At T(K),the solubility product of AgBr is 4 t | vedclass

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(C) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.Let the solubility of $AgBr$ in $0.1 \ M$ $KBr$ be $s$.In

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$4 	imes 10^{-12} \ M$

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(C) The dissociation of $AgBr$ is given by: $AgBr(s) ightleftharpoons Ag^+(aq) + Br^-(aq)$.Let the solubility of $AgBr$ in $0.1 \ M$ $KBr$ be $s$.In the presence of $0.1 \ M$ $KBr$,the concentration of $Br^-$ ions is $[Br^-] = (s + 0.1) \approx 0.1 \ M$ (since $s$ is very small).The concentration of $Ag^+$ ions is $[Ag^+] = s$.The solubility product expression is $K_{sp} = [Ag^+][Br^-]$.Substituting the values: $4 	imes 10^{-13} = s 	imes 0.1$.Solving for $s$: $s = \frac{4 	imes 10^{-13}}{0.1} = 4 	imes 10^{-12} \ M$.

At $T(K)$,the solubility product of $AgBr$ is $4 \times 10^{-13}$. What is its solubility in $0.1 \ M$ $KBr$ solution?

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