Question

The standard free energy of formation for $AgCl$ at $298K$ is $-109.7kJ$ ${mol}^{-1}$.
$\Delta G^o\left({Ag}^{+}\right)=77.2kJ/mol$; $\Delta G^o\left({Cl}^{-}\right)=-131.2kJ/mol$. Find the solubility of $AgCl$ in $0.05M$ $KCl$. Neglect any complication due to complexation.

• A. $K_{sp}=1.723\times {10}^{-10}$, $s=3.446\times {10}^{-9}M$
• B. $K_{sp}=1.723\times {10}^{-16}$, $s=3.446\times {10}^{-7}M$
• C. $K_{sp}=1.723\times {10}^{-5}$, $s=3.446\times {10}^{-5}M$
• D. $K_{sp}=1.723\times {10}^{-5}$, $s=3.446\times {10}^{-9}M$

Answer 1

The correct option is A $K_{sp}=1.723\times {10}^{-10}$, $s=3.446\times {10}^{-9}M$

Given,
$Ag+\frac{1}{2}{Cl}_2\rightleftharpoons AgCl$ $\Delta G_1^o=-109.7kJ/{mole}^{-1}$
$Ag⟶{Ag}^{+}+e^{-}$ $\Delta G_2^o=77.2kJ/{mole}^{-1}$
$e^{-}+\frac{1}{2}{Cl}_2⟶{Cl}^{-}$ $\Delta G_3^o=-131kJ/{mole}^{-1}$
So, for reaction
$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}$

${\Delta }^{o}G=-\Delta G_{1}o+\Delta G_2^o+\Delta G_3^o$
$\Delta G^o=-RT\ln K_{sp}$
$55.7\times {10}^{-3}=-8.314\times 298\ln K_{sp}$
$K_{sp}=1.723\times {10}^{-10}$
$1.723\times {10}^{-10}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]=s\times 0.05$
$s=3.446\times {10}^{-9}M$