Question

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

• A. $4$
• B. $3.32$
• C. $3.01$
• D. $2.5$

Answer 1

Answer: (B)

Suppose $S$ is the molar solubility of silver benzoate in water, then
$C_6{H}_{5}COOA{g}_{(s)}\rightleftharpoons C_6{H}_{5}CO{O}_{(aq)}^{-}+A{g}_{(aq)}^{+}$
$K_{sp}=S^2$
$\therefore S=\sqrt{2.5\times 10^{-13}}$
$=5.0\times 10^{-7}M$
If the solubility of salt of weak add of ionization constant $K_a$ is $S$', then $K_{sp}$, $K_a$ and $S$' are related to each other at $pH=3.19$.
$\therefore \left[H^{+}\right]=6.46\times 10^{-4}M\left(\because pH=3.19\right)$
$K_{sp}=S^{\prime 2}\left[\frac{K_a}{K_a+\left[H^{+}\right]}\right]$
$S^{\prime }={\left\{\frac{2.5\times 10^{-13}}{\left[\frac{6.46\times 10^{-5}}{6.46\times 10^{-5}+6.46\times 10^{-4}}\right]}\right\}}^{1/2}$
$S^{\prime }={\left\{\frac{2.5\times 10^{-13}\times 7.106\times 10^{-4}}{6.46\times 10^{-5}}\right\}}^{1/2}$
$={\left(2.75\times 10^{-12}\right)}^{1/2}$
$=1.658\times 10^{-6}M$
$\therefore$ The ratio of $\frac{S^{\prime }}{S}=\frac{1.658\times 10^{-6}}{5.0\times 10^{-7}}$
$=3.32$
Silver benzoate is $3.32$ times more soluble in buffer of $pH=3.19$ than in pure water.