MX is a salt formed by neutralization of a st | vedclass

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(C) The dissolution of $MX$ is represented as: $MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq)$,with $K_{sp} = [M^{+}][X^{-}]$.In an acidic solution,$

Vedclass · Accepted Answer

$\sqrt{K_{sp} \cdot \left( 1 + \frac{[H^{+}]}{K_a} \right)}$

Vedclass · Answer

(C) The dissolution of $MX$ is represented as: $MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq)$,with $K_{sp} = [M^{+}][X^{-}]$.In an acidic solution,$X^{-}$ reacts with $H^{+}$ to form the weak acid $HX$: $X^{-}(aq) + H^{+}(aq) \rightleftharpoons HX(aq)$.The equilibrium constant for this reaction is $K = \frac{1}{K_a}$.The total concentration of the anion $X$ in the solution is $[X]_{total} = [X^{-}] + [HX]$.From the acid dissociation equilibrium,$[HX] = \frac{[H^{+}][X^{-}]}{K_a}$.Thus,$[X]_{total} = [X^{-}] \left( 1 + \frac{[H^{+}]}{K_a} \right)$.Since $[M^{+}] = [X]_{total} = S$ (solubility),we have $[X^{-}] = \frac{S}{1 + \frac{[H^{+}]}{K_a}}$.Substituting into the $K_{sp}$ expression: $K_{sp} = S \cdot \frac{S}{1 + \frac{[H^{+}]}{K_a}} = \frac{S^2}{1 + \frac{[H^{+}]}{K_a}}$.Solving for $S$,we get $S = \sqrt{K_{sp} \cdot \left( 1 + \frac{[H^{+}]}{K_a} \right)}$.

$MX$ is a salt formed by neutralization of a strong base,$MOH$,and a weak acid,$HX$. If the dissociation constant of $HX$ is $K_a$ and the solubility product of $MX$ is $K_{sp}$,then the solubility of $MX$ in an aqueous acidic solution is given by:

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