JEE Challenger
More from Equilibrium
ChemMedAdv 2023 • Paper 1

Solubility Dependence on pH and Calculation of pKa

On decreasing the pH\text{pH} from 77 to 22, the solubility of a sparingly soluble salt (MX)(\text{MX}) of a weak acid (HX)(\text{HX}) increased from 10−4 mol L−110^{-4}\text{ mol L}^{-1} to 10−3 mol L−110^{-3}\text{ mol L}^{-1}. The pKap\text{K}_a of HX\text{HX} is

Options

A

3

B

4

Correct
C

5

D

2

Topics & Concepts

EquilibriumAcid-Base EquilibriumSolubility Product

Step-by-Step Solution

To determine the pKap\text{K}_a of the weak acid HX\text{HX}, we consider the equilibria present in the solution of the sparingly soluble salt MX\text{MX}:

  1. Dissolution of the salt: MX(s)⇌M+(aq)+X−(aq)\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)

  2. Hydrolysis/protonation of the anion X−\text{X}^-: X−(aq)+H+(aq)⇌HX(aq)\text{X}^-(aq) + \text{H}^+(aq) \rightleftharpoons \text{HX}(aq)

Let SS be the molar solubility of MX\text{MX}. From mass balance, we have: [M+]=S[\text{M}^+] = S [X−]+[HX]=S[\text{X}^-] + [\text{HX}] = S

The acid dissociation constant KaK_a for HX\text{HX} is given by: Ka=[H+][X−][HX]  ⟹  [HX]=[H+][X−]KaK_a = \frac{[\text{H}^+][\text{X}^-]}{[\text{HX}]} \implies [\text{HX}] = \frac{[\text{H}^+][\text{X}^-]}{K_a}

Substituting [HX][\text{HX}] into the mass balance equation for X\text{X}: [X−]+[H+][X−]Ka=S[\text{X}^-] + \frac{[\text{H}^+][\text{X}^-]}{K_a} = S [X−](1+[H+]Ka)=S  ⟹  [X−]=S1+[H+]Ka[\text{X}^-] \left(1 + \frac{[\text{H}^+]}{K_a}\right) = S \implies [\text{X}^-] = \frac{S}{1 + \frac{[\text{H}^+]}{K_a}}

The solubility product constant KspK_{sp} is defined as: Ksp=[M+][X−]=S⋅(S1+[H+]Ka)=S21+[H+]KaK_{sp} = [\text{M}^+][\text{X}^-] = S \cdot \left(\frac{S}{1 + \frac{[\text{H}^+]}{K_a}}\right) = \frac{S^2}{1 + \frac{[\text{H}^+]}{K_a}}

Rearranging for solubility SS: S2=Ksp(1+[H+]Ka)S^2 = K_{sp} \left(1 + \frac{[\text{H}^+]}{K_a}\right)

Now, we apply the given experimental conditions:

At pH1=7\text{pH}_1 = 7: [H+]1=10−7 M,S1=10−4 mol L−1[\text{H}^+]_1 = 10^{-7}\text{ M}, \quad S_1 = 10^{-4}\text{ mol L}^{-1} (10−4)2=Ksp(1+10−7Ka)(10^{-4})^2 = K_{sp} \left(1 + \frac{10^{-7}}{K_a}\right) 10−8=Ksp(1+10−7Ka)— (Equation 1)10^{-8} = K_{sp} \left(1 + \frac{10^{-7}}{K_a}\right) \quad \text{--- (Equation 1)}

At pH2=2\text{pH}_2 = 2: [H+]2=10−2 M,S2=10−3 mol L−1[\text{H}^+]_2 = 10^{-2}\text{ M}, \quad S_2 = 10^{-3}\text{ mol L}^{-1} (10−3)2=Ksp(1+10−2Ka)(10^{-3})^2 = K_{sp} \left(1 + \frac{10^{-2}}{K_a}\right) 10−6=Ksp(1+10−2Ka)— (Equation 2)10^{-6} = K_{sp} \left(1 + \frac{10^{-2}}{K_a}\right) \quad \text{--- (Equation 2)}

Dividing Equation (2) by Equation (1): 10−610−8=1+10−2Ka1+10−7Ka\frac{10^{-6}}{10^{-8}} = \frac{1 + \frac{10^{-2}}{K_a}}{1 + \frac{10^{-7}}{K_a}} 100=1+10−2Ka1+10−7Ka100 = \frac{1 + \frac{10^{-2}}{K_a}}{1 + \frac{10^{-7}}{K_a}}

Since weak acids typically have Ka≫10−7K_a \gg 10^{-7}, the term 10−7Ka≪1\frac{10^{-7}}{K_a} \ll 1, giving 1+10−7Ka≈11 + \frac{10^{-7}}{K_a} \approx 1.

Therefore, the equation simplifies to: 100≈1+10−2Ka100 \approx 1 + \frac{10^{-2}}{K_a} 10−2Ka=99≈102\frac{10^{-2}}{K_a} = 99 \approx 10^2 Ka=10−4K_a = 10^{-4}

Thus, the pKap\text{K}_a of HX\text{HX} is: pKa=−log⁡10(Ka)=−log⁡10(10−4)=4p\text{K}_a = -\log_{10}(K_a) = -\log_{10}(10^{-4}) = 4

Correct Option: (B)

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