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(A) The dissociation of the salt $B_3A_2$ is given by: $B_3A_{2(s)} \rightleftharpoons 3B^{2+} + 2A^{3-}$ Let the solubility be $S \ mol \ L^{-1}$. Th

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$S = \left( \frac{K_{sp}}{108} \right)^{\frac{1}{5}}$

Vedclass · Answer

(A) The dissociation of the salt $B_3A_2$ is given by: $B_3A_{2(s)} ightleftharpoons 3B^{2+} + 2A^{3-}$ Let the solubility be $S \ mol \ L^{-1}$. Then,$[B^{2+}] = 3S$ and $[A^{3-}] = 2S$. The solubility product expression is: $K_{sp} = [B^{2+}]^3 [A^{3-}]^2$ Substituting the values: $K_{sp} = (3S)^3 (2S)^2$ $K_{sp} = (27S^3) 	imes (4S^2)$ $K_{sp} = 108S^5$ Solving for $S$: $S^5 = \frac{K_{sp}}{108}$ $S = \left( \frac{K_{sp}}{108} ight)^{\frac{1}{5}}$

Which from the following equations represents the relation between solubility ($S$ in $mol \ L^{-1}$) and solubility product $(K_{sp})$ for a salt $B_3A_2$?

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