The solubility of a salt A_2B_3 is 10^{-3} M | vedclass

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(A) For a salt of the type $A_2B_3$,the dissociation is given by: $A_2B_3(s) \rightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$.If $S$ is the solubility,th

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$1.08 	imes 10^{-13}$

Vedclass · Answer

(A) For a salt of the type $A_2B_3$,the dissociation is given by: $A_2B_3(s) ightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$.If $S$ is the solubility,then $[A^{3+}] = 2S$ and $[B^{2-}] = 3S$.The solubility product $K_{sp}$ is given by: $K_{sp} = [A^{3+}]^2 [B^{2-}]^3$.$K_{sp} = (2S)^2 	imes (3S)^3 = 4S^2 	imes 27S^3 = 108S^5$.Given $S = 10^{-3} \ M$,substituting the value:$K_{sp} = 108 	imes (10^{-3})^5 = 108 	imes 10^{-15} = 1.08 	imes 10^{-13}$.

The solubility of a salt $A_2B_3$ is $10^{-3} \ M$. What will be its solubility product?

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