The solubility product of a sparingly soluble | vedclass

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(C) The dissociation of the salt is given by: $AX_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$Let $S$ be the solubility of the salt.At equ

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$2.0 	imes 10^{-3}$

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(C) The dissociation of the salt is given by: $AX_{2(s)} ightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$Let $S$ be the solubility of the salt.At equilibrium,$[A^{2+}] = S$ and $[X^{-}] = 2S$.The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2$$K_{sp} = (S)(2S)^2 = 4S^3$Given $K_{sp} = 3.2 	imes 10^{-8}$.$4S^3 = 3.2 	imes 10^{-8}$$S^3 = \frac{3.2}{4} 	imes 10^{-8} = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$$S = \sqrt[3]{8 	imes 10^{-9}} = 2 	imes 10^{-3} \ mol \ dm^{-3}$

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-8}$. What is its solubility (in $mol \ dm^{-3}$)?

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