The solubility product of a sparingly soluble | vedclass

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(A) For a salt of the type $AX_2$,the dissociation is: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ mol/L$.Then $[A^{

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$2.0 	imes 10^{-4}$

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(A) For a salt of the type $AX_2$,the dissociation is: $AX_2(s) ightleftharpoons A^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ mol/L$.Then $[A^{2+}] = S$ and $[X^-] = 2S$.The solubility product expression is $K_{sp} = [A^{2+}][X^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$.So,$4S^3 = 3.2 	imes 10^{-11}$.$S^3 = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.$S = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol/L$.

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility in $mol/L$ is:

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