Solubility of Ca_{3}(PO_{4})_{2} is S mol d | vedclass

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(B) The dissociation of $Ca_{3}(PO_{4})_{2}$ is represented as: $Ca_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_{4}^{3-}(aq)$ If the solu

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$108 \ S^{5}$

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(B) The dissociation of $Ca_{3}(PO_{4})_{2}$ is represented as: $Ca_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_{4}^{3-}(aq)$ If the solubility is $S \ mol \ dm^{-3}$,then the concentration of ions at equilibrium is: $[Ca^{2+}] = 3S$ $[PO_{4}^{3-}] = 2S$ The solubility product expression is: $K_{sp} = [Ca^{2+}]^{3} [PO_{4}^{3-}]^{2}$ Substituting the values: $K_{sp} = (3S)^{3} (2S)^{2}$ $K_{sp} = (27S^{3}) (4S^{2})$ $K_{sp} = 108S^{5}$

Solubility of $Ca_{3}(PO_{4})_{2}$ is $S \ mol \ dm^{-3}$. Find the solubility product $(K_{sp})$.

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