Calculate solubility (mol dm^{-3}) of a spar | vedclass

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Question

(C) For a sparingly soluble salt $AB$,the dissociation equilibrium is: $AB_{(s)} \rightleftharpoons A_{(aq)}^{+} + B_{(aq)}^{-}$ Let the solubility be

Vedclass · Accepted Answer

$4.0 	imes 10^{-3}$

Vedclass · Answer

(C) For a sparingly soluble salt $AB$,the dissociation equilibrium is: $AB_{(s)} ightleftharpoons A_{(aq)}^{+} + B_{(aq)}^{-}$ Let the solubility be $S \ mol \ dm^{-3}$. Then,$[A^{+}] = S$ and $[B^{-}] = S$. The solubility product constant $K_{sp}$ is given by: $K_{sp} = [A^{+}][B^{-}] = S 	imes S = S^2$ Given $K_{sp} = 1.6 	imes 10^{-5}$. Therefore,$S^2 = 1.6 	imes 10^{-5} = 16 	imes 10^{-6}$. Taking the square root on both sides: $S = \sqrt{16 	imes 10^{-6}} = 4.0 	imes 10^{-3} \ mol \ dm^{-3}$.

Calculate solubility $(mol \ dm^{-3})$ of a sparingly soluble electrolyte $AB$ at $298 \ K$ if its solubility product is $1.6 \times 10^{-5}$?

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