The K_{sp} of AgCl is 1.8 times 10^{-10}. The | vedclass

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(A) The precipitation of $AgCl$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.For the reaction $AgCl(s) \rightlefth

Vedclass · Accepted Answer

$4.5 	imes 10^{-8} \ M$

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(A) The precipitation of $AgCl$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.For the reaction $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$,the solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.Given $K_{sp} = 1.8 	imes 10^{-10}$ and $[Ag^+] = 4 	imes 10^{-3} \ M$.To find the minimum concentration of $Cl^-$ required for precipitation,we set the ionic product equal to $K_{sp}$:$[Cl^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.8 	imes 10^{-10}}{4 	imes 10^{-3}} = 0.45 	imes 10^{-7} \ M = 4.5 	imes 10^{-8} \ M$.

The $K_{sp}$ of $AgCl$ is $1.8 \times 10^{-10}$. The concentration of $Cl^-$ required to initiate the precipitation of $AgCl$ in a solution containing $4 \times 10^{-3} \ M \ Ag^+$ is:

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