Chemistry - Ionic Equilibrium Question with Solution | TestHub

Suppose S is the molar solubility of silver benzoate in water, then
${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{A}{\mathrm{g}}_{\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}\mathrm{O}{\mathrm{O}}_{\left(\mathrm{a}\mathrm{q}\right)}^{-}+\mathrm{A}{\mathrm{g}}_{\left(\mathrm{a}\mathrm{q}\right)}^{+}$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}={\mathrm{S}}^2$
So$\mathrm{S}=\sqrt{2.5\times 10^{-13}}=5.0\times 10^{-7}$M
If the solubility of salt of weak acid of ionization constant${\mathrm{K}}_{\mathrm{a}}$is S, then${\mathrm{K}}_{\mathrm{s}\mathrm{p}},{\mathrm{K}}_{\mathrm{a}}$and S' are related to each other at$\mathrm{p}\mathrm{H}=3.19.$
So$\left[{\mathrm{H}}^{+}\right]=6.46\times 10^{-4}\mathrm{M}$$\left(\because \text{ pH}=3.19\right)$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=\mathrm{S}{\mathrm{\prime }}^2\left[\frac{{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{K}}_{\mathrm{a}}+\left[{\mathrm{H}}^{+}\right]}\right]$
$\mathrm{S}\mathrm{\prime }={\left\{\frac{2.5\times 10^{-13}}{\left[\frac{6.46\times 10^{-5}}{6.46\times 10^{-5}+6.46\times 10^{-4}}\right]}\right\}}^{\frac{1}{2}}$
$\mathrm{S}\mathrm{\prime }={\left\{\frac{2.5\times 10^{-13}\times 7.106\times 10^{-4}}{6.46\times 10^{-5}}\right\}}^{\frac{1}{2}}$
$={\left(2.75\times 10^{-12}\right)}^{\frac{1}{2}}$
$=1.658\times 10^{-6}\mathrm{ }\mathrm{M}$
So the ratio of$\frac{\mathrm{S}\mathrm{\prime }}{\mathrm{S}}=\frac{1.658\times 10^{-6}}{5.0\times 10^{-7}}=3.32$
Silver benzoate is 3.32 times more soluble in buffer of pH = 3.19 than in pure water.