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Question

(D) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.The solubility product constant is $K_{sp} = [Ag^+][Cl^-]

Vedclass · Accepted Answer

$1.8 	imes 10^{-8}$

Vedclass · Answer

(D) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.The solubility product constant is $K_{sp} = [Ag^+][Cl^-] = 1.8 	imes 10^{-10}$.In a $0.01 \ M \ HCl$ solution,$HCl$ dissociates completely as $HCl ightarrow H^+ + Cl^-$,so $[Cl^-] = 0.01 \ M = 10^{-2} \ M$.Let the solubility of $AgCl$ be $s$. Then $[Ag^+] = s$ and $[Cl^-] = (s + 0.01) \approx 0.01 \ M$ (since $s$ is very small).Substituting these values into the $K_{sp}$ expression: $1.8 	imes 10^{-10} = s 	imes (0.01)$.$s = \frac{1.8 	imes 10^{-10}}{10^{-2}} = 1.8 	imes 10^{-8} \ M$.

The solubility product of silver chloride at $298 \ K$ is $1.8 \times 10^{-10}$. The solubility of $AgCl$ in $0.01 \ M \ HCl$ solution will be ............

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