The solubility of AgBr with solubility produc | vedclass

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(B) The dissolution of $AgBr$ is represented by the equilibrium: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.The solubility product expressio

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$5 	imes 10^{-12} \ M$

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(B) The dissolution of $AgBr$ is represented by the equilibrium: $AgBr(s) ightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.The solubility product expression is $K_{sp} = [Ag^{+}][Br^{-}] = 5.0 	imes 10^{-13}$.In the presence of $0.1 \ M$ $NaBr$,the concentration of $Br^{-}$ ions is dominated by the strong electrolyte $NaBr$,so $[Br^{-}] \approx 0.1 \ M$.Let the solubility of $AgBr$ be $S$. Then $[Ag^{+}] = S$.Substituting these values into the $K_{sp}$ expression: $S 	imes 0.1 = 5.0 	imes 10^{-13}$.Solving for $S$: $S = (5.0 	imes 10^{-13}) / 0.1 = 5.0 	imes 10^{-12} \ M$.Therefore,the solubility of $AgBr$ in $0.1 \ M$ $NaBr$ is $5.0 	imes 10^{-12} \ M$.

The solubility of $AgBr$ with solubility product $5.0 \times 10^{-13}$ at $298 \ K$ in $0.1 \ M$ $NaBr$ solution would be:

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