The solubility of AgCl in water is 1.43 times | vedclass

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Question

(B) The molar mass of $AgCl$ is $143 \ g \ mol^{-1}$.Solubility in $mol \ L^{-1}$ $(S)$ is calculated as: $S = \frac{	ext{solubility in } g \ L^{-1}}

Vedclass · Accepted Answer

$1 	imes 10^{-10} \ M^2$

Vedclass · Answer

(B) The molar mass of $AgCl$ is $143 \ g \ mol^{-1}$.Solubility in $mol \ L^{-1}$ $(S)$ is calculated as: $S = \frac{	ext{solubility in } g \ L^{-1}}{	ext{molar mass}} = \frac{1.43 	imes 10^{-3} \ g \ L^{-1}}{143 \ g \ mol^{-1}} = 10^{-5} \ mol \ L^{-1}$.For $AgCl$,the dissociation is $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.The solubility product constant is $K_{sp} = [Ag^+][Cl^-] = S 	imes S = S^2$.$K_{sp} = (10^{-5})^2 = 10^{-10} \ M^2$.

The solubility of $AgCl$ in water is $1.43 \times 10^{-3} \ g \ L^{-1}$. What is the value of $K_{sp}$ for $AgCl$ at $25 \ ^oC$?

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