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:$10^{-10} = S 	imes 0.1$.$S = \frac{10^{-10}}{10^{-1}} = 10^{-9}$ $M$.

Salt	$K_{sp}$
$AgCl$	$2 \times 10^{-10}$
$AgBr$	$5 \times 10^{-13}$
$AgI$	$9 \times 10^{-17}$

Formula	Solubility product $(K_{sp})$
$AgBrO_3$	$2.5 \times 10^{-10}$
$MgSO_4$	$2.7 \times 10^{-27}$
$CaCO_3$	$2.7 \times 10^{-18}$
$FeC_2O_4$	$2.4 \times 10^{-15}$

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

Similar Questions

Formula Solubility product $(K_{sp})$
$AgBrO_3$ $2.5 \times 10^{-10}$
$MgSO_4$ $2.7 \times 10^{-27}$
$CaCO_3$ $2.7 \times 10^{-18}$
$FeC_2O_4$ $2.4 \times 10^{-15}$
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