The solubility product of salt B_2A is 3.2 ti | vedclass

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(C) The dissociation of salt $B_2A$ is given by: $B_2A(s) \rightleftharpoons 2B^+(aq) + A^{2-}(aq)$.Let the solubility of $B_2A$ be $s \ mol \ dm^{-3}

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$2.00 	imes 10^{-4} \ mol \ dm^{-3}$

Vedclass · Answer

(C) The dissociation of salt $B_2A$ is given by: $B_2A(s) ightleftharpoons 2B^+(aq) + A^{2-}(aq)$.Let the solubility of $B_2A$ be $s \ mol \ dm^{-3}$.Then,$[B^+] = 2s$ and $[A^{2-}] = s$.The solubility product expression is: $K_{sp} = [B^+]^2 [A^{2-}] = (2s)^2 (s) = 4s^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$.So,$4s^3 = 3.2 	imes 10^{-11}$.$s^3 = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.$s = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol \ dm^{-3}$.

The solubility product of salt $B_2A$ is $3.2 \times 10^{-11}$ at $298 \ K$. What is the solubility of the salt at the same temperature?

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