Calculate the solubility of AgCl(s) in 0.1 M | vedclass

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(C) The dissolution of $AgCl$ is represented by the equilibrium: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-]$.In the prese

Vedclass · Accepted Answer

$2.8 	imes 10^{-9} \ M$

Vedclass · Answer

(C) The dissolution of $AgCl$ is represented by the equilibrium: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-]$.In the presence of $0.1 \ M \ NaCl$,the concentration of $Cl^-$ ions is dominated by the common ion effect from $NaCl$,so $[Cl^-] \approx 0.1 \ M$.Substituting the values: $2.8 	imes 10^{-10} = [Ag^+] 	imes 0.1$.Therefore,the solubility $S = [Ag^+] = \frac{2.8 	imes 10^{-10}}{0.1} = 2.8 	imes 10^{-9} \ M$.

Calculate the solubility of $AgCl(s)$ in $0.1 \ M \ NaCl$ at $25 \ ^oC$. Given ${K_{sp}}(AgCl) = 2.8 \times 10^{-10}$.

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