The K_{sp} of AgCl is 1.0 times 10^{-10}. Cal | vedclass

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(A) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-] = 1.0 	imes 10^{-10}$.In $0.2 \ M

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

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(A) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-] = 1.0 	imes 10^{-10}$.In $0.2 \ M \ AgNO_3$,the concentration of $Ag^+$ ions is $0.2 \ M$ due to complete dissociation of $AgNO_3$.Let the solubility of $AgCl$ be $s$.Then,$[Ag^+] = 0.2 + s \approx 0.2 \ M$ (since $s$ is very small) and $[Cl^-] = s$.Substituting these values into the $K_{sp}$ expression: $1.0 	imes 10^{-10} = (0.2)(s)$.$s = \frac{1.0 	imes 10^{-10}}{0.2} = 5 	imes 10^{-10} \ M$.

The $K_{sp}$ of $AgCl$ is $1.0 \times 10^{-10}$. Calculate the solubility of $AgCl$ in $0.2 \ M \ AgNO_3$.

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