On decreasing the pH from 7 to 2,the solubili | vedclass

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(B) At $pH = 7$,the salt $MX$ dissociates as $MX_{(s)} \rightleftharpoons M^+_{(aq)} + X^-_{(aq)}$. The solubility $S_1 = \sqrt{K_{sp}} = 10^{-4} \ mo

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$4$

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(B) At $pH = 7$,the salt $MX$ dissociates as $MX_{(s)} ightleftharpoons M^+_{(aq)} + X^-_{(aq)}$. The solubility $S_1 = \sqrt{K_{sp}} = 10^{-4} \ mol \ L^{-1}$,so $K_{sp} = 10^{-8}$.At $pH = 2$,$[H^+] = 10^{-2} \ M$. The salt dissociates and $X^-$ reacts with $H^+$: $X^-_{(aq)} + H^+_{(aq)} ightleftharpoons HX_{(aq)}$.The equilibrium constant for this reaction is $K = \frac{1}{K_a}$.The total solubility $S = [M^+] = [X^-] + [HX] = 10^{-3} \ M$.From $K_{sp} = [M^+][X^-]$,we have $[X^-] = \frac{K_{sp}}{[M^+]} = \frac{10^{-8}}{10^{-3}} = 10^{-5} \ M$.Thus,$[HX] = S - [X^-] = 10^{-3} - 10^{-5} \approx 10^{-3} \ M$.Using the equilibrium expression for $HX$ formation: $K_a = \frac{[H^+][X^-]}{[HX]} = \frac{10^{-2} 	imes 10^{-5}}{10^{-3}} = 10^{-4}$.Therefore,$pK_a = -\log(10^{-4}) = 4$.

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:

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