JEE Advanced 2023 Paper 1 Q06 Chemistry Chemical & Ionic Equilibrium Buffers & Solubility Product Hard

JEE Advanced 2023 Paper 1 · Q06 · Buffers & Solubility Product

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

  1. A. $3$
  2. B. $4$
  3. C. $5$
  4. D. $2$
Reveal answer + step-by-step solution

Correct answer:B

Solution

At pH 7, $[X^-] \approx S = 10^{-4}$, so $K_{sp} = S^2 = 10^{-8}$. At pH 2, $S' = 10^{-3}$ and most $X^-$ is protonated to $HX$. Mass balance: $S' = [X^-] + [HX]$. From $K_{sp}$: $[X^-] = K_{sp}/S' = 10^{-5}$. So $[HX] = S' - [X^-] \approx 10^{-3}$. From $K_a = [H^+][X^-]/[HX] = 10^{-2}\cdot 10^{-5}/10^{-3} = 10^{-4}$. Hence $pK_a = 4$.

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