The ionisation constant of benzoic acid is $6.46 imes 10^{-5}$ and $K_{s p}$ for silver benzoate i

Question

The ionisation constant of benzoic acid is $6.46 	imes 10^{-5}$ and $K_{s p}$ for silver benzoate is $2.5 	imes 10^{-13}$. How many times is silver benzoate more soluble in a buffer of $\mathrm{pH}=3.19$ compared to its solubility in pure water?

Professor PC Thomas Classes · Answer

(b) : Suppose $S$ is the molar solubility of silver benzoate in water, then
$
\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}_{(s)} ightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}_{(a q)}^{-}+\mathrm{Ag}_{(a q)}^{+} \
& K_{s p}=S^2 	herefore S=\sqrt{2.5 	imes 10^{-13}}=5.0 	imes 10^{-7} \mathrm{M}
\end{aligned}
$
If the solubility of salt of weak acid of ionisation constant $K_a$ is $S^{\prime}$, then $K_{s p}, K_a$ and $S^{\prime}$ are related to each other at $\mathrm{pH}=3.19$.
$
\begin{array}{rlr}
	herefore & {\left[\mathrm{H}^{+}ight]}  =6.46 	imes 10^{-4} \mathrm{M} \
& K_{s p}  =S^{\prime 2}\left[\frac{K_a}{K_a+\left[\mathrm{H}^{+}ight]}ight] \
& S^{\prime}  =\left\{\frac{2.5 	imes 10^{-13}}{\left[\frac{6.46 	imes 10^{-5}}{6.46 	imes 10^{-5}+6.46 	imes 10^{-4}}ight]}ight\}^{1 / 2} \
& S^{\prime}  =\left\{\frac{2.5 	imes 10^{-13} 	imes 7.106 	imes 10^{-4}}{6.46 	imes 10^{-5}}ight\}^{1 / 2} \
&  =\left(2.75 	imes 10^{-12}ight)^{1 / 2}=1.658 	imes 10^{-6} \mathrm{M}
\end{array}
$
$	herefore \quad$ The ratio of $\frac{S^{\prime}}{S}=\frac{1.658 	imes 10^{-6}}{5.0 	imes 10^{-7}}=3.32$
Silver benzoate is 3.32 times more soluble in buffer of $\mathrm{pH}=3.19$ than in pure water.