If the solubility product of AB_2 is 3.20 tim | vedclass

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(C) The dissociation of $AB_2$ is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$Let the solubility be $s \ mol \ L^{-1}$. Then

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(C) The dissociation of $AB_2$ is given by: $AB_{2(s)} ightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$Let the solubility be $s \ mol \ L^{-1}$. Then $[A^{2+}] = s$ and $[B^{-}] = 2s$.The solubility product expression is: $K_{sp} = [A^{2+}][B^{-}]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 3.20 	imes 10^{-11}$,we have $4s^3 = 3.20 	imes 10^{-11}$.$s^3 = \frac{3.20 	imes 10^{-11}}{4} = 0.80 	imes 10^{-11} = 8.0 	imes 10^{-12}$.Taking the cube root: $s = \sqrt[3]{8.0 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol \ L^{-1}$.Thus,the value is $2 	imes 10^{-4} \ mol \ L^{-1}$.

If the solubility product of $AB_2$ is $3.20 \times 10^{-11} \ mol^3 L^{-3}$,then the solubility of $AB_2$ in pure water is......... $\times 10^{-4} \ mol \ L^{-1}$. [Assuming that neither kind of ion reacts with water]

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