1. Dissolution equilibrium of the salt

For the sparingly soluble salt $\mathrm{MX}$:

$\mathrm{MX(s)} \rightleftharpoons \mathrm{M^+} + \mathrm{X^-}$

If the solubility is $s$, then:

$K_{sp} = [\mathrm{M^+}][\mathrm{X^-}]$

But since $\mathrm{X^-}$ is the conjugate base of weak acid $\mathrm{HX}$, in acidic medium it gets protonated:

$\mathrm{H^+} + \mathrm{X^-} \rightleftharpoons \mathrm{HX}$

So total dissolved salt increases at lower pH.

2. Relation between total solubility and pH

Let total solubility be $s$. Then:

$[\mathrm{M^+}] = s$

Out of total anion from dissolved salt, only a fraction remains as $\mathrm{X^-}$. Let that free concentration be $[\mathrm{X^-}]$. Then:

$K_{sp} = s[\mathrm{X^-}]$

From acid dissociation of $\mathrm{HX}$:

$K_a = \frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]}$

Using mass balance on anion:

$s = [\mathrm{X^-}] + [\mathrm{HX}]$

From Henderson-type relation:

$\frac{[\mathrm{HX}]}{[\mathrm{X^-}]} = \frac{[\mathrm{H^+}]}{K_a}$

Hence:

$s = [\mathrm{X^-}]\left(1 + \frac{[\mathrm{H^+}]}{K_a}\right)$

So:

$[\mathrm{X^-}] = \frac{s}{1 + [\mathrm{H^+}]/K_a}$

Substitute into $K_{sp}$:

$K_{sp} = s \cdot \frac{s}{1 + [\mathrm{H^+}]/K_a} = \frac{s^2}{1 + [\mathrm{H^+}]/K_a}$

Thus for a given salt:

$s^2 \propto 1 + \frac{[\mathrm{H^+}]}{K_a}$

3. Use the two given solubilities

At $\mathrm{pH}=7$:

$[\mathrm{H^+}]_1 = 10^{-7}, \qquad s_1 = 10^{-4}$

At $\mathrm{pH}=2$:

$[\mathrm{H^+}]_2 = 10^{-2}, \qquad s_2 = 10^{-3}$

Since $K_{sp}$ is same:

$\frac{s_1^2}{1 + 10^{-7}/K_a} = \frac{s_2^2}{1 + 10^{-2}/K_a}$

Now:

$s_1^2 = 10^{-8}, \qquad s_2^2 = 10^{-6}$

So:

$\frac{10^{-8}}{1 + 10^{-7}/K_a} = \frac{10^{-6}}{1 + 10^{-2}/K_a}$

Cross-multiply:

$10^{-8}\left(1 + \frac{10^{-2}}{K_a}\right) = 10^{-6}\left(1 + \frac{10^{-7}}{K_a}\right)$

Divide by $10^{-8}$:

$1 + \frac{10^{-2}}{K_a} = 100\left(1 + \frac{10^{-7}}{K_a}\right)$

$1 + \frac{10^{-2}}{K_a} = 100 + \frac{10^{-5}}{K_a}$

$\frac{10^{-2}}{K_a} - \frac{10^{-5}}{K_a} = 99$

$\frac{10^{-2} - 10^{-5}}{K_a} = 99$

$\frac{0.00999}{K_a} = 99$

$K_a = \frac{0.00999}{99} \approx 1.01 \times 10^{-4}$

Therefore:

$pK_a = -\log K_a \approx 4$

4. Option check

• A: $3$ ❌
• B: $4$ ✅
• C: $5$ ❌
• D: $2$ ❌

So the correct answer is B.