Solubility of a salt A_2B_3 is 1 times 10^{-3 | vedclass

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Question

(A) The dissociation of the salt is given by: $A_2B_{3(s)} \rightleftharpoons 2A^{3+}_{(aq)} + 3B^{2-}_{(aq)}$Here,the stoichiometric coefficients are

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$1.08 	imes 10^{-13}$

Vedclass · Answer

(A) The dissociation of the salt is given by: $A_2B_{3(s)} ightleftharpoons 2A^{3+}_{(aq)} + 3B^{2-}_{(aq)}$Here,the stoichiometric coefficients are $x=2$ and $y=3$.The solubility product expression is $K_{sp} = [A^{3+}]^2 [B^{2-}]^3 = (2S)^2 (3S)^3$.$K_{sp} = 4S^2 	imes 27S^3 = 108S^5$.Given solubility $S = 1 	imes 10^{-3} \ mol \ dm^{-3}$.$K_{sp} = 108 	imes (1 	imes 10^{-3})^5 = 108 	imes 10^{-15} = 1.08 	imes 10^{-13}$.

Solubility of a salt $A_2B_3$ is $1 \times 10^{-3} \ mol \ dm^{-3}$. What is the value of its solubility product?

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