Answer:
Solubility
of silver benzoate in pure water
\[=\sqrt{{{K}_{sp}}}=\sqrt{2.5\times
{{10}^{-13}}}\]
\[=5\times
{{10}^{-7}}mol{{L}^{-1}}\]
\[[{{H}^{+}}]\]Buffer =Antilog
\[(-3.19)=6.457\times {{10}^{-4}}M\]
\[{{K}_{a}}=\frac{[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}][{{H}^{+}}]}{[{{C}_{6}}{{H}_{5}}COOH]}\]
\[\frac{[{{C}_{6}}{{H}_{5}}COOH]}{[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]}=\frac{[{{H}^{+}}]}{{{K}_{a}}}=\frac{6.457\times
{{10}^{-4}}}{6.46\times {{10}^{-5}}}=10\]Let solubility in buffer is \['x'\]
\[[A{{g}^{+}}]=x=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]+[{{C}_{6}}{{H}_{5}}COOH]\]
\[=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]+10[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]\]
\[=11[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]\]
\[[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]=\frac{x}{11}\]
Most of benzoate ion is converted to benzoic acid
\[{{K}_{sp}}=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}][A{{g}^{+}}]\]
\[=x\times \frac{x}{11}=\frac{{{x}^{2}}}{11}\]
\[x=1.66\times {{10}^{-6}}\]
Ratio of solubility will be = \[\frac{1.66\times
{{10}^{-6}}}{5\times {{10}^{-7}}}=3.32\]
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