The solubility of AgCl at 20,^oC is 1.435 tim | vedclass

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(A) $1$. Calculate the molar mass of $AgCl$: $M(AgCl) = 108 + 35.5 = 143.5 \, g/mol$. $2$. Convert solubility from $g/L$ to $mol/L$ (molar solubility,

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$1.0 	imes 10^{-10}$

Vedclass · Answer

(A) $1$. Calculate the molar mass of $AgCl$: $M(AgCl) = 108 + 35.5 = 143.5 \, g/mol$. $2$. Convert solubility from $g/L$ to $mol/L$ (molar solubility,$s$): $s = \frac{1.435 	imes 10^{-3} \, g/L}{143.5 \, g/mol} = 1.0 	imes 10^{-5} \, mol/L$. $3$. For $AgCl$,the dissociation is $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$. $4$. The solubility product constant $(K_{sp})$ is given by $K_{sp} = [Ag^+][Cl^-] = s 	imes s = s^2$. $5$. $K_{sp} = (1.0 	imes 10^{-5})^2 = 1.0 	imes 10^{-10}$.

The solubility of $AgCl$ at $20\,^oC$ is $1.435 \times 10^{-3} \, g/L$. The solubility product of $AgCl$ is .......

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