Calculate the solubility of sparingly soluble | vedclass

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Question

(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$Let the solubility of the salt be $S \ mol

Vedclass · Accepted Answer

$7.00 	imes 10^{-5}$

Vedclass · Answer

(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} ightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$Let the solubility of the salt 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 $K_{sp} = 4.9 	imes 10^{-9}$.Therefore,$S^2 = 4.9 	imes 10^{-9} = 49 	imes 10^{-10}$.Taking the square root on both sides: $S = \sqrt{49 	imes 10^{-10}} = 7.00 	imes 10^{-5} \ mol \ dm^{-3}$.

Calculate the solubility of sparingly soluble salt $BA$ in $mol \ dm^{-3}$ at $300 \ K$ if its solubility product is $4.9 \times 10^{-9}$ at same temperature.

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