The solubility product of the sparingly solub | vedclass

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(C) The dissociation of the salt $AB_2$ is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.Let the solubility of $AB_2$ be $s \ mol \ dm^

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$4 	imes 10^{-4}$

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(C) The dissociation of the salt $AB_2$ is given by: $AB_2(s) ightleftharpoons A^{2+}(aq) + 2B^-(aq)$.Let the solubility of $AB_2$ be $s \ mol \ dm^{-3}$.Then,$[A^{2+}] = s$ and $[B^-] = 2s$.The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2$.Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 2.56 	imes 10^{-10}$.So,$4s^3 = 2.56 	imes 10^{-10}$.$s^3 = 0.64 	imes 10^{-10} = 64 	imes 10^{-12}$.$s = \sqrt[3]{64 	imes 10^{-12}} = 4 	imes 10^{-4} \ mol \ dm^{-3}$.

The solubility product of the sparingly soluble salt $AB_2$ is $2.56 \times 10^{-10}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

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