If 0.001435 g of AgCl is present in 100 mL | vedclass

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(B) The solubility $s$ is the concentration in $mol/L$. Given mass of $AgCl = 0.001435 \ g$ in $100 \ mL$. Mass in $1000 \ mL (1 \ L) = 0.001435 	ime

Vedclass · Accepted Answer

$10^{-8}$

Vedclass · Answer

(B) The solubility $s$ is the concentration in $mol/L$. Given mass of $AgCl = 0.001435 \ g$ in $100 \ mL$. Mass in $1000 \ mL (1 \ L) = 0.001435 	imes 10 = 0.01435 \ g/L$. Molar mass of $AgCl = 107.87 + 35.45 \approx 143.5 \ g/mol$. Solubility $s = \frac{0.01435 \ g/L}{143.5 \ g/mol} = 10^{-4} \ mol/L$. For $AgCl ightleftharpoons Ag^+ + Cl^-$,$K_{sp} = [Ag^+][Cl^-] = s^2$. $K_{sp} = (10^{-4})^2 = 10^{-8}$.

If $0.001435 \ g$ of $AgCl$ is present in $100 \ mL$ of water,calculate the $K_{sp}$.

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