The solubility of salt BA_2 is 4 times 10^{-4 | vedclass

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Question

(B) The dissociation of the salt $BA_2$ is given by: $BA_2 (s) ightleftharpoons B^{2+} (aq) + 2A^- (aq)$.Let the solubility of the salt be $s = 4

Vedclass · Accepted Answer

$2.56 	imes 10^{-10}$

Vedclass · Answer

(B) The dissociation of the salt $BA_2$ is given by: $BA_2 (s) ightleftharpoons B^{2+} (aq) + 2A^- (aq)$.Let the solubility of the salt be $s = 4 	imes 10^{-4} \ mol \ dm^{-3}$.Then,$[B^{2+}] = s$ and $[A^-] = 2s$.The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [B^{2+}][A^-]^2$.Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.Now,substitute $s = 4 	imes 10^{-4}$:$K_{sp} = 4 	imes (4 	imes 10^{-4})^3 = 4 	imes (64 	imes 10^{-12}) = 256 	imes 10^{-12} = 2.56 	imes 10^{-10}$.

The solubility of salt $BA_2$ is $4 \times 10^{-4} \ mol \ dm^{-3}$. What is the solubility product of the salt?

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