The solubility of AgBr_{(s)},having solubilit | vedclass

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Question

(B) The solubility product $(K_{sp})$ of $AgBr$ is $5 	imes 10^{-10}$.For the dissociation reaction: $AgBr_{(s)} ightleftharpoons Ag^{+}_{(aq)} + B

Vedclass · Accepted Answer

$25 	imes 10^{-10} \ M$

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(B) The solubility product $(K_{sp})$ of $AgBr$ is $5 	imes 10^{-10}$.For the dissociation reaction: $AgBr_{(s)} ightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.The expression for $K_{sp}$ is $K_{sp} = [Ag^{+}][Br^{-}]$.In $0.2 \ M$ $NaBr$ solution,the concentration of $Br^{-}$ ions is $0.2 \ M$ due to the common ion effect.Let the solubility of $AgBr$ be $s$.Then $[Ag^{+}] = s$ and $[Br^{-}] = 0.2 + s \approx 0.2$ (since $s$ is very small).$5 	imes 10^{-10} = s 	imes 0.2$.$s = \frac{5 	imes 10^{-10}}{0.2} = 25 	imes 10^{-10} \ M$.

The solubility of $AgBr_{(s)}$,having solubility product $5 \times 10^{-10}$ in $0.2 \ M$ $NaBr$ solution,equals

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