The solubility of AgCl is 1.435 times 10^{-5} | vedclass

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(B) Step $1$: Calculate molar solubility $(S)$ in $mol \ L^{-1}$.$S = \frac{	ext{solubility in } g \ L^{-1}}{	ext{Molar mass of } AgCl} = \frac{1.43

Vedclass · Accepted Answer

$1.0 	imes 10^{-14}$

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(B) Step $1$: Calculate molar solubility $(S)$ in $mol \ L^{-1}$.$S = \frac{	ext{solubility in } g \ L^{-1}}{	ext{Molar mass of } AgCl} = \frac{1.435 	imes 10^{-5} \ g \ L^{-1}}{143.5 \ g \ mol^{-1}} = 1.0 	imes 10^{-7} \ mol \ L^{-1}$.Step $2$: Write the dissociation equilibrium for $AgCl$.$AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.Step $3$: Calculate $K_{sp}$.$K_{sp} = [Ag^+][Cl^-] = S 	imes S = S^2$.$K_{sp} = (1.0 	imes 10^{-7})^2 = 1.0 	imes 10^{-14}$.

The solubility of $AgCl$ is $1.435 \times 10^{-5} \ g \ L^{-1}$ at $30^{\circ} C$. Calculate its solubility product $(K_{sp})$. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

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