The solubility of CuBr is 2 times 10^{-4} mo | vedclass

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(A) The dissociation of $CuBr$ is given by: $CuBr(s) \rightleftharpoons Cu^{+}(aq) + Br^{-}(aq)$Let $S$ be the solubility of $CuBr$ in $mol/L$.Then,$[

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$4 	imes 10^{-8} \ mol^2 \ L^{-2}$

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(A) The dissociation of $CuBr$ is given by: $CuBr(s) ightleftharpoons Cu^{+}(aq) + Br^{-}(aq)$Let $S$ be the solubility of $CuBr$ in $mol/L$.Then,$[Cu^{+}] = S$ and $[Br^{-}] = S$.The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Cu^{+}][Br^{-}] = S 	imes S = S^2$.Given $S = 2 	imes 10^{-4} \ mol/L$,we have:$K_{sp} = (2 	imes 10^{-4})^2 = 4 	imes 10^{-8} \ mol^2 \ L^{-2}$.

The solubility of $CuBr$ is $2 \times 10^{-4} \ mol/L$ at $25 \ ^\circ C$. The $K_{sp}$ value for $CuBr$ is

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