Solubility of a binary sparingly soluble salt | vedclass

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(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$): $S = \frac{1.12 	imes 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} =

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$1 	imes 10^{-12}$

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(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$): $S = \frac{1.12 	imes 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} = 1.0 	imes 10^{-6} \ mol \ dm^{-3}$. $2$. For a binary salt $(AB)$,the dissociation is $AB ightleftharpoons A^+ + B^-$. $3$. The solubility product expression is $K_{sp} = [A^+][B^-] = S 	imes S = S^2$. $4$. Substitute the value of $S$: $K_{sp} = (1.0 	imes 10^{-6})^2 = 1.0 	imes 10^{-12} \ mol^2 \ dm^{-6}$.

Solubility of a binary sparingly soluble salt is $1.12 \times 10^{-4} \ g \ dm^{-3}$. Calculate its solubility product $(K_{sp})$ if the molar mass of the salt is $112 \ g \ mol^{-1}$.

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