The solubility product of a sparingly soluble | vedclass

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Question

(C) For a sparingly soluble salt of the type $AX$,the dissociation is given by: $AX(s) \rightleftharpoons A^+(aq) + X^-(aq)$.Let the solubility of the

Vedclass · Accepted Answer

$7.0 	imes 10^{-7}$

Vedclass · Answer

(C) For a sparingly soluble salt of the type $AX$,the dissociation is given by: $AX(s) ightleftharpoons A^+(aq) + X^-(aq)$.Let the solubility of the salt be $S \ mol \ dm^{-3}$.Then,$[A^+] = S$ and $[X^-] = S$.The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [A^+][X^-] = S 	imes S = S^2$.Given $K_{sp} = 4.9 	imes 10^{-13}$.Therefore,$S^2 = 4.9 	imes 10^{-13} = 49 	imes 10^{-14}$.Taking the square root on both sides: $S = \sqrt{49 	imes 10^{-14}} = 7.0 	imes 10^{-7} \ mol \ dm^{-3}$.

The solubility product of a sparingly soluble salt $AX$ is $4.9 \times 10^{-13}$. What is its solubility in $mol \ dm^{-3}$?

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