For a sparingly soluble salt AB_2,the equilib | vedclass

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(B) The dissociation of the salt is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$The expression for the solubility product co

Vedclass · Accepted Answer

$6.91 	imes 10^{-12}$

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(B) The dissociation of the salt is given by: $AB_{2(s)} ightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$The expression for the solubility product constant $(K_{sp})$ is: $K_{sp} = [A^{2+}] [B^{-}]^2$Given concentrations: $[A^{2+}] = 1.2 	imes 10^{-4} \ M$ and $[B^{-}] = 0.24 	imes 10^{-3} \ M = 2.4 	imes 10^{-4} \ M$Substituting the values: $K_{sp} = (1.2 	imes 10^{-4}) 	imes (2.4 	imes 10^{-4})^2$$K_{sp} = (1.2 	imes 10^{-4}) 	imes (5.76 	imes 10^{-8})$$K_{sp} = 6.912 	imes 10^{-12} \ M^3$Thus,the correct option is $B$.

For a sparingly soluble salt $AB_2$,the equilibrium concentrations of $A^{2+}$ ions and $B^{-}$ ions are $1.2 \times 10^{-4} \ M$ and $0.24 \times 10^{-3} \ M$,respectively. The solubility product of $AB_2$ is :

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