Calculate the solubility in mol dm^{-3} of s | vedclass

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Question

(A) For a sparingly soluble salt $BA$,the dissociation equilibrium is represented as:$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$Let the

Vedclass · Accepted Answer

$7.0 	imes 10^{-7}$

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(A) For a sparingly soluble salt $BA$,the dissociation equilibrium is represented as:$BA_{(s)} ightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$Let the solubility of $BA$ be $S \ mol \ dm^{-3}$.Then,$[B^{+}] = S$ and $[A^{-}] = S$.The solubility product constant $K_{sp}$ is given by:$K_{sp} = [B^{+}][A^{-}] = S 	imes S = S^2$Given $K_{sp} = 4.9 	imes 10^{-13}$.Therefore,$S^2 = 4.9 	imes 10^{-13} = 49 	imes 10^{-14}$.Taking the square root on both sides:$S = \sqrt{49 	imes 10^{-14}} = 7.0 	imes 10^{-7} \ mol \ dm^{-3}$.

Calculate the solubility in $mol \ dm^{-3}$ of sparingly soluble salt $BA$ if its solubility product is $4.9 \times 10^{-13}$ at the same temperature.

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