The solubility (in mol L^{-1}) of AgCl (K_{s | vedclass

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(A) Let the solubility of $AgCl$ be $x \ mol \ L^{-1}$.The dissociation equilibrium is: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.The solub

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$1.0 	imes 10^{-9}$

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(A) Let the solubility of $AgCl$ be $x \ mol \ L^{-1}$.The dissociation equilibrium is: $AgCl(s) ightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.The solubility product expression is: $K_{sp} = [Ag^{+}][Cl^{-}]$.In a $0.1 \ M \ KCl$ solution,$KCl$ dissociates completely: $KCl(aq) ightarrow K^{+}(aq) + Cl^{-}(aq)$.Thus,the concentration of $Cl^{-}$ from $KCl$ is $0.1 \ M$.The total concentration of $Cl^{-}$ is $[Cl^{-}] = (x + 0.1) \ M \approx 0.1 \ M$ (since $x$ is very small).Substituting the values into the $K_{sp}$ expression: $1.0 	imes 10^{-10} = (x)(0.1)$.Solving for $x$: $x = \frac{1.0 	imes 10^{-10}}{0.1} = 1.0 	imes 10^{-9} \ mol \ L^{-1}$.

The solubility (in $mol \ L^{-1}$) of $AgCl$ $(K_{sp} = 1.0 \times 10^{-10})$ in a $0.1 \ M \ KCl$ solution will be

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