The solubility in water of a sparingly solubl | vedclass

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Question

(A) The dissociation of the salt is given by: $AB_2(s) ightleftharpoons A^{2+}(aq) + 2B^{-}(aq)$Let the solubility be $s = 1.0 	imes 10^{-5} \ mol

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$4 	imes 10^{-15}$

Vedclass · Answer

(A) The dissociation of the salt is given by: $AB_2(s) ightleftharpoons A^{2+}(aq) + 2B^{-}(aq)$Let the solubility be $s = 1.0 	imes 10^{-5} \ mol \ L^{-1}$.The concentrations of the ions are $[A^{2+}] = s$ and $[B^{-}] = 2s$.The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [A^{2+}][B^{-}]^2$Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$Calculating the value: $K_{sp} = 4 	imes (1.0 	imes 10^{-5})^3 = 4 	imes 10^{-15}$

The solubility in water of a sparingly soluble salt $AB_2$ is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product constant will be

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