At T K,the solubility product of AX is 10^{-1 | vedclass

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(C) The dissolution of $AX$ is given by: $AX(s) \rightleftharpoons A^+(aq) + X^-(aq)$.$K_{sp} = [A^+][X^-] = 10^{-10}$.In $0.1$ $M$ $HX$ solution,$HX$

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$10^{-9}$

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(C) The dissolution of $AX$ is given by: $AX(s) ightleftharpoons A^+(aq) + X^-(aq)$.$K_{sp} = [A^+][X^-] = 10^{-10}$.In $0.1$ $M$ $HX$ solution,$HX$ is a strong acid (assuming complete dissociation),so $[H^+] = 0.1$ $M$ and $[X^-] = 0.1$ $M$.Let $S$ be the molar solubility of $AX$ in the presence of $0.1$ $M$ $HX$.Then $[A^+] = S$ and $[X^-] = 0.1 + S \approx 0.1$ (since $S$ is very small compared to $0.1$).Substituting these into the $K_{sp}$ expression:$10^{-10} = S 	imes 0.1$.$S = \frac{10^{-10}}{10^{-1}} = 10^{-9}$ $M$.

At $T$ $K$,the solubility product of $AX$ is $10^{-10}$. What is the molar solubility of $AX$ in $0.1$ $M$ $HX$ solution?

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