On decreasing the pH from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from $$10^{-4}$$ mol L$$^{-1}$$ to $$10^{-3}$$ mol L$$^{-1}$$. The pK$$_a$$ of HX is:

Let the sparingly soluble salt be $$MX$$ and its anion be the conjugate base of the weak acid $$HX$$.

Dissolution equilibrium of the salt: $$MX(s) \rightleftharpoons M^{+} + X^{-}$$
Solubility in the given medium = $$S$$ mol L$$^{-1}$$ ⇒ $$[M^{+}] = S$$.

The anion $$X^{-}$$ is protonated by the hydronium ions present in solution:
$$X^{-} + H^{+} \rightleftharpoons HX$$ with $$K_{a} = \dfrac{[H^{+}][X^{-}]}{[HX]}$$.

Out of the total $$S$$ moles of $$X^{-}$$ produced, only a fraction remains as $$X^{-}$$; let that fraction be $$\alpha$$. Then
$$[X^{-}] = \alpha S,\qquad [HX] = (1-\alpha)S$$.

Applying the definition of $$K_{a}$$:
$$K_{a}= \dfrac{[H^{+}]\,\alpha S}{(1-\alpha)S}= [H^{+}]\,\dfrac{\alpha}{1-\alpha} \; -(1)$$
⇒ $$\dfrac{\alpha}{1-\alpha}= \dfrac{K_{a}}{[H^{+}]} \; -(2)$$
⇒ $$\alpha = \dfrac{K_{a}}{K_{a} + [H^{+}]} \; -(3)$$.

Solubility‐product of the salt:
$$K_{sp} = [M^{+}][X^{-}] = S \times (\alpha S)= S^{2}\,\dfrac{K_{a}}{K_{a} + [H^{+}]} \; -(4)$$.

The same $$K_{sp}$$ must hold at every pH. So for two different pH values,

Case 1: pH 7 ⇒ $$[H^{+}]_{1}=10^{-7}$$ mol L$$^{-1}$$, given $$S_{1}=10^{-4}$$ mol L$$^{-1}$$.
$$K_{sp}= S_{1}^{2}\,\dfrac{K_{a}}{K_{a}+10^{-7}} \; -(5)$$.

Case 2: pH 2 ⇒ $$[H^{+}]_{2}=10^{-2}$$ mol L$$^{-1}$$, given $$S_{2}=10^{-3}$$ mol L$$^{-1}$$.
$$K_{sp}= S_{2}^{2}\,\dfrac{K_{a}}{K_{a}+10^{-2}} \; -(6)$$.

Equate $$(5)$$ and $$(6)$$ because $$K_{sp}$$ is constant:

$$\dfrac{S_{1}^{2}}{K_{a}+10^{-7}} = \dfrac{S_{2}^{2}}{K_{a}+10^{-2}}$$

Insert the numerical solubilities:

$$\dfrac{(10^{-4})^{2}}{K_{a}+10^{-7}} = \dfrac{(10^{-3})^{2}}{K_{a}+10^{-2}}$$

Simplify the solubility ratio:
$$(10^{-4})^{2} = 10^{-8}, \quad (10^{-3})^{2} = 10^{-6}$$
$$\dfrac{10^{-8}}{K_{a}+10^{-7}} = \dfrac{10^{-6}}{K_{a}+10^{-2}}$$

Cross-multiply:

$$10^{-8}(K_{a}+10^{-2}) = 10^{-6}(K_{a}+10^{-7})$$

Divide both sides by $$10^{-8}$$ to clear the powers of 10:

$$(K_{a}+10^{-2}) = 100\,(K_{a}+10^{-7})$$

Expand the right side:

$$K_{a}+10^{-2} = 100K_{a} + 10^{-5}$$

Collect like terms:

$$10^{-2} - 10^{-5} = 100K_{a} - K_{a} = 99K_{a}$$

Evaluate the left side:
$$10^{-2} - 10^{-5} = 0.01 - 0.00001 = 0.00999 = 9.99\times10^{-3}$$

Hence
$$K_{a} = \dfrac{9.99\times10^{-3}}{99} \approx 1.0\times10^{-4}$$

Therefore
$$pK_{a} = -\log_{10}(K_{a}) \approx -\log_{10}(1.0\times10^{-4}) = 4$$

So, the pK$$_a$$ of the weak acid $$HX$$ is 4.

Option B which is: 4