NCERT Solutions for Class 11 Chemistry Chapter 7: Equilibrium

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) โ‡Œ 2SO3(g)

Ans. The equilibrium constant (Kc) 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 Mโ€“1 (approx)

Therefore, Kc for the equilibrium is 12.23 Mโ€“1

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

I2(g) โ‡Œ 2I(g)

Calculate Kp for the equilibrium.

Ans. Total pressure = Ptotal = 105 Pa

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

โˆด 60% are I2 molecules (gaseous)

Partial pressure of I(g) atoms,

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

= 4 ร— 104 Pa

Partial pressure of I2 molecules,

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

= 6 ร— 104 Pa

Now, for the given reaction,

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

= 2.67 ร— 104 Pa

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

(i) 2NOCl (g) โ‡Œ 2NO (g) + Cl2 (g)

(ii) 2Cu(NO3)2(s) โ‡Œ 2CuO(s) + 4NO2(g) + O2(g)

(iii) CH3COOC2H5(aq) + H2O(l) โ‡Œ CH3COOH
(aq) + C2H5OH(aq)

(iv) Fe3+(aq) + 3OHโ€“(aq) โ‡Œ Fe(OH)3(s)

(v) I2(s) + 5F2 โ‡Œ 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 Kp:

(i) 2NOCl(g) โ‡Œ 2NO(g) + Cl2(g); Kp = 1.8 ร— 10โ€“2 at 500 K

(ii) CaCO3(s) โ‡Œ CaO(s) + CO2(g); Kp = 167 at 1073 K

Ans. (i) 2NOCl(g) โ‡Œ 2NO(g) + Cl2(g)

The relation between Kp and Kc given as:

Kp = Kc (RT)ฮ”n

In this reaction,

ฮ”n = 3 โ€“ 2 = 1

Given values:

R = 0.0831 bar L molโ€“1 Kโ€“1

T = 500 K

Kp = 1.8 ร— 10โ€“2

Kp = Kc(RT)ฮ”n

1.8 ร— 10โ€“2 = Kc (0.0831 ร— 500)1

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

= 4.33 ร— 10โ€“4 (approx)

(ii) In this reaction,

CaCO3(s) โ‡Œ CaO(s) + CO2(g)

ฮ”n = 2 โ€“ 1 = 1

Given values:

R = 0.0831 bar L molโ€“1 Kโ€“1

T = 1073 K,

Kp = 167

Kp = Kc (RT)ฮ”n

โ‡’ 167 = Kc (0.0831 ร— 1073 )1

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

7.6. For the following equilibrium, Kc = 6.3 ร— 1014 at 1000 K
NO(g) + O3(g) โ‡Œ NO2(g) + O2(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is โ€ฒKc for the reverse reaction?

Ans. It is given that โ€ฒKc for the forward reaction is
6.3 ร— 1014 .

Then, Kc for the reverse reaction will be,

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

= 1.59 ร— 10โ€“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,

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

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) โ‡Œ 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 ร— 10โ€“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) โ‡Œ 2N2O (g)

Initial moles 0.482 0.933 0
Eqm. moles (0.482 โ€“ 2x) (0.933 โ€“ 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:

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

The value of equilibrium constant i.e. Kc = 2.0 ร— 10โˆ’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,

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

โ‡’ x2 = 43.35 ร— 10โ€“40

โ‡’ x = 6.6 ร— 10โ€“20

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

The composition of equilibrium mixture = 6.6 ร— 10โ€“21.

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

2NO(g) + Br2(g) โ‡Œย  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:

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

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 โ€“ 0.0518

= 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 โ€“ 0.0259

= 0.0178 mol

7.10. At 450K, Kp = 2.0 ร— 1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) โ‡Œ 2SO3(g)

What is Kc at this temperature?

Ans. For the given reaction,

ฮ”n = 2 โ€“ 3 = โ€“ 1

T = 450 K

R = 0.0831 bar L molโ€“1 Kโ€“1

Kp = 2.0 ร— 1010 bar โ€“1

We know that,

Kp = Kc(RT)ฮ”n

โ‡’ (2.0 ร— 1010) barโ€“1 = Kc(0.0831 L bar Kโ€“1molโ€“1 ร— 450 K)โ€“1

โ‡’ Kc = 2.0 ร— 1010 barโ€“1 ร— 0.0831 L bar Kโ€“1 molโ€“1 ร— 450 K

= 74.79 ร— 1010 L molโ€“1

= 7.48 ร— 1011 L molโ€“1

= 7.48 ร— 1011 Mโ€“1

Thus, the value of Kc at 450 K is 7.48 ร— 1011 Mโ€“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 Kp for the given equilibrium?

2HI(g) โ‡Œ 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 โ€“ 0.04 = 0.16.

The given reaction is:

2HI(g)ย  โ‡Œย 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

$$\text{Therefore,}\space \text{K}_p = \text{PH}_2ร—{\text{P}}_{\text{H}_2}ร—\frac{\text{P}_{I_{2}}}{\text{P}^{2}_{\text{HI}}}\\=0.08ร—\frac{0.08}{(0.04)^{2}}\\=\frac{0.0064}{0.0016}\\=4.0 $$

Thus, the value of Kp 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) โ‡Œ 2NH3(g) is 1.7 ร— 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) โ‡Œ 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 Qc 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 ร— 103

Since, Qc โ‰  Kc (1.7 ร— 102) , the reaction mixture is not at equilibrium. Again, Qc (2.4 ร— 103) > Kc (1.7 ร— 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) โ‡Œ 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) โ‡Œ H2(g) + CO2(g)

Calculate the equilibrium constant for the reaction.

Ans. The given reaction is:

H2O(g) + CO(g) โ‡Œ 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,

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

= 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) โ‡Œ 2HI(g)

is 54.8. If 0.5 mol Lโ€“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 Kc for the given reaction is 54.8.

H2(g) + I2(g) โ‡Œ 2HI(g)

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

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

[HI] = 0.5 mol Lโ€“1

Let the concentrations of hydrogen and iodine at equilibrium be x mol Lโ€“1

[H2] = [I2] = x mol Lโ€“1

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

โ‡’ x = 0.06754

x = 0.068 mol Lโ€“1 (approx)

Hence, at equilibrium, [H2] = [I2] = 0.068 mol Lโ€“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) โ‡Œย  I2(g) + Cl2(g); Kc = 0.14

Ans. The given reaction is:

2ICl(g) โ‡Œย I2(g) + Cl2(g)

Initial concen. 0.78 M 0 0
At equilibrium (0.78 โ€“ 2x)M xM xM

Now, we can write,

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

โ‡’ x = 0.292 โ€“ 0.748x

โ‡’ 1.748 x = 0.292

โ‡’ x = 0.167

Hence, at equilibrium,

[I2] = [Cl2] = 0.167 M

[ICl] = (0.78 โ€“ 2 ร— 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) โ‡Œ 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) โ‡Œ C2H4(g) + H2(g)

Initial concen. 4 atm 0 0
At equilibrium (4 โ€“ p)M pM pM

Now, we can write,

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

โ‡’ p2 = 0.16 ร— 0.04 p

โ‡’ p2 + 0.04p โ€“ 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 โ€“ p

= 4 โ€“ 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) โ‡Œ CH3COOC2H5(l) + H2O(l)

(i) Write the concentration ratio (reaction quotient), Qc, 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) โ‡Œ 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:

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

= 3.919

= 3.92 (approx)

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

CH3COOH(l) +ย  C2H5OH(l) โ‡Œ 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,

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

= 0.2037

= 0.204 (approx)

Since Qc< Kc, 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 ร— 10โ€“1 mol Lโ€“1. If value of Kc is 8.3 ร— 10โ€“3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5(g) โ‡Œ PCl3(g) + Cl2(g)

Ans. Let the concentration of both PCl3 and Cl2 at equilibrium be ร— mol Lโ€“1. The given reaction is:

PCl5(g) โ‡Œ PCl3(g) + Cl2(g)

At equilibrium 0.5 ร— 10โ€“1 x mol x mol
mol Lโ€“1 Lโ€“1 Lโ€“1

Given that the value of equilibrium constant, Kc is 8.3 ร— 10โ€“3.

Now we can write the expression for equilibrium as:

$$\text{K}_c=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\\\Rarr\space 8.3 ร— 10^{\normalsizeโ€“3} =\frac{xร—x}{0.5ร—10^{\normalsize-1}}$$

โ‡’ x2 = 4.15 ร— 10โ€“4

โ‡’ x = 2.04 ร— 10โ€“2

=ย  0.0204

= 0.02 (approx)

Therefore, at equilibrium, [PCl3] = [Cl2] = 0.02 mol Lโ€“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) โ‡Œ Fe(s) + CO2(g);

Kp = 0.265 atm at 1050K

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: PCO = 1.4 atm and PCO2 = 0.80 atm.

Ans. For the given reaction,

FeO(s) + CO(g) โ‡Œ Fe(s) + CO2(g)

Initially, 1.4 atm 0.80 atm
At eq. 1.4 + p 0.8 โ€“ 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 Kp = 0.265

Since Qp > Kp, the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.

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

So,

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

โ‡’ 0.371 + 0.265 p = 0.80 โ€“ p

โ‡’ 1.265 p = 0.429

โ‡’ p = 0.339 atm

Therefore, equilibrium partial pressure of CO2, PCO2 is 0.80 โ€“ 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) โ‡Œ 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 Lโ€“1N2, 2.0 mol Lโ€“1 H2 and 0.5 mol Lโ€“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) โ‡Œ 2NH3(g)

At particular time 3.0 mol Lโ€“1ย  2.0 mol Lโ€“1ย  0.5 mol Lโ€“1ย time

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 Kc = 0.061.

Since Qc โ‰  Kc, the reaction is not at equilibrium.

Since Qc < Kc, 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) โ‡Œ Br2(g) + Cl2(g) for which Kc = 32 at 500 K.

If initially pure BrCl is present at a concentration of 3.3 ร— 10โ€“3 mol Lโ€“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) โ‡Œ Br2(g) + Cl2(g)

Initial con. 3.3 ร— 10โ€“3 0 0
At equilibrium (3.3 ร— 10โ€“3 โ€“ 2x) x x

Now,

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

โ‡’ x = 18.678 ร— 10โ€“3 โ€“ 11.32x

โ‡’ 12.32x = 18.67 ร— 10โ€“3

โ‡’ x = 1.5 ร— 10โ€“3

Therefore, at equilibrium,

[BrCl] = (3.3ร—10โ€“3 โ€“ 2x)

= 3.3 ร— 10โ€“3 โ€“ (2 ร— 1.5 ร— 10โ€“3)

= 3.3 ร— 10โ€“3โ€“ 3.0 ร— 10โ€“3

= 0.33 ร— 10โ€“3

= 3.3 ร— 10โ€“4 mol Lโ€“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) โ‡Œ 2CO(g)

Calculate Kc 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 โ€“ 90.55) = 9.45 g

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

= 0.938 atm

Partial pressure of CO2,ย 

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

= 0.062 atm

Kp for the reaction C(s) + CO2(g) โ‡Œ 2CO(g)

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

Now โˆ†ng = 2 โ€“ 1 = 1, Kp = Kc(RT)โˆ†ng

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

7.24. Calculate

(A) โˆ†Gยฐ and

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

NO(g) + ยฝO2(g) โ‡Œ NO2(g)

where โˆ†fGยฐ (NO2) = 52.0 kJ/mol

โˆ†fGยฐ (NO) = 87.0 kJ/mol

โˆ†fGยฐ (O2) = 0 kJ/mol
Ans. (A) For the given reaction,

โˆ†fGยฐ = โˆ†Gยฐ(Products) โ€“ โˆ†Gยฐ(Reactants)

โˆ†rGยฐ = 52.0 โ€“ [87 + 0]

= โ€“ 35.0 kJ molโ€“1

(B) We know that,

โˆ†Gยฐ = โ€“ RT ln Kc

โˆ†Gยฐ = โ€“ 2.303 RT log Kc

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

= 6.134

โˆด Kc = antilog (6.134)

= 1.36 ร— 106

= 1.36 ร— 106

Hence, the equilibrium constant for the given reaction Kc i s 1.36 ร— 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) โ‡Œย  PCl3(g) + Cl2(g)

(b) CaO(s) + CO2(g) โ‡Œย  CaCO3(s)

(c) 3Fe(s) + 4H2O(g) โ‡Œย  Fe3O4(s) + 4H2(g)

Ans. (a) According to Le Chatelierโ€™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 Chatelierโ€™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 Chatelierโ€™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) โ‡Œย  CO(g) + Cl2(g)

(ii) CH4(g) + 2S2(g) โ‡Œย  CS2(g) + 2H2S(g)

(iii) CO2(g) + C(s) โ‡Œย  2CO (g)

(iv) 2H2(g) + CO(g) โ‡Œย  CH3OH (g)

(v) CaCO3(s) โ‡Œย  CaO(s) + CO2(g)

(vi) 4NH3(g) + 5O2 (g) โ‡Œย  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 Chatelierโ€™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 ร—105 at 1024 K

H2(g) + Br2(g) โ‡Œ 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,

Kp for the reaction,

H2(g) + Br2(g) โ‡Œ 2HBr(g) is 1.6 ร—105at 1024K Therefore, for the reaction,

2HBr(g) โ‡Œ H2(g) + Br2(g), the equilibrium constant will be,

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

= 6.25 ร— 10โ€“6

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

2HBr(g) โ‡Œ H2(g) + Br2(g)

Initial concentration 10 0 0
At equilibrium 10 โ€“ 2p p p

Now,

$$\frac{P_{\text{H}_2ร—P_{\text{Br}_2}}}{P^{2}_{\text{HBr}}}=\text{K}'p\\=\frac{p^{2}}{(10-2p)^{2}}=6.25ร—10^{\normalsize-6}\\\frac{p}{(10-2p)}=2.5ร—10^{\normalsize-3}$$

p = 2.5 ร— 10โ€“3 (10 โ€“ 2p)

p = 2.5 ร— 10โ€“2 โ€“ (5.0ร—10โ€“3)p

p + (5.0ร—10โ€“3)p = 2.5 ร— 10โ€“2

1.005p = 2.5 ร— 10โ€“2

p = 2.49 ร— 10โ€“2 bar

= 2.5 ร— 10โ€“2 bar (approx)

Therefore, at equilibrium,

[H2] = [Br2] = 2.49 ร— 10โ€“2 bar

[HBr] = 10 โ€“ 2 ร— (2.49 ร— 10โ€“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) โ‡Œย  CO(g) + 3H2(g)

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

(b) How will the values of Kpย  and 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,

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

(b) (i) According to Le Chatelierโ€™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 Chatelierโ€™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) โ‡Œ CH3OH(g)

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

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

(c) According to Le Chatelierโ€™s Principle, on removal of CO, equilibrium will shift in the backward direction.

(d) According to Le Chatelierโ€™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 ร— 10โ€“3. If decomposition is depicted as,

PCl5 (g) โ‡Œย  PCl3 (g) + Cl2 (g)

ฮ”rHโŠ– ย  = 124.0 kJ molโ€“1

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

(b) what is the value of Kc for the 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) 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 Kc for the reverse reaction at the same temperature is:

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

= 1.2048 ร— 102

(c) (i) Kc would remain the same because here, the temperature remains the same.

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

(iii) In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.

7.31. Dihydrogen gas used in Haberโ€™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) โ‡Œ CO2(g) + H2(g)

If a reaction vessel at 400ยฐC is charged with an equimolar mixture of CO and steam such that PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400ยฐC.

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

The given reaction is:

CO(g) + H2O (g) โ‡Œย  CO2(g) + H2(g)

Initial con. 4.0 bar 4.0 bar 0 0
At eqil. 4.0 โ€“ p 4.0 โ€“ p p p

It is given that,

Kp = 10.1

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

โ‡’ p = 12.712 โ€“ 3.178 p

โ‡’ 4.178 p = 12.712

โ‡’ 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) โ‡Œย  2Cl (g) Kc = 5 ร—10โ€“39

(b) Cl2(g) + 2NO(g) โ‡Œย  2NOCl(g) Kc = 3.7 ร— 108

(c) Cl2(g) + 2NO2(g) โ‡Œย  2NO2Cl (g) Kc = 1.8

Ans. If the value of Kc lies between 10โ€“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) โ‡Œย  2O3 (g) is 2.0 ร—10โ€“50 at 25ยฐC.

If the equilibrium concentration of O2 in air at 25ยฐC is 1.6 ร—10โ€“2, what is the concentration of O3?

Ans. Given:

Kc = 2.0 ร— 10โ€“50

[O2(g)] = 1.6 ร—10โ€“2

For the reaction,

3O2(g) โ‡Œย  2O3(g)

Equilibriumย  ย  ย  1.6 ร— 10โ€“2ย  ย  x

concentrations/M

$$\text{K}_c=\frac{[\text{O}_3]^{2}}{[\text{O}_3]^{3}}\\2.0 ร— 10^{\normalsizeโ€“50} =\frac{x^{2}}{(1.6ร—10^{\normalsize-2})^3}$$

x2 = 2.0 ร— 10โ€“50 ร— (1.6 ร— 10โ€“2)3

= 8.192 ร— 10โ€“56

orย  ย  ย x = 2.86 ร— 10โ€“28 M

Hence, the concentration of O3 is 2.86 ร— 10โ€“28 M

7.34. The reaction,

CO(g) + 3H2(g) โ‡Œย  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, Kc for the reaction at the given temperature is 3.90?

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

CO(g) + 3H2(g) โ‡Œ 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:

Kc = 3.90.

Therefore,

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

= 0.0585 M

= 5.85 ร— 10โ€“2 M

Therefore, the concentration of CH4 at equilibrium is 5.85 ร— 10โ€“2 M.

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

HNO2, CNโ€“, HClO4, Fโ€“, OHโ€“, CO32โ€“, and S2โ€“

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 NO2โ€“1 (base)
CNโ€“ HCN (acid)
HClO4 ClO4โ€“ (base)
Fโ€“ HF (acid)
OHโ€“ H2O (acid) / O2โ€“ (base)
CO32โ€“ HCO3โ€“ (acid)
S2โ€“ HSโ€“ (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 Brรถnsted acids: HF, H2SO4 and HCO3โ€“ ?

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

Bronsted Acid Conjugate Base
HF F-
H2SO4 HSO4โ€“
HCO3โ€“ CO32-

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

NH2โ€“ , NH3and HCOOโ€“.

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

Bronsted Base Conjugate Acid
NH2โ€“ NH3
NH3 NH4+
HCOOโ€“ HCOOH

7.39. The species: H2O, HCO3โ€“, HSO4โ€“ and NH3 can act both as Brรถnsted 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+ OHโ€“
HCO3โ€“ H2CO3 CO32โ€“
HSO4โ€“ H2SO4 SO42โ€“
NH3 NH4+ NH2โ€“

7.40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

(a) OHโ€“

(b) Fโ€“

(c) H+

(d) BCl3.

Ans. (a) OHโ€“ can donate its lone pair of electrons. So, it is a Lewis base.

(b) Fโ€“ can donate its lone pair of electrons. So, it is a Lewis base.

(c) H+ can accept a lone pair of electrons. So, it is a Lewis acid.

(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 ร— 10โ€“3 M. what is its pH?
Ans. Given,

Concentration of hydrogen ion [H+] in a sample of soft drink = 3.8 ร— 10โ€“3 M

โˆดย  pH Value of soft drink = โ€“ log [H+]

= โ€“ log (3.8 ร— 10โ€“3)

= โ€“ log 3.8 โ€“ log 10โ€“3

= โ€“ log 3.8 + 3

= โ€“ 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 = โ€“ log [H+]

โ‡’ log [H+] = โ€“ pH

โ‡’ [H+] = antilog (โ€“pH)

= antilog (โ€“ 3.76)

= 1.74 ร— 10โ€“4 M

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 ร— 10โ€“4 M.

7.43. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 ร— 10โ€“4, 1.8 ร— 10โ€“4 and 4.8 ร— 10โ€“9 respectively. Calculate the ionization constants of the corresponding conjugate base?

Ans. It is known that,

$$\text{K}_b=\frac{\text{K}_{w}}{\text{K}_{a}}\\\text{Given, K}_a \text{of HF} = 6.8 ร— 10^{โ€“4}\\\text{Hence, K}_\text{b}\space\text{of its conjugate base Fโ€“ =}\frac{\text{K}_{w}}{\text{K}_{a}}\\=\frac{1ร—10^{\normalsize-14}}{6.8ร—10^{\normalsize-4}}$$

= 1.5 ร— 10โ€“11

Given, Ka of HCOOH = 1.8 ร— 10โ€“4

$$\text{Hence, K}_\text{b}\space\text{of its conjugate base HCOOโ€“ =}\frac{\text{K}_w}{\text{K}_a}\\=\frac{1ร—10^{\normalsize-14}}{4.8ร—10^{\normalsize-9}}$$

= 2.08 ร— 10โ€“6

7.44. The ionization constant of phenol is 1.0 ร— 10โ€“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

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

Initial con. 0.05 0 0
At equilibrium 0.05 โ€“ x x x

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

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

$$\\\text{K}_a = \frac{xร—x}{0.05-x}$$

โˆด x2 = 1 ร—10โ€“10 ร— 0.05

= 5 ร— 10โ€“12

= 2.2 ร— 10โ€“6 M = [H3O+]

Since [H3O+] = [C6H5Oโ€“],

[C6H5Oโ€“] = 2.2 ร— 10โ€“6 M

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

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

Concentration

Also,

C6H5OH + H2O โ†’ C6H5Oโ€“ + H3O+

Concentration 0.05 โ€“ 0.05 ฮฑ 0.01+0.05ฮฑ 0.05 ฮฑ

[C6H5OH] = 0.05 โ€“ 0.05 ฮฑ ; 0.05 M

[C6H5Oโ€“] = 0.01 + 0.05 ฮฑ ; 0.01 M

[H3O+] = 0.05 ฮฑ

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

1.0 ร— 10 โ€“10 = 0.01ฮฑ

ฮฑ = 1 ร— 10 โ€“8

7.45. The first ionization constant of H2S is 9.1 ร— 10โ€“8. Calculate the concentration of HSโ€“ 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 ร— 10โ€“13, calculate the concentration of S2โ€“ under both conditions.

Ans. To calculate the concentration of HS โ€“ ion:

Case I (in the absence of HCl):

Let the concentration of HS be x M.

H2S โ†’ H+ + HS

Cฮด 0.1 0 0
Cf 0.1-x x x

Then,

$$\text{K}_{a_1}=\frac{[\text{H}^{+}][\text{HS}^{-}]}{[\text{H}_2\text{S}]}\\9.1 ร— 10^{โ€“8} =\frac{(x)(x)}{0.1-x}$$

(9.1 ร— 10โ€“8) (0.1 โ€“ x) = x2

Taking (0.1 โ€“ x) M as 0.1 M

We have,

(9.1 ร— 10โ€“8) (0.1) = x2

(9.1 ร— 10โ€“8) = x2

x = (9.1 ร— 10โ€“9)1/2

x = 9.54 ร— 10โ€“5 M

โ‡’ [HSโ€“] = 9.54 ร— 10โ€“5 M

Case II (in the presence of HCl): In the presence of 0.1 M of HCl, let [HSโ€“] be y M.

Then,

$$\text{H}_2\text{S} \xrightarrow{} \text{HS}^โ€“ + \text{H}^โ€“$$

Cยด 0.1 0 0
Cf 0.1 โ€“ y y y

Also,

$$\text{HCl} โ‡Œ \underset{0.1}{\text{H}^{+}}+\underset{0.1}{\text{Cl}^{-}}$$

Now,

$$\text{K}_{a_1}=\frac{[\text{HS}^{-}][\text{HS}^{+}]}{[\text{H}_2\text{S}]}\\\text{K}_{\text{a}_1}=\frac{[y](0.1+y)}{((0.1-y))}\\9.1 ร— 10^{โ€“8} =\frac{yร—0.1}{0.1}$$

(โˆต 0.1 โ€“ y = 0.1 M and 0.1 + y = 0.1 M )

9.1 ร— 10โ€“8 = y

โ‡’ [HSโ€“] = 9.1 ร— 10โ€“8

Case III : To calculate [S2โ€“] in absence of 0.1 M HCl

H2Sโ€“ โ‡Œ H+ + S2โ€“

From case I [HSโ€“] = [H+] = 9.54 ร— 10โ€“5 N

[S2โ€“] = x

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

x = 1.2 ร— 10โ€“13 M = [S2โ€“]

Case IV : To calculate [S2โ€“] in presence of 0.1 M HCl

HSโ€“ โ‡Œ H+ + S2โ€“

From case II [HSโ€“] = 9.1 ร— 10โ€“8M

[H+] = 0.1 M (from HCl)

[S2โ€“] = x

$$\text{K}_{a_2}=\frac{[\text{H}^{+}]+[\text{S}^{2-}]}{\text{[HS}^{-}]}=\frac{0.1ร—x}{9.1ร—10^{\normalsize -8}}\\1.2 ร— 10^{โ€“13} =\frac{0.1ร—x}{9.1ร—10^{\normalsize-8}}$$

x = 1.09 ร— 10โ€“19 M = [S2โ€“]

7.46. The ionization constant of acetic acid is 1.74 ร— 10โ€“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 โ‡Œ CH3COOโ€“ + H+

Ka = 1.74 ร— 10โ€“5

(2) H2O + H2O โ‡Œ H3O+ + OHโ€“ Kw = 1.0 ร— 10โ€“14

Since Ka >> Kw :

CH3COOH + H2O โ‡Œ CH3COO + H3O+

Cf 0.05 0 0
0.05 โ€“0.5ฮฑ 0.05ฮฑ 0.05ฮฑ

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

1.74 ร— 10โ€“5 โ€“ 1.74 ร— 10โ€“5ฮฑย = 0.05 ฮฑ2

0.05ฮฑ2 + 1.74 + 10โ€“5ฮฑ โ€“ 1.74 ร— 10โ€“5

D = b2 โ€“ 4ac

= (1.74 ร— 10โ€“5)2 โ€“ 4(0.05) (โ€“ 1.74 ร— 10โ€“5)

= 3.02 ร— 10โ€“25 + 0.348 ร— 10โ€“5 = 0

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

= (3.48 ร— 10โ€“4)1/2

= 1.86 ร— 10โ€“2

CH3COOH ย โ‡Œ CH3COOโ€“ + H+

CH3COOโ€“ = C.a

= 0.05 ร— 1.86 ร— 10โ€“3

= 9.3 ร— 10โ€“4 = 0.00093

pH = โ€“ log [H+]

= โ€“ 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 pKa.

Ans. Let the organic acid be HA.

โ‡’ HA ย โ‡Œ H+ + Aโ€“

Concentration of HA = 0.01 M

pH = 4.15 = โ€“ log[H+]

[H+] = 7.08 ร— 10โ€“5

[HA] = 0.01

Then,

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

Ka = 5.01 ร— 10โ€“7

pKa = โ€“log Ka

= โ€“ log (5.01 ร— 10โ€“7)

pKa= 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) โ‡Œ H3O+ + Clโ€“

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

โ‡’ [H3O+] = 0.003

Now,

pH = โ€“ log [H3O+]

= โ€“ log(0.003)

= 2.52

Hence, the pH of the solution is 2.52.

(b) 0.005 M NaOH

NaOH(aq) โ‡Œ Na+(aq) + OHโ€“(aq)

[OH โ€“] = [NaOH]
โ‡’ [OHโ€“] = 0.005

pOH = โ€“ log [OHโ€“]

= โ€“ log (0.005 )

pOH = 2.30

โ‡’ pH = 14 โ€“ 2.30

= 11.70

Hence, the pH of the solution is 11.70.

(c) 0.002 HBr

HBr(aq) ย โ‡Œ H+ + Brโ€“

[H+] = [HBr ]

โ‡’ [H+] = 0.002

โ‡’ pH = โ€“ log [H+]

= โ€“ log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

(d) 0.002 M KOH

KOH (aq) โ‡Œ K+(aq) + OHโ€“(aq)

[OHโ€“ ] = [KOH]

โ‡’ [OH โ€“] = 0.002

Now, pOH = โ€“ log[OHโ€“ ]

= 2.69

โ‡’ pH = 14 โ€“ 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,

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

TlOH(aq) โ†’ Tl(aq)+ + OH(aq)โ€“

[OH(aq)โ€“] = [TlOH(aq)]

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

221 ร— 10โ€“14 = [H+]

โ‡’ pH = โ€“ log[H+]

= โ€“ log(221 ร— 10โ€“14)

= โ€“ log (2.21 ร— 10โ€“12)

= 11.65

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

Ca(OH)2 โ†’ Ca2 + 2OHโ€“

[Ca(OH)2] = 0.3 ร— 1000 / 500

= 0.6 M

[OHโ€“aq] = 2 ร— [Ca(OH)2aq]

= 2 ร— 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 ร— 10โ€“14

pH = โ€“ log (0.833 ร— 10โ€“14)

= โ€“ log (8.33 ร— 10โ€“13)

= (โ€“ 0.902 + 13 )

= 12.098

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

NaOH โ†’ Na+(aq) + OHโ€“(aq)

[OHโ€“(aq)] = 1.5 M

Then,

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

= 6.66 ร— 10โ€“13

pH = โ€“ log(6.66 ร— 10โ€“13)

= 12.18

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

13.6 ร— 1 mL = M2 ร— 1000 mL (Before dilution) (After dilution)

13.6 ร— 10โ€“3 = M2

M2 = 1.36 ร— 10โ€“2

[H+] = 1.36 ร— 10โ€“2

pH = โ€“ log (1.36 ร— 10โ€“2)

โ‰ˆ 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 pKa 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 ย โ‡Œ BrCH2COOโ€“ + H+

Initial conc. C 0 0
Equilibrium conc. (C โ€“ Cฮฑ) Cฮฑ Cฮฑ

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

= 0.1 ร— (0.132)2

= 1.74 ร— 10โ€“3

โ‡’ pKa = โ€“ log Ka

= โ€“ log (1.74 ร— 10โ€“3)

= 3 โ€“ 0.2405 = 2.76

โ‡’ (H+) = Ca = 0.1 ร— 0.132

= 1.32 ร— 10โ€“2 M

โ‡’ pH = โ€“ log (1.32 ร— 10โ€“2)

= 2 โ€“ 0.1206 = 1.88.

Hence, the pH of the solution is 1.88 and the pKa 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 pKb.

Ans. Given,

pH of 0.005 M codeine solution = 9.95

Now,

pOH = pKw โ€“ pH

= 14.0 โ€“ 9.95 = 4.05

[OHโ€“] = antilog (โ€“ 4.05)

= 8.91 ร— 10โ€“5 M

Since, [OHโ€“] = Cฮฑ;

$$\alpha=\frac{[\text{OH}^{-}]}{c}=\frac{8.91ร—10^{\normalsize-5}}{0.005}=0.0178\space \text{or}\space 1.78 ร— 10^{\normalsizeโ€“2}\\\text{K}_b=\frac{c\alpha^{2}}{1-\alpha}\\=\frac{0.005ร—0.0178^{2}}{1-0.0178}$$

= 1.6 ร— 10โ€“6

pKb = โ€“ log Kb

= โ€“ log 1.6 ร— 10โ€“6 = 5.8.

Hence, the ionization constant is 1.78 ร— 10โ€“2 and the pKb value is 5.8

7.52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 ร— 10โ€“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,

Kb = 4.27 ร— 10โ€“10

c = 0.001 M

pH = ?

ฮฑ = ?
Kb = Cฮฑ2

4.27 x 10โ€“10 = 0.001 ร— ฮฑ2

4270 ร— 10โ€“10 = ฮฑ2

= 65.34 ร— 10โ€“5 = ฮฑ

= 6.53 ร— 10โ€“4

Then,

[anion] = ca = .001 ร— 65.34 ร— 10โ€“5

= .065 ร— 10โ€“5

Now,

pOH = โ€“ log (0.065 ร— 10โ€“5)

= 6.187

pH = 7.183

Now,

Ka ร— Kb = Kw

โˆด 4.27 ร— 10โ€“10 ร— Ka = Kw

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

= 2.34 ร— 10โ€“5

Thus, the ionization constant of the conjugate acid of aniline is 2.34 ร— 10โ€“5.

7.53. Calculate the degree of ionization of 0.05 M acetic acid if its pKa 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; pKa = 4.74

pKa = โ€“ log Ka

Ka = antilog (โ€“4.74) = 1.82 ร— 10โ€“5

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

In presence of 0.01 M HCl

Equilibria and equilibrium concentrations are

CH3COOH โ‡Œ H+ + CH3COOโ€“

Initial conc. 0.05 M 0 0
After dissociation 0.05(1 โ€“ ฮฑ) 0.05ฮฑ 0.05ฮฑ

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

[H+] = 0.01 + 0.05a โ‰ƒ 0.01

[HAc] = 0.05(1 โ€“ a) โ‰ƒ 0.01

$$\therefore\space 1.82 ร— 10^{\normalsizeโ€“5} =\frac{0.01ร—0.05a}{0.05}\\\alpha=\frac{1.82ร—10^{\normalsize-5}}{0.01}$$

= 1.82 ร— 10โ€“3

In presence of 0.1 M HCl

HAc ย โ‡Œ H+ + Acโ€“

$$\underset{0.05(1 โ€“ ฮฑ)}{\text{HAc}} โ‡Œ\underset{0.05 ฮฑ}{\text{H}^{+}}+\underset{0.05 ฮฑ}{\text{Ac}^{-}}\\\underset{0}{\text{HCL}}\xrightarrow{}\underset{0.1\text{M}}{\text{H}^{+}} + \underset{0.1\text{M}}{\text{Cl}^{-}}$$

[H+] = 0.1 + 0.05a โ‰ƒ 0.1 M

[HAc] = 0.05(1 โ€“ a) โ‰ƒ 0.05 M

Ka = 1.82 ร— 10โ€“5

$$\text{K}_a=\frac{[\text{H}^{+}][\text{Ac}^{-}]}{\text{HAc}}\\1.82 ร— 10^{โ€“5} =\frac{0.1ร—0.05ฮฑ}{0.05}\\\alpha=\frac{1.82ร—10^{\normalsize-5}}{0.1}=1.82ร—10^{\normalsize-4}$$

7.54. The ionization constant of dimethylamine is 5.4 ร— 10โ€“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,

Kb = 5.4 ร— 10โ€“4

c = 0.02 M

then,

$$ฮฑ=\sqrt\frac{\text{K}_b}{c}\\=\sqrt\frac{5.4ร—10^{\normalsize-4}}{0.02}$$

= 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

[OHโ€“] = x + 0.1; 0.1

$$\Rarr\space\text{K}_b=\frac{[(\text{CH}_3)_2 \text{NH}_2^+][\text{OH}^{-}]}{[(\text{CH}_3)_2\text{NH}^{-}]}\\5.4 ร— 10^{\normalsizeโ€“4} =\frac{xร—0.1}{0.02}$$

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

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 = โ€“ log [H+]

therefore, 6.83 = โ€“ log [H+]

[H+] = antilog (โ€“6.83)

[H+] =1.48 ร— 10โ€“7 M

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

1.2 = โ€“ log [H+]

[H+] = antilog (โ€“1.2)

[H+] = 0.063 or 6.30 ร— 10โ€“2 M

(c) Human blood, 7.38 :

pH = 7.38 = โ€“ log [H+]

[H+] = antilog (โ€“7.38)

[H+] = 4.17 ร— 10โ€“8 M

(d) Human saliva, 6.4 :

pH = 6.4

6.4 = โ€“ log [H+]

[H+] = antilog โ€“6.4

[H+] = 3.98 ร— 10โ€“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 = โ€“log [H+]

(i) pH of milk = 6.8

Since, pH = โ€“log [H+]

6.8 = โ€“log [H+]

[H+] = antilog (โ€“6.8)

= 1.58 ร— 10โ€“7 M

(ii) pH of black coffee = 5.0

Since, pH = โ€“log [H+]

5.0 = โ€“log [H+]

[H+] = antilog (โ€“5.0)

= 1 ร— 10โ€“5 M

(iii) pH of tomato juice = 4.2

Since, pH = โ€“log [H+]

4.2 = โ€“log [H+]

[H+] = antilog (โ€“4.2)

= 6.31ร—10โ€“5 M

(iv) pH of lemon juice = 2.2

pH = โ€“log [H+]

2.2 = โ€“log [H+]

[H+] = antilog(โ€“2.2)

= 6.31 ร— 10โ€“3 M

(v) pH of egg white = 7.8

Since, pH = โ€“log [H+]

7.8 = โ€“log [H+]

[H+] = antilog(โ€“7.8)

= 1.58ร—10โ€“8 M

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,

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

= 0.05 M Now, [H+][OHโ€“] = Kw

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

= 2 ร— 10โ€“13

pH = โ€“ log [H+]

= โ€“ log (2 ร— 10โ€“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) โ†’ Sr2+(aq) + 2(OHโ€“)(aq)
\ [Sr2+] = 0.1581 M
[OHโ€“] = 2 ร— 0.1581 M
M = 0.3126 M
And,
[H+][OHโ€“] = Kw

$$\text{Sr(OH)}_2\text{(aq)} \xrightarrow{} \text{Sr}^{2+}\text{(aq)} + 2(\text{OH}^{โ€“})(\text{aq})$$

โˆด [Sr2+] = 0.1581 M

[OHโ€“] = 2 ร— 0.1581 M

M = 0.3126 M

And,

[H+][OHโ€“] = Kw

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

= 3.2 ร— 10โ€“14

pH = โ€“log [H+]

= โ€“ log (3.2 ร— 10โ€“14)

= 13.49

7.59. The ionization constant of propanoic acid is 1.32 ร— 10โ€“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 ฮฑ.

Then, representing propionic acid as HA, we have:

$$\underset{(0.05 โ€“ 0.05a)โ‰ˆ0.05}{\text{HA}}+\text{H}_2\text{O} โ‡Œ\underset{.05ฮฑ}{\text{H}_3\text{O}^{+}}+\underset{.05ฮฑ}{\text{A}^{-}}\\\text{K}_a=\frac{[\text{H}_3\text{O}^{+}][\text{A}^{-}]}{[\text{HA}]}\\=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^{2}\\\alpha=\sqrt{\frac{\text{K}_a}{.05}}=1.63ร—10^{\normalsize-2}$$

Then, [H3O+] = .05ฮฑ = .05 ร— 1.63 ร— 10โ€“2

โˆด pH = 3.09 = Kb, 15 ร— 10โ€“4 M

In the presence of 0.1 M of HCl, let aโ€ฒ be the degree of ionization.

Then [H3O+] = 0.01

[Aโ€“] = 0.05ฮฑโ€ฒ

[HA] = 0.05

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

1.32 ร— 10โ€“5 = 0.01 ร— ฮฑโ€ฒ

ฮฑโ€ฒ = 1.32 ร— 10โ€“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 (โ€“2.34) = 4.57 ร— 10โ€“3 M

Ionisation equilibrium and equilibrium concentrations are:

HCNO (aq) ย โ‡Œ H+(aq) + CNOโ€“ (aq)
0.1(1 โ€“ a) 0.1a 0.1a
โˆด [H+] = 0.1a = 4.57 ร— 10โ€“3

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

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

= 4.57 ร— 10โ€“2 = 0.0457

Since a is small

Ka = ca2 = 0.1 ร— (4.57 ร— 10โ€“2)2

= 2.09 ร— 10โ€“4

7.61. The ionization constant of nitrous acid is 4.5 ร— 10โ€“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).

NO2โ€“ + H2O โ‡Œ HNO2 + OHโ€“

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

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

[NO2โ€“] = 0.04 โ€“ x; 0.04

[HNO2] = x

[OHโ€“] = xย 

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

x2 = 0.0088 ร— 10โ€“10

x = 0.093 ร— 10โ€“5

โˆด [OHโ€“] = 0.093 ร— 10โ€“5 M

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

โ‡’ pH = โ€“ log(10.75 ร— 10โ€“9)

= 7.96

ย Therefore, degree of hydrolysis

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

= 2.325 ร— 10โ€“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 = โ€“ log [H+]

[H+] = antilog (โ€“3.44)

[H+] = 3.63 ร— 10โ€“4

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

Kh = 6.6 ร— 10โ€“6

Pyridinium hydrochloride is a weak acid and the

[H+] = cฮฑ

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

Since it is weak acid, Ostwaldโ€™s dilution law is applicable and

Ka = cฮฑ2 = 0.02 ร— (0.0182)2

= 6.63 ร— 10โ€“6

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

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

Hence, the ionisation constant of pyridine is 1.5 ร— 10โ€“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 ร— 10โ€“3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Ans. It is given that Ka for ClCH2COOH is 1.35 ร— 10โ€“3.

Since the acid is weak, Ostwaldโ€™s dilution law can be used as :

โ‡’ Ka = cฮฑ2

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

= 0.116

โˆด [H+] = cฮฑ = 0.1 ร— 0.116

= 0.0116

โ‡’ pH = โ€“ log [H+] = 1.94

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

ClCH2COOโ€“ + H2O โ‡Œ ClCH2COOH + OHโ€“

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

= 0.740 ร— 10โ€“11

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

0.074 ร— 10โ€“11 = x2

โ‡’ x2 = 0.74 ร— 10โ€“12

x = 0.86 ร— 10โ€“6

[OHโ€“] = 0.86 ร— 10โ€“6

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

[H+] = 1.162 ร— 10โ€“8

pH = โ€“log [H+]

= โ€“ log (1.162 ร— 10โ€“18)

= 7.94

7.65. Ionic product of water at 310 K is 2.7 ร— 10โ€“14. What is the pH of neutral water at this temperature?

Ans. Ionic product is given as:

Kw = [H+][OHโ€“]

Let [H+] = x.

Since [H+] = [OHโ€“], Kw = x2

โ‡’ Kw at 310 K is 2.7 ร— 10โ€“14,

โˆด 2.7 ร— 10โ€“14 = x2

$$=\sqrt{2.7ร—10^{\normalsize-14}}$$

โ‡’ x = 1.643 ร— 10โ€“7

โ‡’ x = 1.643 ร— 10โ€“7 mol Lโ€“1

โ‡’ [H+] = 1.643 ร— 10โ€“7

โ‡’ pH = โ€“ log [H+]

= โ€“ log(1.643 ร— 10โ€“7)

= 7 โ€“ log 1.643

= 7 โ€“ 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

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

= 0.0428

Now,

pOH = โ€“ log [OH] = โ€“ log [0.0428]

= 1.36

Now, pH = 14 โ€“ 1.36

= 12.63

$$\text{(b) Moles of H}_3\text{O}^+ =\frac{2ร—10ร—0.01}{1000}\\= 0.0002\space\text{mol}\\\text{and Moles of OH}^{\normalsizeโ€“}=\frac{2ร—10ร—0.01}{1000}$$

= 0.0002 mol

Both H3O+and OHโ€“ are equal, the solution is neutral and pH = 7.

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

= 0.05

Now,

pH = โ€“ 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)KSPAgCrO().,24111012=ร—โˆ’ KSPBaCrO().,4121010=ร—โˆ’ KSPFeOH[()].,3101038=ร—โˆ’ KSPPbCl().,216105=ร—โˆ’ KSPHgI()..22451029=ร—โˆ’ Ans. (1) Silver Chromate

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

For Ag2CrO4,

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

Ksp = 1.1 ร— 10โ€“12

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

= (0.275 ร— 10โ€“12)1/3

= 6.5 ร— 10โ€“5 M

Molarity of [Ag+] = 2[Ag2CrO4]

= 2 ร— s = 2 ร— 6.5 ร— 10โ€“5

= 1.3 ร— 10โ€“4 M

Molarity of [CrO42โ€“] = 2[Ag2CrO4]

= s = 6.5 ร— 10โ€“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, Ksp = 1.2 ร— 10โ€“10

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

= (1.2ร—10-10)1/2

= 1.1 ร— 10โ€“5 M

From the formula,

Molarity of [Ba2+] = [CrO42โ€“] = s = 1.1 ร— 10โ€“5 M

(3) Ferric Hydroxide

Fe(OH)3 โ†’ Fe2+ + 3OHโ€“

Ksp = [Fe2+][OHโ€“]3

Let s be the solubility of Fe(OH)3

Thus, [Fe3+] = s and [OHโ€“] = 3s

โ‡’ Ksp = s.(3s)3

= s.27 s3

Ksp = 27s4

1.0 ร— 10โ€“38 = 27s4

.037 ร— 10โ€“38 = s4

.00037 ร— 10โ€“36 = s4

โ‡’ 1.39 ร— 10โ€“10 M = S

Molarity of Fe3+ = s = 1.39 ร— 10โ€“10 M

Molarity of OHโ€“ = 3s = 4.17 ร— 10โ€“10 M

BaCrO4 โ†’ Ba2+ + CrO42โ€“

Then, Ksp = [Ba2+][CrO42โ€“]

Let s be the solubility of BaCrO4.

Thus, [Ba2+] = s and [CrO42โ€“] = s

โ‡’ Ksp = s2

โ‡’ 1.2 ร— 10โ€“10 = s2

โ‡’ s = 1.09 ร— 10โ€“5 M

(4) Lead chloride

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

Ksp = [Pb2+][Clโ€“]2

Let Ksp be the solubility of PbCl2.

[PB2+] = s and [Clโ€“] = 2s

Thus, Ksp = s.(2s)2

= 4s3

โ‡’ 1.6 ร— 10โ€“5 = 4s3

โ‡’ 0.4 ร—10โ€“5 = s3

4 ร— 10โ€“6 = s3

โ‡’ 1.58 ร— 10โ€“2M = S.1

Molarity of Pb2+ = s = 1.58 ร— 10โ€“2 M

Molarity of chloride = 2s
= 3.16 ร— 10โ€“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

โ‡’ [Hg22+] = s and [Iโ€“] = 2s

Thus, Ksp = s.(2s)2

โ‡’ Ksp = 4s3

โ‡’ 4.5 ร— 10โ€“29 = 4s3

โ‡’ 1.125 ร—10โ€“29 = s3

โ‡’ s = 2.24 ร— 10โ€“10 M

Molarity of Hg22+ = s = 2.24 ร— 10โ€“10 M

Molarity of Iโ€“ = 2s = 4.48 ร— 10โ€“10 M

7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1 ร— 10โ€“12 and 5.0 ร— 10โ€“13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans. Let s be the solubility of Ag2CrO4.

Then, Ag2CrO4 โ‡Œ Ag2+ + 2CrO4โ€“

Ksp = (2s)2.s = 4s3

1.1 ร— 10โ€“12 = 4s3

s = 6.5 ร— 10โ€“5 M

Let s be the solubility of AgBr.

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

Ksp = sโ€ฒ2 = 5.0 ร— 10โ€“13

โˆด sโ€ฒ = 7.07 ร— 10โ€“7 M

Now, ratio of their Molarities

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

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 Ksp = 7.4 ร— 10โ€“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+][IO3โ€“]2

= (0.001)(0.001)2

= 1 ร— 10โ€“9

Since the ionic product (1 ร— 10โ€“9) is less than Ksp (7.4 ร— 10โ€“9), precipitation will not occur.

7.70. The ionization constant of benzoic acid is 6.46 ร— 10โ€“5 and Ksp for silver benzoate is 2.5 ร— 10โ€“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 ร— 10โ€“4 M

C6H5COOH + H2O โ‡Œ C6H5COOโ€“ + H3O

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

Let the solubility of C6H5COOAg be x mol/L.

Then,

[Ag+] = x

[C6H5COOH] + [C6H5COOโ€“] = x

10 [C6H5COOโ€“] + [C6H5COOโ€“] = x

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

x = 1.66 ร— 10โ€“6 mol/L

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 ร— 10โ€“6 mol/L. Now, let the solubility of C6H5COOAg be xโ€™ mol/L.

Then, [Ag+] = xโ€ฒ M and [CH3COOโ€“] = xโ€ฒ M

Ksp = [Ag+][CH3COOโ€“]

Ksp = (xโ€ฒ)2

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

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, Ksp = 6.3 ร— 10โ€“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.

$$\therefore\space[\text{FeSO}_4]=[\text{Na}_2\text{S}]=\frac{x}{2}\text{M}\\\text{Then},\space[\text{Fe}^{2+}]=[\text{FeSO}_4]=\frac{x}{2}\text{M}\\\text{Also},\space[\text{S}^{2-}]=[\text{Na}_2\text{S}]=\frac{x}{2}\text{M}\\\text{FeS(s)} โ‡Œ \text{Fe}^{2+}\text{(aq)} + \text{S}^{2\normalsizeโ€“}\text{(aq)}\\\text{K}_{sp}=[\text{Fe}^{2+}][\text{S}^{2-}]\\6.3ร—10^{\normalsize-18}=\bigg(\frac{x}{2}\bigg)\bigg(\frac{x}{2}\bigg)\\\frac{x^{2}}{4}=6.3ร—10^{\normalsize-18}$$

โ‡’ x = 5.02 ร— 10โ€“9

Thus, if the concentrations of both solutions are equal to or less than 5.02 ร— 10โ€“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, Ksp is 9.1 ร— 10โ€“6).

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

CaSO4(s) โ‡Œ Ca2+ (aq) + SO42โ€“ (aq)

Ksp = [Ca2+][SO42โ€“]

Let the solubility of CaSO4 be s.

Then, Ksp = s2

9.1 ร— 10โ€“6 = s2

s = 3.02 ร— 10โ€“3 mol/L

Molecular mass of CaSO4 = 136 g/mol

Solubility of CaSO4 in gram/L

= 3.02 ร— 10โ€“3 ร— 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 ร— 10โ€“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 Ksp value.

Before mixing:
[S2โ€“] = 1.0 ร— 10โ€“19 M [M2+] = 0.04 M

Volume = 10 mL volume = 5 mL

After Mixing

[S2โ€“] = ? [M2+] = ?

Volume = (10 + 5) = 15 mL

Volume = 15 mL

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

= 1.33 ร— 10โ€“2 M
Ionic product = [M2+][S2โ€“]
= (1.33 ร— 10โ€“2)(6.67 ร— 10โ€“20)
= 8.87 ร— 10โ€“22
This ionic product exceeds the Ksp of ZnS and CdS. Therefore, precipitation will occur in CdCl2 and ZnCl2 solutions.