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 (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 M–1 (approx)
Therefore, Kc 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 Kp for the equilibrium.
Ans. Total pressure = Ptotal = 105 Pa
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
7.5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl(g) ⇌ 2NO(g) + Cl2(g); Kp = 1.8 × 10–2 at 500 K
(ii) CaCO3(s) ⇌ CaO(s) + CO2(g); Kp = 167 at 1073 K
Ans. (i) 2NOCl(g) ⇌ 2NO(g) + Cl2(g)
The relation between Kp and Kc given as:
Kp = Kc (RT)Δn
In this reaction,
Δn = 3 – 2 = 1
Given values:
R = 0.0831 bar L mol–1 K–1
T = 500 K
Kp = 1.8 × 10–2
Kp = Kc(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,
Kp = 167
Kp = Kc (RT)Δn
⇒ 167 = Kc (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 ′Kc for the reverse reaction?
Ans. It is given that ′Kc for the forward reaction is
6.3 × 1014 .
Then, Kc 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. Kc = 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, Kp = 2.0 × 1010/bar for the given reaction at equilibrium.
2SO2(g) + O2(g) ⇌ 2SO3(g)
What is Kc at this temperature?
Ans. For the given reaction,
Δn = 2 – 3 = – 1
T = 450 K
R = 0.0831 bar L mol–1 K–1
Kp = 2.0 × 1010 bar –1
We know that,
Kp = Kc(RT)Δn
⇒ (2.0 × 1010) bar–1 = Kc(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 Kp 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 |
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 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 × 103
Since, Qc ≠ Kc (1.7 × 102) , the reaction mixture is not at equilibrium. Again, Qc (2.4 × 103) > Kc (1.7 × 102).
Hence, the reaction will proceed in the reverse direction.
7.13. The equilibrium constant expression for a gas reaction is,
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 Kc for the given reaction is 54.8.
H2(g) + I2(g) ⇌ 2HI(g)
Therefore, at equilibrium, the equilibrium constant Kc 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); Kc = 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), 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) ⇌ 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 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 × 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);
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) ⇌ 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 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,
$$\text{K}_p=\frac{\text{P}_{\text{CO}_2}}{\text{P}_\text{CO}}\\\Rarr 0.265 =\frac{0.80-p}{1.4+p}$$
⇒ 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–1 2.0 mol L–1 0.5 mol L–1 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 ≠ 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) ⇌ Br2(g) + Cl2(g) for which Kc = 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 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 – 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
Kp 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 ∆ng = 2 – 1 = 1, Kp = Kc(RT)∆ng
$$\text{or K}_c=\frac{\text{K}_P}{\text{RT}}=\frac{14.19}{0.0821×1127}=0.153$$
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 Kc
∆G° = – 2.303 RT log Kc
$$\text{log K}_c =\frac{-35.0×10^{3}}{-2.303×8.314×298}$$
= 6.134
∴ Kc = antilog (6.134)
= 1.36 × 106
= 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,
Kp 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 Kp for the above reaction.
(b) How will the values of Kp 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,
$$\text{K}_p=\frac{\text{P}_{\text{CO}}×\text{P}^{3}_{\text{H}_2}}{\text{P}_{\text{CH}_4}×\text{P}_{\text{H}_2\text{O}}}$$
(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.
(c) According to Le Chatelier’s Principle, on removal of CO, 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)
ΔrH⊖ = 124.0 kJ mol–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:
$$\text{K}_c'=\frac{1}{\text{K}_c}=\frac{1}{8.3×10^{\normalsize-3}}$$
= 1.2048 × 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 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 PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 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,
Kp = 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) Kc = 5 ×10–39
(b) Cl2(g) + 2NO(g) ⇌ 2NOCl(g) Kc = 3.7 × 108
(c) Cl2(g) + 2NO2(g) ⇌ 2NO2Cl (g) Kc = 1.8
Ans. If the value of Kc 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:
Kc = 2.0 × 10–50
[O2(g)] = 1.6 ×10–2
For the reaction,
3O2(g) ⇌ 2O3(g)
Equilibrium 1.6 × 10–2 x
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, 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) ⇌ 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,
$$\frac{[\text{CH}_4(\text{g})][\text{H}_2\text{O}(\text{g})]}{[\text{Co(g)}][\text{H}_2\text{(g)}]^{3}}=\text{K}_c\\\Rarr\space\frac{x×0.02}{0.3×(0.1)^{3}}=3.90\\\Rarr\space x=\frac{3.90×0.3×(0.1)^{3}}{0.02}\\=\frac{0.00117}{0.02}$$
= 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–, CO32–, and S2–
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 | ClO4– (base) |
| F– | HF (acid) |
| OH– | H2O (acid) / O2– (base) |
| CO32– | HCO3– (acid) |
| S2– | 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– | CO32- |
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 | CO32– |
| HSO4– | H2SO4 | SO42– |
| 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.
(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 × 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 S2– 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
| Cδ | 0.1 | 0 | 0 |
| Cf | 0.1-x | x | x |
Then,
$$\text{K}_{a_1}=\frac{[\text{H}^{+}][\text{HS}^{-}]}{[\text{H}_2\text{S}]}\\9.1 × 10^{–8} =\frac{(x)(x)}{0.1-x}$$
(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 |
| Cf | 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 [S2–] in absence of 0.1 M HCl
H2S– ⇌ H+ + S2–
From case I [HS–] = [H+] = 9.54 × 10–5 N
[S2–] = 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 = [S2–]
Case IV : To calculate [S2–] in presence of 0.1 M HCl
HS– ⇌ H+ + S2–
From case II [HS–] = 9.1 × 10–8M
[H+] = 0.1 M (from HCl)
[S2–] = 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 = [S2–]
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+
Ka = 1.74 × 10–5
(2) H2O + H2O ⇌ H3O+ + OH– Kw = 1.0 × 10–14
Since Ka >> Kw :
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 pKa.
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,
$$\text{K}_a=\frac{(7.08×10^{\normalsize-5})(7.08×10^{\normalsize-5})}{0.01}$$
Ka = 5.01 × 10–7
pKa = –log Ka
= – log (5.01 × 10–7)
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) ⇌ 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)2 dissolved 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
(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
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 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 ⇌ BrCH2COO– + H+
| Initial conc. | C | 0 | 0 |
| Equilibrium conc. | (C – Cα) | Cα | Cα |
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 – 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
pKb = – log Kb
= – log 1.6 × 10–6 = 5.8.
Hence, the ionization constant is 1.78 × 10–2 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 × 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,
Kb = 4.27 × 10–10
c = 0.001 M
pH = ?
α = ?
Kb = 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,
Ka × Kb = Kw
∴ 4.27 × 10–10 × Ka = Kw
$$\text{K}_a=\frac{10^{\normalsize-14}}{4.27×10^{\normalsize-10}}$$
= 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 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 = – log Ka
Ka = 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 |
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,
Kb = 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
(c) Human blood, 7.38 :
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)
= 1.58×10–8 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,
= 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–] = Kw
$$\text{[H}^{+}]=\frac{\text{K}_w}{[\text{OH}^{-}]}\\=\frac{10^{\normalsize-14}}{0.312}$$
= 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 = Kb, 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
Ka = 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
$$\text{And,} \space \text{K}_\text{h} =\frac{(3.63×10^{\normalsize-4})^{2}}{0.02}$$
Kh = 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
Ka = 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 Ka for ClCH2COOH is 1.35 × 10–3.
Since the acid is weak, Ostwald’s dilution law can be used as :
⇒ Ka = 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:
Kw = [H+][OH–]
Let [H+] = x.
Since [H+] = [OH–], Kw = x2
⇒ Kw 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
$$\text{Ag}_2\text{CrO}_4\xrightarrow{} 2\text{Ag}^{\normalsize+}+\text{CrO}_4^{2-}\\\text{Then},\\\text{K}_{\text{sp}}=[\text{Ag}^{\normalsize+}]^{2}[\text{CrO}_4^{2-}]\\\text{Let ‘s’ be the solubility of Ag}_2\text{CrO}_4\\s=\bigg(\frac{\text{K}_{sp}}{x^{x}y^{y}}\bigg)^{\frac{1}{x+y}}$$
For Ag2CrO4,
x = 2, y = 1 and x + y = 3;
Ksp = 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 [CrO42–] = 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, Ksp = 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+] = [CrO42–] = s = 1.1 × 10–5 M
(3) Ferric Hydroxide
Fe(OH)3 → Fe2+ + 3OH–
Ksp = [Fe2+][OH–]3
Let s be the solubility of Fe(OH)3
Thus, [Fe3+] = s and [OH–] = 3s
⇒ Ksp = s.(3s)3
= s.27 s3
Ksp = 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+ + CrO42–
Then, Ksp = [Ba2+][CrO42–]
Let s be the solubility of BaCrO4.
Thus, [Ba2+] = s and [CrO42–] = s
⇒ Ksp = s2
⇒ 1.2 × 10–10 = s2
⇒ s = 1.09 × 10–5 M
(4) Lead chloride
$$\text{PbCl}_2\xrightarrow{}\text{Pb}^{2\normalsize+}+2\text{Cl}^{\normalsize-}$$
Ksp = [Pb2+][Cl–]2
Let Ksp be the solubility of PbCl2.
[PB2+] = s and [Cl–] = 2s
Thus, Ksp = 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, Ksp = s.(2s)2
⇒ Ksp = 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 × 10–12 and 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–
Ksp = (2s)2.s = 4s3
1.1 × 10–12 = 4s3
s = 6.5 × 10–5 M
Let s be the solubility of AgBr.
$$\text{AgBr(s)}\xrightarrow{}\text{Ag}^{\normalsize+}+\text{Br}^{\normalsize -}$$
Ksp = 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 Ksp = 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 Ksp (7.4 × 10–9), precipitation will not occur.
7.70. The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp 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
Ksp = [Ag+][CH3COO–]
Ksp = (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, Ksp = 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, Ksp is 9.1 × 10–6).
Ans. The dissociation of calcium sulphate can be given as:
CaSO4(s) ⇌ Ca2+ (aq) + SO42– (aq)
Ksp = [Ca2+][SO42–]
Let the solubility of CaSO4 be s.
Then, Ksp = 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 Ksp value.
Before mixing:
[S2–] = 1.0 × 10–19 M [M2+] = 0.04 M
Volume = 10 mL volume = 5 mL
After Mixing
[S2–] = ? [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+][S2–]
= (1.33 × 10–2)(6.67 × 10–20)
= 8.87 × 10–22
This ionic product exceeds the Ksp of ZnS and CdS. Therefore, precipitation will occur in CdCl2 and ZnCl2 solutions.
