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

The Correct Option is B

Suppose $S$ is the molar solubility of silver benzoate in water, then $ {C6H5COOAg_{(s)}<=> C6H5COO^{-}_{(aq)} +Ag^+_{(aq)}}$ $K_{sp}=S^{2}$ $\therefore S=\sqrt{2.5\times10^{-13}}$ $=5.0\times10^{-7}\,M$ 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 $pH = 3.19$. $\therefore \left[H^{+}\right]=6.46\times10^{-4}\,M\quad\left(\because pH=3.19\right)$ $K_{sp}=S'^{2}\left[\frac{K_{a}}{K_{a}+\left[H^{+}\right]}\right]$ $S'=\left\{\frac{2.5\times10^{-13}}{\left[\frac{6.46\times10^{-5}}{6.46\times10^{-5}+6.46\times10^{-4}}\right]}\right\}^{1/2}$ $S'=\left\{\frac{2.5\times10^{-13}\times7.106\times10^{-4}}{6.46\times10^{-5}}\right\}^{1/2}$ $=\left(2.75\times10^{-12}\right)^{1/2}$ $=1.658\times10^{-6}\,M$ $\therefore$ The ratio of $\frac{S'}{S}=\frac{1.658\times10^{-6}}{5.0\times10^{-7}}$ $=3.32$ Silver benzoate is $3.32$ times more soluble in buffer of $pH = 3.19$ than in pure water.