If the solubility product K_{sp} of a sparing | vedclass

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(B) For a sparingly soluble salt $MX_2$,the dissociation equilibrium is: $MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^{-}(aq)$Let the solubility of the

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$1.36 	imes 10^{-4}$

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(B) For a sparingly soluble salt $MX_2$,the dissociation equilibrium is: $MX_2(s) ightleftharpoons M^{2+}(aq) + 2X^{-}(aq)$Let the solubility of the salt be $S \, 	ext{mol L}^{-1}$. Then,$[M^{2+}] = S$ and $[X^{-}] = 2S$.The solubility product expression is: $K_{sp} = [M^{2+}][X^{-}]^2 = (S)(2S)^2 = 4S^3$Given $K_{sp} = 1.0 	imes 10^{-11}$,we have:$4S^3 = 1.0 	imes 10^{-11}$$S^3 = 0.25 	imes 10^{-11} = 2.5 	imes 10^{-12}$$S = \sqrt[3]{2.5 	imes 10^{-12}} \approx 1.357 	imes 10^{-4} \, 	ext{mol L}^{-1}$Rounding to the nearest value,$S \approx 1.36 	imes 10^{-4} \, 	ext{mol L}^{-1}$.

If the solubility product $K_{sp}$ of a sparingly soluble salt $MX_2$ at $25\,^{\circ}C$ is $1.0 \times 10^{-11}$,the solubility of the salt in $\text{mol L}^{-1}$ at this temperature will be:

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