{log K}_c =\frac{-35.0×10^{3}}{-2.303×8.314×298}$$

= 6.134

∴ K_c = antilog (6.134)

= 1.36 × 10⁶

Hence, the equilibrium constant for the given reaction Kc i s 1.36 × 10⁶.

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) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s)

(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)

Ans. (a) According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is less than that of gaseous reactants. Thus, the reaction will proceed in the backward direction. As a result, the number of moles of reaction products will decrease.

(c) According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is equal to that of gaseous reactants. Thus, the decreasing of pressure does not affect the equilibrium. As a result, number of moles of reaction products remains the same.

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) COCl₂(g) ⇌ CO(g) + Cl₂(g)

(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)

(iii) CO₂(g) + C(s) ⇌ 2CO (g)

(iv) 2H₂(g) + CO(g) ⇌ CH₃OH (g)

(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)

(vi) 4NH₃(g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O(g)

Ans. The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure. According to Le Chatelier’s principle, if pressure is increased, then the equilibrium shifts in the direction in which the number of moles of gases is less.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

7.27. The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024 K

H₂(g) + Br₂(g) ⇌ 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Ans. Given,

K_p for the reaction,

H₂(g) + Br₂(g) ⇌ 2HBr(g) is 1.6 ×10⁵at 1024K Therefore, for the reaction,

2HBr(g) ⇌ H₂(g) + Br₂(g), the equilibrium constant will be,

$$\text{K}_p'=\frac{1}{\text{K}_\text{p}}\\=\frac{1}{1.6×10^{5}}$$

= 6.25 × 10^–6

Now, let p be the pressure of both H2 and Br2 at equilibrium.

2HBr(g) ⇌ H₂(g) + Br₂(g)


Initial concentration	10	0	0
At equilibrium	10 – 2p	p	p

Now,

$$\frac{P_{\text{H}_2×P_{\text{Br}_2}}}{P^{2}_{\text{HBr}}}=\text{K}'p\\=\frac{p^{2}}{(10-2p)^{2}}=6.25×10^{\normalsize-6}\\\frac{p}{(10-2p)}=2.5×10^{\normalsize-3}$$

p = 2.5 × 10^–3 (10 – 2p)

p = 2.5 × 10^–2 – (5.0×10^–3)p

p + (5.0×10^–3)p = 2.5 × 10^–2

1.005p = 2.5 × 10^–2

p = 2.49 × 10^–2 bar

= 2.5 × 10^–2 bar (approx)

Therefore, at equilibrium,

[H₂] = [Br₂] = 2.49 × 10^–2 bar

[HBr] = 10 – 2 × (2.49 × 10^–2) bar

= 9.95 bar

= 10 bar (approx)

7.28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)

(a) Write as expression for K_p for the above reaction.

(b) How will the values of K_pand composition of equilibrium mixture be affected by :

(i) increasing the pressure

(ii) increasing the temperature

(iii) using a catalyst?

Ans. (a) For the given reaction,

$$\text{K}_p=\frac{\text{P}_{\text{CO}}×\text{P}^{3}_{\text{H}_2}}{\text{P}_{\text{CH}_4}×\text{P}_{\text{H}_2\text{O}}}$$

(b) (i) According to Le Chatelier’s principle, if pressure is increased, then the equilibrium shifts in the direction in which the number of moles of gases is less. Thus, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction with increasing temperature.

(iii) The equilibrium composition of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

7.29. Describe the effect of :

(a) addition of H₂

(b) addition of CH₃OH

(c) removal of CO

(d) removal of CH3OH on the equilibrium of the reaction:

2H₂(g) + CO(g) ⇌ CH₃OH(g)

Ans. (a) According to Le Chatelier’s Principle, on addition of H₂, the equilibrium of the given reaction will shift in the forward direction.

(b) According to Le Chatelier’s Principle, on addition of CH₃OH, the equilibrium will shift in the backward direction.

(d) According to Le Chatelier’s Principle, on removal of CH₃OH, the equilibrium will shift in the forward direction.

7.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl₅ is 8.3 × 10^–3. If decomposition is depicted as,

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

Δ_rH^⊖ = 124.0 kJ mol^–1

(a) write an expression for K_c for the reaction.

(b) what is the value of K_c for the reverse reaction at the same temperature?

(c) what would be the effect on K_c if (i) more PCl₅ is added (ii) pressure is increased (iii) the temperature is increased?

$$\textbf{Ans.}\space \text{(a) K}_c =\frac{[\text{PCl}_3(\text{g})][\text{Cl}_2(\text{g})]}{[\text{PCl}_5(\text{g})]}$$

(b) Value of K_c for the reverse reaction at the same temperature is:

$$\text{K}_c'=\frac{1}{\text{K}_c}=\frac{1}{8.3×10^{\normalsize-3}}$$

= 1.2048 × 10²

(ii) K_c is constant at constant temperature. Thus, in this case, K_c would not change.

(iii) In an endothermic reaction, the value of K_c increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of K_c will increase if 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 H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g) + H₂O (g) ⇌ CO₂(g) + H₂(g)

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that P_CO = P_H₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p = 10.1 at 400°C.

Ans. Let the partial pressure of both carbon dioxide and hydrogen gas be p.

The given reaction is:

CO(g) + H₂O (g) ⇌ CO₂(g) + H₂(g)


Initial con.	4.0 bar	4.0 bar	0	0
At eqil.	4.0 – p	4.0 – p	p	p

It is given that,

K_p = 10.1

$$\text{Now},\space\frac{\text{P}_{\text{CO}_2}×\text{P}_{\text{H}_2}}{\text{P}_{\text{CO}}×\text{P}_{\text{H}_2\text{O}}}=\text{K}_p\\\Rarr\space\frac{p×p}{(4.0-p)(4.0-p)}=10.1\\\Rarr\frac{p^{2}}{(4.0-P)^{2}}=10.1\\\Rarr \frac{p}{4.0-p}=3.178$$

⇒ p = 12.712 – 3.178 p

⇒ 4.178 p = 12.712

⇒ p = 3.04

Hence, at equilibrium, the partial pressure of H₂ will be 3.04 bar.

7.32. Predict which of the following reaction will have appreciable concentration of reactants and products:

(a) Cl₂(g) ⇌ 2Cl (g) K_c = 5 ×10^–39

(b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g) K_c = 3.7 × 10⁸

(c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl (g) K_c = 1.8

Ans. If the value of K_c lies between 10^–3 and 10³, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

7.33. The value of Kc for the reaction

3O₂(g) ⇌ 2O₃ (g) is 2.0 ×10^–50 at 25°C.

If the equilibrium concentration of O₂ in air at 25°C is 1.6 ×10^–2, what is the concentration of O3?

Ans. Given:

K_c = 2.0 × 10^–50

[O₂(g)] = 1.6 ×10^–2

For the reaction,

3O₂(g) ⇌ 2O₃(g)

Equilibrium 1.6 × 10^–2x

concentrations/M

$$\text{K}_c=\frac{[\text{O}_3]^{2}}{[\text{O}_3]^{3}}\\2.0 × 10^{\normalsize–50} =\frac{x^{2}}{(1.6×10^{\normalsize-2})^3}$$

x² = 2.0 × 10^–50 × (1.6 × 10^–2)³

= 8.192 × 10^–56

or x = 2.86 × 10^–28 M

Hence, the concentration of O₃ is 2.86 × 10^–28 M

7.34. The reaction,

CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

is at equilibrium at 1300 K in a 1 L flask.

It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90?

Ans. Let the concentration of methane at equilibrium be X.

CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

$$\text{At equilibrium}\space\frac{0.3}{1}= 0.3\text{M}\space\frac{0.1}{1}= 0.1\text{M}\space\space x\space\frac{0.02}{1}= 0.02\text{m} $$

Given:

K_c = 3.90.

Therefore,

$$\frac{[\text{CH}_4(\text{g})][\text{H}_2\text{O}(\text{g})]}{[\text{Co(g)}][\text{H}_2\text{(g)}]^{3}}=\text{K}_c\\\Rarr\space\frac{x×0.02}{0.3×(0.1)^{3}}=3.90\\\Rarr\space x=\frac{3.90×0.3×(0.1)^{3}}{0.02}\\=\frac{0.00117}{0.02}$$

= 0.0585 M

= 5.85 × 10^–2 M

Therefore, the concentration of CH4 at equilibrium is 5.85 × 10^–2 M.

7.35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

HNO₂, CN^–, HClO₄, F^–, OH^–, CO₃^2–, and S^2–

Ans. A conjugate acid-base pair is a pair that differs only by one proton. The conjugate acid-base for the given species are as follows.

Species	Conjugated acid-base
HNO₂	NO₂^–1 (base)
CN^–	HCN (acid)
HClO₄	ClO^4– (base)
F^–	HF (acid)
OH^–	H₂O (acid) / O^2– (base)
CO₃^2–	HCO₃– (acid)
S^2–	HS^– (acid)

7.36. Which of the followings are Lewis acids?

H₂O, BF₃, H⁺, and NH₄⁺.

Ans. Lewis acids are those acids, which can accept a pair of electrons. From the given options, BF₃, H⁺ , and NH₄⁺ are examples of Lewis acids, because they are electron deficient and can accept a pair of electrons.

7.37. What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄ and HCO₃^– ?

Ans. The conjugate bases for the given Brönsted acids are as follows:

Bronsted Acid	Conjugate Base
HF	F^-
H₂SO₄	HSO₄^–
HCO₃^–	CO₃^2-

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

NH₂^– , NH₃and HCOO^–.

Ans: The conjugate acids for the Brönsted bases are as follows:

Bronsted Base	Conjugate Acid
NH₂^–	NH₃
NH₃	NH₄⁺
HCOO^–	HCOOH

7.39. The species: H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Ans. The corresponding conjugate acid and base for each of the given species are as follows:

Species	Conjugate Acid	Conjugate Base
H₂O	H₃O⁺	OH^–
HCO₃^–	H₂CO₃	CO₃^2–
HSO₄^–	H₂SO₄	SO₄^2–
NH₃	NH₄⁺	NH₂^–

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^–

(c) H⁺

(d) BCl₃.

Ans. (a) OH^– can donate its lone pair of electrons. So, it is a Lewis base.

(b) F^– can donate its lone pair of electrons. So, it is a Lewis base.

(d) BCl₃ can accept a lone pair of electrons. So, it is a Lewis acid.

7.41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^–3 M. what is its pH?
Ans. Given,

Concentration of hydrogen ion [H⁺] in a sample of soft drink = 3.8 × 10^–3 M

∴ pH Value of soft drink = – log [H⁺]

= – log (3.8 × 10^–3)

= – log 3.8 – log 10^–3

= – log 3.8 + 3

= – 0.58 + 3

= 2.42

Hence, the pH value of given sample of soft drink is 2.42.

7.42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it?

Ans. Given,

pH of a sample of vinegar = 3.76

Now, the concentration of hydrogen ion in vinegar can be calculated using,

pH = – log [H⁺]

⇒ log [H⁺] = – pH

⇒ [H⁺] = antilog (–pH)

= antilog (– 3.76)

= 1.74 × 10^–4 M

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10^–4 M.

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?

Ans. It is known that,

$$\text{K}_b=\frac{\text{K}_{w}}{\text{K}_{a}}\\\text{Given, K}_a \text{of HF} = 6.8 × 10^{–4}\\\text{Hence, K}_\text{b}\space\text{of its conjugate base F– =}\frac{\text{K}_{w}}{\text{K}_{a}}\\=\frac{1×10^{\normalsize-14}}{6.8×10^{\normalsize-4}}$$

= 1.5 × 10^–11

Given, Ka of HCOOH = 1.8 × 10^–4

$$\text{Hence, K}_\text{b}\space\text{of its conjugate base HCOO– =}\frac{\text{K}_w}{\text{K}_a}\\=\frac{1×10^{\normalsize-14}}{4.8×10^{\normalsize-9}}$$

= 2.08 × 10^–6

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?

Ans. Ionization of phenol

$$\text{C}_6\text{H}_5\text{OH} + \text{H}_2\text{O}\xrightarrow{}\text{C}_6\text{H}_5\text{O}^– + \text{H}_3\text{O}^+$$

Initial con.	0.05	0	0
At equilibrium	0.05 – x	x	x

$$\text{K}_a=\frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}\\\text{K}_a = \frac{x×x}{0.05-x}$$

As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator

$$\\\text{K}_a = \frac{x×x}{0.05-x}$$

∴ x² = 1 ×10^–10 × 0.05

= 5 × 10^–12

= 2.2 × 10^–6 M = [H₃O⁺]

Since [H₃O⁺] = [C₆H₅O^–],

[C₆H₅O^–] = 2.2 × 10^–6 M

Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa.

$$\underset{0.01}{\text{C}_6\text{H}_5\text{ONa}}\xrightarrow{}\text{C}_6\text{H}_5\text{O}^– + \text{Na}^+$$

Concentration

Also,

C₆H₅OH + H₂O → C₆H₅O– + H₃O⁺

Concentration 0.05 – 0.05 α 0.01+0.05α 0.05 α

[C₆H₅OH] = 0.05 – 0.05 α ; 0.05 M

[C₆H₅O^–] = 0.01 + 0.05 α ; 0.01 M

[H₃O⁺] = 0.05 α

$$\text{K}_\text{a}= \frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}\\=\frac{(0.01)(0.05\alpha)}{0.05}$$

1.0 × 10 ^–10 = 0.01α

α = 1 × 10 ^–8

7.45. The first ionization constant of H₂S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also ? If the second dissociation constant of H₂S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.

Ans. To calculate the concentration of HS – ion:

Case I (in the absence of HCl):

Let the concentration of HS be x M.

H₂S → H⁺ + HS

C_δ	0.1	0	0
C_f	0.1-x	x	x

Then,

$$\text{K}_{a_1}=\frac{[\text{H}^{+}][\text{HS}^{-}]}{[\text{H}_2\text{S}]}\\9.1 × 10^{–8} =\frac{(x)(x)}{0.1-x}$$

(9.1 × 10^–8) (0.1 – x) = x²

Taking (0.1 – x) M as 0.1 M

We have,

(9.1 × 10^–8) (0.1) = x²

(9.1 × 10^–8) = x²

x = (9.1 × 10^–9)^1/2

x = 9.54 × 10^–5 M

⇒ [HS^–] = 9.54 × 10^–5 M

Case II (in the presence of HCl): In the presence of 0.1 M of HCl, let [HS^–] be y M.

Then,

$$\text{H}_2\text{S} \xrightarrow{} \text{HS}^– + \text{H}^–$$

C´	0.1	0	0
C_f	0.1 – y	y	y

Also,

$$\text{HCl} ⇌ \underset{0.1}{\text{H}^{+}}+\underset{0.1}{\text{Cl}^{-}}$$

Now,

$$\text{K}_{a_1}=\frac{[\text{HS}^{-}][\text{HS}^{+}]}{[\text{H}_2\text{S}]}\\\text{K}_{\text{a}_1}=\frac{[y](0.1+y)}{((0.1-y))}\\9.1 × 10^{–8} =\frac{y×0.1}{0.1}$$

(∵ 0.1 – y = 0.1 M and 0.1 + y = 0.1 M )

9.1 × 10^–8 = y

⇒ [HS^–] = 9.1 × 10^–8

Case III : To calculate [S^2–] in absence of 0.1 M HCl

H₂S^– ⇌ H⁺ + S^2–

From case I [HS^–] = [H⁺] = 9.54 × 10^–5 N

[S^2–] = x

$$\text{K}_{a_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{\text{HS}^{-}}\\1.2×10^{\normalsize-13}=\frac{[9.54×10^{\normalsize-5}][x]}{[9.54×10^{\normalsize-5}]}$$

x = 1.2 × 10^–13 M = [S^2–]

Case IV : To calculate [S^2–] in presence of 0.1 M HCl

HS^– ⇌ H⁺ + S^2–

From case II [HS^–] = 9.1 × 10^–8M

[H⁺] = 0.1 M (from HCl)

[S^2–] = x

$$\text{K}_{a_2}=\frac{[\text{H}^{+}]+[\text{S}^{2-}]}{\text{[HS}^{-}]}=\frac{0.1×x}{9.1×10^{\normalsize -8}}\\1.2 × 10^{–13} =\frac{0.1×x}{9.1×10^{\normalsize-8}}$$

x = 1.09 × 10^–19 M = [S^2–]

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.

Ans. (1) CH₃COOH ⇌ CH₃COO^– + H⁺

K_a = 1.74 × 10^–5

(2) H₂O + H₂O ⇌ H₃O⁺ + OH^– K_w = 1.0 × 10^–14

Since K_a >> K_w :

CH₃COOH + H₂O ⇌ CH₃COO + H₃O⁺

Cf	0.05	0	0
	0.05 –0.5α	0.05α	0.05α

$$\text{K}_a=\frac{(0.05α)(0.05α)}{(0.05-0.05α)}\\=\frac{(0.05α)(0.05α)}{0.05(1-α)}\\=\frac{(0.05α)^{2}}{1-α}\\=1.74×10^{\normalsize-5}=\frac{(0.05α)^{2}}{1-α}$$

1.74 × 10^–5 – 1.74 × 10^–5α = 0.05 α²

0.05α² + 1.74 + 10^–5α – 1.74 × 10^–5

D = b² – 4ac

= (1.74 × 10^–5)² – 4(0.05) (– 1.74 × 10^–5)

= 3.02 × 10^–25 + 0.348 × 10^–5 = 0

$$\alpha=\bigg(\frac{\text{K}_a}{\text{C}}\bigg)^{\frac{1}{2}}\\\alpha=\bigg(\frac{1.74×10^{\normalsize-5}}{0.05}\bigg)^{\frac{1}{2}}$$

= (3.48 × 10^–4)^1/2

= 1.86 × 10^–2

CH₃COOH ⇌ CH₃COO^– + H⁺

CH₃COO^– = C.a

= 0.05 × 1.86 × 10^–3

= 9.3 × 10^–4 = 0.00093

pH = – log [H⁺]

= – log [0.00093]

= 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 and pH is 3.03.

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_a.

Ans. Let the organic acid be HA.

⇒ HA ⇌ H⁺ + A^–

Concentration of HA = 0.01 M

pH = 4.15 = – log[H⁺]

[H⁺] = 7.08 × 10^–5

[HA] = 0.01

Then,

$$\text{K}_a=\frac{(7.08×10^{\normalsize-5})(7.08×10^{\normalsize-5})}{0.01}$$

K_a = 5.01 × 10^–7

pK_a = –log K_a

= – log (5.01 × 10^–7)

pK_a= 6.3001

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

Ans. (a) 0.003 M HCl

H₂O + HCl (aq) ⇌ H₃O⁺ + Cl^–

Since HCl is completely ionized,
[H₃O⁺] = [HCl]

⇒ [H₃O⁺] = 0.003

Now,

pH = – log [H₃O⁺]

= – log(0.003)

= 2.52

Hence, the pH of the solution is 2.52.

(b) 0.005 M NaOH

NaOH(aq) ⇌ Na⁺(aq) + OH^–(aq)

[OH ^–] = [NaOH]
⇒ [OH^–] = 0.005

pOH = – log [OH^–]

= – log (0.005 )

pOH = 2.30

⇒ pH = 14 – 2.30

= 11.70

Hence, the pH of the solution is 11.70.

(c) 0.002 HBr

HBr(aq) ⇌ H⁺ + Br^–

[H⁺] = [HBr ]

⇒ [H⁺] = 0.002

⇒ pH = – log [H⁺]

= – log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

(d) 0.002 M KOH

KOH (aq) ⇌ K⁺(aq) + OH^–(aq)

[OH^– ] = [KOH]

⇒ [OH ^–] = 0.002

Now, pOH = – log[OH^– ]

= 2.69

⇒ pH = 14 – 2.69

= 11.31

Hence, the pH of the solution is 11.31.

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)2 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.

Ans. (a) For 2g of TlOH dissolved in water to give 2 L of solution,

$$\text{[TlOH(aq)] =}\frac{29}{2\text{L}}\\=\frac{2}{2}×\frac{1}{221}\text{M}\\=\frac{1}{221}\text{M}$$

TlOH(aq) → Tl(aq)⁺ + OH(aq)^–

[OH(aq)^–] = [TlOH(aq)]

$$=\frac{1}{221}\text{M}\\\text{K}_w =[\text{H}^+] [\text{OH}^–]\\10^{–14} = [\text{H}^+]\bigg(\frac{1}{221}\bigg)$$

221 × 10^–14 = [H⁺]

⇒ pH = – log[H⁺]

= – log(221 × 10^–14)

= – log (2.21 × 10^–12)

= 11.65

(b) For 0.3 g of Ca(OH)₂dissolved in water to give 500 mL of solution:

Ca(OH)₂ → Ca₂+ 2OH^–

[Ca(OH)₂] = 0.3 × 1000 / 500

= 0.6 M

[OH^–aq] = 2 × [Ca(OH)²aq]

= 2 × 0.6

= 1.2 M

$$\text{[H}^+]=\frac{\text{K}_w}{[\text{OH}^{-}(\text{aq})]}\\\frac{10^{\normalsize-14}}{1.2}\text{M}$$

= 0.833 × 10^–14

pH = – log (0.833 × 10–¹⁴)

= – log (8.33 × 10^–13)

= (– 0.902 + 13 )

= 12.098

NaOH → Na⁺(aq) + OH^–(aq)

[OH^–(aq)] = 1.5 M

Then,

$$\text{[H}^{+}]=\frac{10^{\normalsize-14}}{1.5}$$

= 6.66 × 10^–13

pH = – log(6.66 × 10^–13)

= 12.18

(d) For 1 mL of 13.6 M HCl diluted with water to give 1 L of solution:

13.6 × 1 mL = M₂ × 1000 mL (Before dilution) (After dilution)

13.6 × 10^–3 = M₂

M₂ = 1.36 × 10^–2

[H⁺] = 1.36 × 10^–2

pH = – log (1.36 × 10^–2)

≈ 1.87

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_a of bromoacetic acid.

Ans. Given,

Degree of ionization, a = 0.132

Concentration, c = 0.1 M

Now, the ionization of bromoacetic acid can be given as,

BrCH₂COOH ⇌ BrCH₂COO^– + H⁺

Initial conc.	C	0	0
Equilibrium conc.	(C – Cα)	Cα	Cα

$$\Rarr\space\text{K}_a=\frac{\text{C}\alpha^{2}}{(1-\alpha)}=\text{C}\alpha^{2}$$

= 0.1 × (0.132)²

= 1.74 × 10^–3

⇒ pK_a = – log K_a

= – log (1.74 × 10^–3)

= 3 – 0.2405 = 2.76

⇒ (H⁺) = Ca = 0.1 × 0.132

= 1.32 × 10^–2 M

⇒ pH = – log (1.32 × 10^–2)

= 2 – 0.1206 = 1.88.

Hence, the pH of the solution is 1.88 and the pK_a value of bromoacetic acid is 2.76.

7.51. The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

Ans. Given,

pH of 0.005 M codeine solution = 9.95

Now,

pOH = pK_w – pH

= 14.0 – 9.95 = 4.05

[OH^–] = antilog (– 4.05)

= 8.91 × 10^–5 M

Since, [OH^–] = Cα;

$$\alpha=\frac{[\text{OH}^{-}]}{c}=\frac{8.91×10^{\normalsize-5}}{0.005}=0.0178\space \text{or}\space 1.78 × 10^{\normalsize–2}\\\text{K}_b=\frac{c\alpha^{2}}{1-\alpha}\\=\frac{0.005×0.0178^{2}}{1-0.0178}$$

= 1.6 × 10^–6

pK_b = – log K_b

= – log 1.6 × 10^–6 = 5.8.

Hence, the ionization constant is 1.78 × 10^–2 and the pK_b value is 5.8

7.52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 × 10^–4. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Ans. Given that,

K_b = 4.27 × 10^–10

c = 0.001 M

pH = ?

α = ?
K_b = Cα²

4.27 x 10^–10 = 0.001 × α²

4270 × 10^–10 = α²

= 65.34 × 10^–5 = α

= 6.53 × 10^–4

Then,

[anion] = ca = .001 × 65.34 × 10^–5

= .065 × 10^–5

Now,

pOH = – log (0.065 × 10^–5)

= 6.187

pH = 7.183

Now,

K_a × K_b = K_w

∴ 4.27 × 10^–10 × K_a= K_w

$$\text{K}_a=\frac{10^{\normalsize-14}}{4.27×10^{\normalsize-10}}$$

= 2.34 × 10^–5

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.

7.53. Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74.

How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

Ans. Given.

c = 0.05 M; pK_a = 4.74

pK_a= – log K_a

K_a = antilog (–4.74) = 1.82 × 10^–5

$$\alpha=\sqrt{\frac{\text{K}_a}{c}}=\sqrt{\frac{1.82×10^{\normalsize-5}}{0.05}}=0.019=1.9×10^{\normalsize-2}$$

In presence of 0.01 M HCl

Equilibria and equilibrium concentrations are

CH₃COOH ⇌ H⁺ + CH₃COO^–

Initial conc.	0.05 M	0	0
After dissociation	0.05(1 – α)	0.05α	0.05α

$$\underset{0}{\text{HCl}}\xrightarrow{}\underset{0.01\text{M}}{\text{H}^{+}}+\underset{0.01\text{M}}{\text{Cl}^{-}}\\\text{K}_a=\frac{[\text{CH}_3\text{COO}^{-}][\text{H}^{+}]}{[\text{CH}_3\text{COOH}]}$$

[H⁺] = 0.01 + 0.05a ≃ 0.01

[HAc] = 0.05(1 – a) ≃ 0.01

$$\therefore\space 1.82 × 10^{\normalsize–5} =\frac{0.01×0.05a}{0.05}\\\alpha=\frac{1.82×10^{\normalsize-5}}{0.01}$$

= 1.82 × 10^–3

In presence of 0.1 M HCl

HAc ⇌ H⁺ + Ac^–

$$\underset{0.05(1 – α)}{\text{HAc}} ⇌\underset{0.05 α}{\text{H}^{+}}+\underset{0.05 α}{\text{Ac}^{-}}\\\underset{0}{\text{HCL}}\xrightarrow{}\underset{0.1\text{M}}{\text{H}^{+}} + \underset{0.1\text{M}}{\text{Cl}^{-}}$$

[H⁺] = 0.1 + 0.05a ≃ 0.1 M

[HAc] = 0.05(1 – a) ≃ 0.05 M

K_a = 1.82 × 10^–5

$$\text{K}_a=\frac{[\text{H}^{+}][\text{Ac}^{-}]}{\text{HAc}}\\1.82 × 10^{–5} =\frac{0.1×0.05α}{0.05}\\\alpha=\frac{1.82×10^{\normalsize-5}}{0.1}=1.82×10^{\normalsize-4}$$

7.54. The ionization constant of dimethylamine is 5.4 × 10^–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Ans. Given that,

K_b = 5.4 × 10^–4

c = 0.02 M

then,

$$α=\sqrt\frac{\text{K}_b}{c}\\=\sqrt\frac{5.4×10^{\normalsize-4}}{0.02}$$

= 0.164

Now, if 0.1 M of NaOH is added to the solution, as NaOH is a strong base, it undergoes complete ionization as :

[(CH₃)₂NH₂⁺] = x

[OH^–] = x + 0.1; 0.1

$$\Rarr\space\text{K}_b=\frac{[(\text{CH}_3)_2 \text{NH}_2^+][\text{OH}^{-}]}{[(\text{CH}_3)_2\text{NH}^{-}]}\\5.4 × 10^{\normalsize–4} =\frac{x×0.1}{0.02}$$

$$\underset{;0.02M}{\underset{(0.02-x)}{(\text{CH}_3)_2\text{NH + H}_2\text{O}}⇌}\underset{}{\underset{x}{(\text{CH}_3)_2\text{NH}_2^+}}\underset{;0.1M}{\underset{x}{\text{OH}}^{-}}$$

x = 0.0054

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

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

(c) Human blood, 7.38

(d) Human saliva, 6.4.

Ans. (a) Human muscle fluid 6.83:

pH = 6.83

As, pH = – log [H⁺]

therefore, 6.83 = – log [H⁺]

[H⁺] = antilog (–6.83)

[H⁺] =1.48 × 10^–7 M

(b) Human stomach fluid, 1.2 :
pH =1.2

1.2 = – log [H⁺]

[H⁺] = antilog (–1.2)

[H⁺] = 0.063 or 6.30 × 10^–2 M

pH = 7.38 = – log [H⁺]

[H⁺] = antilog (–7.38)

[H⁺] = 4.17 × 10^–8 M

(d) Human saliva, 6.4 :

pH = 6.4

6.4 = – log [H⁺]

[H⁺] = antilog –6.4

[H⁺] = 3.98 × 10^–7

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.

Ans. The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = –log [H⁺]

(i) pH of milk = 6.8

Since, pH = –log [H⁺]

6.8 = –log [H⁺]

[H⁺] = antilog (–6.8)

= 1.58 × 10^–7 M

(ii) pH of black coffee = 5.0

Since, pH = –log [H⁺]

5.0 = –log [H⁺]

[H⁺] = antilog (–5.0)

= 1 × 10^–5 M

(iii) pH of tomato juice = 4.2

Since, pH = –log [H⁺]

4.2 = –log [H⁺]

[H⁺] = antilog (–4.2)

= 6.31×10^–5 M

(iv) pH of lemon juice = 2.2

pH = –log [H⁺]

2.2 = –log [H⁺]

[H⁺] = antilog(–2.2)

= 6.31 × 10^–3 M

(v) pH of egg white = 7.8

Since, pH = –log [H⁺]

7.8 = –log [H⁺]

[H⁺] = antilog(–7.8)

= 1.58×10^–8M

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?

Ans. We know that,

$$\text{[KOH(aq)] =}\frac{0.561}{\frac{1}{5}}\text{g/L}\\\text{= 2.805 g/L}\\= 2.805×\frac{1}{56.11}\text{M}$$

= 0.05 M Now, [H⁺][OH^–] = K_w

$$\text{or},\space[\text{H}^{\normalsize+}]=\frac{\text{K}_w}{[\text{OH}^{\normalsize-}]}\\=\frac{10^{\normalsize-14}}{0.05}$$

= 2 × 10^–13

pH = – log [H⁺]

= – log (2 × 10^–13) = 12.70

7.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Ans. Given that,

Solubility of Sr(OH)₂ = 19.23 g/L

Then, concentration of Sr(OH)₂

$$=\frac{19.23}{121.63}\text{M}$$

= 0.158 M

Now,

Sr(OH)₂(aq) → Sr2+(aq) + 2(OH–)(aq)
\ [Sr2+] = 0.1581 M
[OH–] = 2 × 0.1581 M
M = 0.3126 M
And,
[H+][OH–] = Kw

$$\text{Sr(OH)}_2\text{(aq)} \xrightarrow{} \text{Sr}^{2+}\text{(aq)} + 2(\text{OH}^{–})(\text{aq})$$

∴ [Sr²⁺] = 0.1581 M

[OH^–] = 2 × 0.1581 M

M = 0.3126 M

And,

[H⁺][OH^–] = K_w

$$\text{[H}^{+}]=\frac{\text{K}_w}{[\text{OH}^{-}]}\\=\frac{10^{\normalsize-14}}{0.312}$$

= 3.2 × 10^–14

pH = –log [H⁺]

= – log (3.2 × 10^–14)

= 13.49

7.59. The ionization constant of propanoic acid is 1.32 × 10^–5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Ans. Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

$$\underset{(0.05 – 0.05a)≈0.05}{\text{HA}}+\text{H}_2\text{O} ⇌\underset{.05α}{\text{H}_3\text{O}^{+}}+\underset{.05α}{\text{A}^{-}}\\\text{K}_a=\frac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]}\\=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^{2}\\\alpha=\sqrt{\frac{\text{K}_a}{.05}}=1.63×10^{\normalsize-2}$$

Then, [H₃O⁺] = .05α = .05 × 1.63 × 10^–2

∴ pH = 3.09 = K_b, 15 × 10^–4 M

In the presence of 0.1 M of HCl, let a′ be the degree of ionization.

Then [H₃O⁺] = 0.01

[A^–] = 0.05α′

[HA] = 0.05

$$\text{K}_a=0.01×\frac{0.5\alpha'}{0.5}$$

1.32 × 10^–5 = 0.01 × α′

α′ = 1.32 × 10^–3

7.60. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Ans. Given: pH = 2.34 and concentration of acid, c = 0.1 M

[H+] = antilog (–2.34) = 4.57 × 10^–3 M

Ionisation equilibrium and equilibrium concentrations are:

HCNO (aq) ⇌ H⁺(aq) + CNO^– (aq)
0.1(1 – a) 0.1a 0.1a
∴ [H+] = 0.1a = 4.57 × 10–3

$$\underset{0.1(1-\alpha)}{\text{HCNO}(\text{aq})} ⇌\underset{0.1\alpha}{\text{H}^{+}\text{(aq)}}+\underset{0.1\alpha}{\text{CNO}^{-}\text{(aq)}}$$

$$\therefore\space[\text{H}^{+}]=0.1\alpha=457×10^{\normalsize-3}\\\alpha=\frac{4.57×10^{\normalsize -3}}{0.1}$$

= 4.57 × 10^–2 = 0.0457

Since a is small

K_a = ca² = 0.1 × (4.57 × 10^–2)²

= 2.09 × 10^–4

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.

Ans. NaNO₂ is the salt of a strong base (NaOH) and a weak acid (HNO₂).

NO₂^– + H₂O ⇌ HNO₂ + OH^–

$$\text{K}_b=\frac{[\text{HNO}_2][\text{OH}^{-}]}{[\text{NO}_2^{-}]}\\\Rarr\space\frac{\text{K}_w}{\text{K}_a}=\frac{10^{\normalsize-14}}{4.5×10^{\normalsize-4}}=0.22×10^{\normalsize-10}$$

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

[NO₂^–] = 0.04 – x; 0.04

[HNO₂] = x

[OH^–] = x

$$\text{K}_b=\frac{x^{2}}{0.04}=0.22×10^{\normalsize-10}$$

x² = 0.0088 × 10^–10

x = 0.093 × 10^–5

∴ [OH^–] = 0.093 × 10^–5 M

$$[\text{H}_3\text{O}^{\normalsize+}]=\frac{10^{\normalsize-14}}{0.093×10^{\normalsize-5}}=10.75×10^{\normalsize-9}\space\text{M}$$

⇒ pH = – log(10.75 × 10^–9)

= 7.96

Therefore, degree of hydrolysis

$$=\frac{x}{0.04}=\frac{0.093×10^{\normalsize-5}}{0.04}$$

= 2.325 × 10^–5

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

Ans. Given, pH of pyridinium hydrochloride = 3.44

Concentration, C = 0.02 m

We know that,

pH = – log [H⁺]

[H⁺] = antilog (–3.44)

[H⁺] = 3.63 × 10^–4

$$\text{And,} \space \text{K}_\text{h} =\frac{(3.63×10^{\normalsize-4})^{2}}{0.02}$$

K_h = 6.6 × 10^–6

Pyridinium hydrochloride is a weak acid and the

[H⁺] = cα

$$\alpha=\frac{[\text{H}^{+}]}{c}=\frac{(3.63×10^{\normalsize-4})^{2}}{0.02}=0.0182$$

Since it is weak acid, Ostwald’s dilution law is applicable and

K_a = cα² = 0.02 × (0.0182)²

= 6.63 × 10^–6

Pyridine is conjugate base of pyridinium hydrochloride and its dissociation constant is related to that of the latter as:

$$\text{K}_b=\frac{\text{K}_w}{\text{K}_a}=\frac{1×10^{\normalsize-14}}{6.63×10^{\normalsize-6}}=1.5×10^{\normalsize-9}$$

Hence, the ionisation constant of pyridine is 1.5 × 10^–9.

7.63. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF.

Ans. If the salt gives strong acid and strong base, it is a neutral solution, if a salt gives weak base and strong acid, then the solution is acidic and if a salt gives strong base and weak acid, then the solution is basic.

(i) NaCl

$$\text{NaCl} + \text{H}_2\text{O}\xrightarrow{}\space\underset{\text{Strong}\space \text{base}}{\text{NaOH}}+\underset{\text{Strong acid}}{\text{HCl}}$$

Thus, it is a neutral solution.

(ii) KBr

$$\text{KBr} + \text{H}_2\text{O}\xrightarrow{}\space\underset{\text{Strong}\space \text{base}}{\text{KOH}}+\underset{\text{Strong acid}}{\text{HBr}}$$

Thus, it is a neutral solution.

(iii) NaCN

$$\text{NaCN} + \text{H}_2\text{O}\xrightarrow{}\space\underset{\text{weak}\space \text{acid}}{\text{HCN}}+\underset{\text{Strong base}}{\text{NaOH}}$$

Thus, it is a basic solution.

(iv) NH₄NO₃

$$\text{NH}_4\text{NO}_3 + \text{H}_2\text{O}\xrightarrow{}\underset{\text{Weak base}}{\text{NH}_4\text{OH}}+\underset{\text{Strong acid}}{\text{HNO}_3}$$

Thus, it is an acidic solution.

(v) NaNO₂

$$\text{NaNO}_2 + \text{H}_2\text{O}\xrightarrow{}\underset{\text{Strong base}}{\text{NaOH}}+\underset{\text{Weak acid}}{\text{HNO}_2}$$

Thus, it is a basic solution.

(vi) KG

$$\text{KF} + \text{H}_2\text{O}\xrightarrow{}\underset{\text{Strong base}}{\text{KOH}}+\underset{\text{Weak acid}}{\text{HF}}$$

Thus, it is a basic solution.

7.64. The ionization constant of chloroacetic acid is 1.35 × 10^–3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Ans. It is given that K_a for ClCH₂COOH is 1.35 × 10^–3.

Since the acid is weak, Ostwald’s dilution law can be used as :

⇒ K_a = cα²

$$\therefore\space\alpha=\sqrt{\frac{\text{K}_a}{c}}\\=\sqrt{\frac{1.35×10^{\normalsize-3}}{0.1}}\\\text{[}\therefore \text{concentration of acid = 0.1 m]}\\\alpha=\sqrt{1.35×10^{\normalsize-2}}$$

= 0.116

∴ [H⁺] = cα = 0.1 × 0.116

= 0.0116

⇒ pH = – log [H⁺] = 1.94

ClCH₂COONa is the salt of a weak acid i.e., ClCH₂COOH and a strong base i.e., NaOH.

ClCH₂COO– + H₂O ⇌ ClCH₂COOH + OH^–

$$\text{K}_b=\frac{[\text{ClCH}_2\text{COOH}][\text{OH}^{-}]}{[\text{ClCH}_2\text{COO}^{-}]}\\\text{K}_b=\frac{\text{K}_w}{\text{K}_a}\\\text{K}_b=\frac{10^{\normalsize-14}}{1.35×10^{\normalsize-3}}$$

= 0.740 × 10^–11

$$\text{Also},\space\space\text{K}_b=\frac{x^{2}}{0.1}\\\text{(where x is the concentration of OH}^– \text{and ClCH}_2\text{COOH})\\0.740×10^{\normalsize-11}=\frac{x^{2}}{0.1}$$

0.074 × 10^–11 = x²

⇒ x² = 0.74 × 10^–12

x = 0.86 × 10^–6

[OH^–] = 0.86 × 10^–6

$$\therefore\space\text{[H}^{+}]=\frac{\text{K}_w}{0.86×10^{\normalsize-4}}\\=\frac{10^{\normalsize-14}}{0.86×10^{\normalsize-6}}$$

[H⁺] = 1.162 × 10^–8

pH = –log [H⁺]

= – log (1.162 × 10^–18)

= 7.94

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

Ans. Ionic product is given as:

K_w = [H⁺][OH^–]

Let [H⁺] = x.

Since [H⁺] = [OH^–], K_w = x²

⇒ K_w at 310 K is 2.7 × 10^–14,

∴ 2.7 × 10^–14 = x²

$$=\sqrt{2.7×10^{\normalsize-14}}$$

⇒ x = 1.643 × 10^–7

⇒ x = 1.643 × 10^–7 mol L^–1

⇒ [H⁺] = 1.643 × 10^–7

⇒ pH = – log [H⁺]

= – log(1.643 × 10^–7)

= 7 – log 1.643

= 7 – 0.2156

= 6.78

Hence, the pH of neutral water is 6.78.

7.66. Calculate the pH of the resultant mixtures:

(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl

(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂

(c) 10 mL of 0.1 MH₂SO₄ + 10 mL of 0.1 M KOH

$$\textbf{Ans.}\space\text{(a) Moles of H}_3\text{O}^+=25×\frac{0.1}{1000}=0.0025\space\text{mol}\\\text{and Moles of OH}^{\normalsize-} = 10 × 0.2×\frac{2}{1000}\\= 0.004 \space\text{mol}\\\text{Thus excess of OH}^{\normalsize–} = 0.0015 \text{mol}\\\text{[OH}^{\normalsize-}]=\frac{0.0015}{35×10^{\normalsize-3}\text{mol/L}}$$

= 0.0428

Now,

pOH = – log [OH] = – log [0.0428]

= 1.36

Now, pH = 14 – 1.36

= 12.63

$$\text{(b) Moles of H}_3\text{O}^+ =\frac{2×10×0.01}{1000}\\= 0.0002\space\text{mol}\\\text{and Moles of OH}^{\normalsize–}=\frac{2×10×0.01}{1000}$$

= 0.0002 mol

Both H₃O⁺and OH^– are equal, the solution is neutral and pH = 7.

$$\text{(c) Moles of H}_3\text{O}^{\normalsize+} =\frac{2×10×0.1}{1000}=0.002\space\text{mol}\\\text{and Moles of OH}^{\normalsize–} =\frac{10×0.1}{1000}=0.001\space\text{mol}\\\text{Thus excess of H}_3\text{O}^{\normalsize+}=0.001\space\text{mol}\\\text{[\text{H}}_3\text{O}^{\normalsize+}]=\frac{0.001}{20×10^{\normalsize-3}\text{mol/L}}$$

= 0.05

Now,

pH = – log (0.05)

= 1.30

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 below. Determine also the molarities of individual ions. (i)K_{SPAgCrO().,24111012=}×− KSPBaCrO().,4121010=×− KSPFeOH[()].,3101038=×− KSPPbCl().,216105=×− KSPHgI()..22451029=×− Ans. (1) Silver Chromate

$$\text{Ag}_2\text{CrO}_4\xrightarrow{} 2\text{Ag}^{\normalsize+}+\text{CrO}_4^{2-}\\\text{Then},\\\text{K}_{\text{sp}}=[\text{Ag}^{\normalsize+}]^{2}[\text{CrO}_4^{2-}]\\\text{Let ‘s’ be the solubility of Ag}_2\text{CrO}_4\\s=\bigg(\frac{\text{K}_{sp}}{x^{x}y^{y}}\bigg)^{\frac{1}{x+y}}$$

For Ag₂CrO₄,

x = 2, y = 1 and x + y = 3;

K_sp = 1.1 × 10^–12

$$\therefore\space s=\bigg(\frac{1.1×10^{\normalsize-12}}{2^{2}×1^{2}}\bigg)^{\frac{1}{3}}\\=\bigg(\frac{1.1×10^{\normalsize-12}}{4}\bigg)^{\frac{1}{3}}$$

= (0.275 × 10^–12)^1/3

= 6.5 × 10^–5 M

Molarity of [Ag⁺] = 2[Ag₂CrO₄]

= 2 × s = 2 × 6.5 × 10^–5

= 1.3 × 10^–4 M

Molarity of [CrO₄^2–] = 2[Ag₂CrO₄]

= s = 6.5 × 10^–5 M

(2) Barium Chromate

$$\text{BaCrO}_4\xrightarrow{}\text{Ba}^{2+}+\text{CrO}_4^{2-}$$

Let s be the solubility of BaCrO₄

x = 1, y = 1, x + y = 2, K_sp = 1.2 × 10^–10

$$s=\bigg(\frac{\text{K}_{sp}}{x^{x}y^{y}}\bigg)^{\frac{1}{x+y}}$$

= (1.2×10^-10)^1/2

= 1.1 × 10^–5 M

From the formula,

Molarity of [Ba²⁺] = [CrO₄^2–] = s = 1.1 × 10^–5 M

(3) Ferric Hydroxide

Fe(OH)₃ → Fe²⁺ + 3OH^–

K_sp = [Fe²⁺][OH^–]³

Let s be the solubility of Fe(OH)₃

Thus, [Fe³⁺] = s and [OH^–] = 3s

⇒ K_sp = s.(3s)³

= s.27 s³

K_sp = 27s⁴

1.0 × 10^–38 = 27s⁴

.037 × 10^–38 = s⁴

.00037 × 10^–36 = s⁴

⇒ 1.39 × 10^–10 M = S

Molarity of Fe³⁺ = s = 1.39 × 10^–10 M

Molarity of OH^– = 3s = 4.17 × 10^–10 M

BaCrO₄ → Ba²⁺ + CrO₄^2–

Then, K_sp = [Ba²⁺][CrO₄^2–]

Let s be the solubility of BaCrO₄.

Thus, [Ba²⁺] = s and [CrO₄^2–] = s

⇒ K_sp = s²

⇒ 1.2 × 10^–10 = s²

⇒ s = 1.09 × 10^–5 M

(4) Lead chloride

$$\text{PbCl}_2\xrightarrow{}\text{Pb}^{2\normalsize+}+2\text{Cl}^{\normalsize-}$$

K_sp = [Pb²⁺][Cl^–]²

Let Ksp be the solubility of PbCl₂.

[PB²⁺] = s and [Cl^–] = 2s

Thus, K_sp = s.(2s)²

= 4s³

⇒ 1.6 × 10^–5 = 4s³

⇒ 0.4 ×10^–5 = s³

4 × 10^–6 = s³

⇒ 1.58 × 10^–2M = S.1

Molarity of Pb²⁺ = s = 1.58 × 10^–2 M

Molarity of chloride = 2s
= 3.16 × 10^–2 M

(5) Mercurous Iodide

$$\text{Hg}_2\text{I}_2\xrightarrow{}\text{Hg}^{2+}+2\text{I}^{-}\\\text{K}_{sp}=[\text{Hg}_2^{2+}][\text{I}^{-}]^{2}$$

Let s be the solubility of Hg₂I₂

⇒ [Hg₂²⁺] = s and [I^–] = 2s

Thus, K_sp = s.(2s)²

⇒ K_sp = 4s³

⇒ 4.5 × 10^–29 = 4s³

⇒ 1.125 ×10^–29 = s³

⇒ s = 2.24 × 10^–10 M

Molarity of Hg₂²⁺ = s = 2.24 × 10^–10 M

Molarity of I^– = 2s = 4.48 × 10^–10 M

7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10^–12and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans. Let s be the solubility of Ag²CrO⁴.

Then, Ag₂CrO₄ ⇌ Ag²⁺ + 2CrO₄^–

K_sp = (2s)².s = 4s³

1.1 × 10^–12 = 4s³

s = 6.5 × 10^–5 M

Let s be the solubility of AgBr.

$$\text{AgBr(s)}\xrightarrow{}\text{Ag}^{\normalsize+}+\text{Br}^{\normalsize -}$$

K_sp = s′² = 5.0 × 10^–13

∴ s′ = 7.07 × 10^–7 M

Now, ratio of their Molarities

$$\frac{s}{s'}=\frac{6.5×10^{\normalsize-5}\text{M}}{7.07×10^{\normalsize-7}\text{M}}=91.9$$

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 K_sp = 7.4 × 10^–8).

Ans. When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Then,

$$\underset{0.001\text{M}}{\text{NaIO}_3}\xrightarrow{}\text{Na}^{\normalsize+}+\underset{0.001\text{M}}{\text{IO}_3^{\normalsize-}}\\\underset{0.001\text{M}}{\text{Cu}(\text{ClO}_3)_2}\xrightarrow{}\underset{0.001\text{M}}{\text{Cu}^{2+}+2\text{ClO}_3^{-}}$$

Now, the solubility equilibrium for copper iodate can be written as:

$$\text{Cu(IO}_3)_2\xrightarrow{}\text{Cu}^{2+}(\text{aq})+\text{2IO}_3^{\normalsize -}\text{(aq)}$$

Ionic product of copper iodate

= [Cu²⁺][IO₃^–]²

= (0.001)(0.001)²

= 1 × 10^–9

Since the ionic product (1 × 10^–9) is less than K_sp (7.4 × 10^–9), precipitation will not occur.

7.70. The ionization constant of benzoic acid is 6.46 × 10^–5 and K_sp 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?

Ans. Since pH = 3.19,

[H₃O⁺] = 6.46 × 10^–4 M

C₆H₅COOH + H₂O ⇌ C₆H₅COO^– + H₃O

$$\text{K}_a=\frac{[\text{C}_6\text{H}_5\text{COO}^{-}][\text{H}_3\text{O}^{+0}]}{[\text{C}_6\text{H}_5\text{COOH}]}\\\frac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^{-}]}=\frac{[\text{H}_3\text{O}^{+}]}{\text{K}_a}\\=\frac{6.46×10^{\normalsize-4}}{6.46×10^{\normalsize-5}}=10$$

Let the solubility of C₆H₅COOAg be x mol/L.

Then,

[Ag⁺] = x

[C₆H₅COOH] + [C₆H₅COO^–] = x

10 [C₆H₅COO^–] + [C₆H₅COO^–] = x

$$\text{[C}_6\text{H}_5\text{COOH]} =\frac{x}{11}\\\text{K}_{sp}[\text{Ag}^{+}][\text{C}_6\text{H}_5\text{COO}^{-}]\\2.5×10^{\normalsize-13}=x\bigg(\frac{x}{11}\bigg)$$

x = 1.66 × 10^–6 mol/L

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10^–6 mol/L. Now, let the solubility of C₆H₅COOAg be x’ mol/L.

Then, [Ag⁺] = x′ M and [CH₃COO^–] = x′ M

K_sp = [Ag⁺][CH₃COO^–]

K_sp = (x′)²

$$x=\sqrt{\text{K}_{sp}}=\sqrt{2.5×10^{\normalsize-13}}\\= 5 × 10^{\normalsize–7} \text{mol/L}\\\therefore\space\frac{x}{x'}=\frac{1.66×10^{\normalsize-6}}{5×10^{\normalsize-7}}=3.32$$

Hence, C₆H₅COOAg is approximately 3.317 times more soluble in a low pH solution.

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_sp = 6.3 × 10^–18).

Ans. Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2.

$$\therefore\space[\text{FeSO}_4]=[\text{Na}_2\text{S}]=\frac{x}{2}\text{M}\\\text{Then},\space[\text{Fe}^{2+}]=[\text{FeSO}_4]=\frac{x}{2}\text{M}\\\text{Also},\space[\text{S}^{2-}]=[\text{Na}_2\text{S}]=\frac{x}{2}\text{M}\\\text{FeS(s)} ⇌ \text{Fe}^{2+}\text{(aq)} + \text{S}^{2\normalsize–}\text{(aq)}\\\text{K}_{sp}=[\text{Fe}^{2+}][\text{S}^{2-}]\\6.3×10^{\normalsize-18}=\bigg(\frac{x}{2}\bigg)\bigg(\frac{x}{2}\bigg)\\\frac{x^{2}}{4}=6.3×10^{\normalsize-18}$$

⇒ x = 5.02 × 10^–9

Thus, if the concentrations of both solutions are equal to or less than 5.02 × 10^–9 M, then there will be no precipitation of iron sulphide.

7.72. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10^–6).

Ans. The dissociation of calcium sulphate can be given as:

CaSO₄(s) ⇌ Ca²⁺ (aq) + SO₄^2– (aq)

K_sp = [Ca²⁺][SO₄^2–]

Let the solubility of CaSO₄ be s.

Then, K_sp = s²

9.1 × 10^–6 = s²

s = 3.02 × 10^–3 mol/L

Molecular mass of CaSO₄ = 136 g/mol

Solubility of CaSO₄ in gram/L

= 3.02 × 10^–3 × 136

= 0.41 g/L

This means that we need 1 L of water to dissolve 0.41g of CaSO₄.

$$\text{Therefore, to dissolve 1 g of CaSO}_4 \text{we require water} =\frac{1}{0.41}\text{L}=2.44\text{L of water}$$

7.73. The concentration of sulphide ion in 0.1 M 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: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

Ans. For precipitation to take place, it is required that the calculated ionic product exceeds the K_sp value.

Before mixing:
[S^2–] = 1.0 × 10^–19 M [M²⁺] = 0.04 M

Volume = 10 mL volume = 5 mL

After Mixing

[S^2–] = ? [M²⁺] = ?

Volume = (10 + 5) = 15 mL

Volume = 15 mL

$$[\text{S}^{2-}]=\frac{1.0×10^{\normalsize-19}×10}{15}\\=6.67×10^{-20}\text{M}\\\text{[M}^{2+}]=\frac{0.04×5}{15}$$

= 1.33 × 10^–2 M
Ionic product = [M²⁺][S^2–]
= (1.33 × 10^–2)(6.67 × 10^–20)
= 8.87 × 10^–22
This ionic product exceeds the K_sp of ZnS and CdS. Therefore, precipitation will occur in CdCl₂ and ZnCl₂ solutions.