Question

The ionization constant of benzoic acid is $6.46\times 10^{- 5}$ and $K_{s p}$ for silver benzoic 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
$C _6 H _5 COOAg _{ s } \rightleftharpoons C _6 H _5 COO _{ aq }^{-}+ Ag _{ aq }^{+}$ So $S=\sqrt{2.5 \times 1 0^{- 13}}=5.0\times 10^{- 7}$ M
If the solubility of salt of weak acid of ionization constant $K_{a}$ is S, then $K_{s p},K_{a}$ and S' are related to each other at $pH=3.19.$
So $\left[\right.H^{+}\left]\right.=6.46\times 10^{- 4}M$ $\left(\because \text{ pH} = 3.19\right)$
$K_{s p}=S′^{2}\left[\frac{K_{a}}{K_{a} + \left[H^{+}\right]}\right]$
$S′=\left\{\frac{2.5 \times 1 0^{- 13}}{\left[\frac{6.46 \times 1 0^{- 5}}{6.46 \times 1 0^{- 5} + 6.46 \times 1 0^{- 4}}\right]}\right\}^{\frac{1}{2}}$
$S′=\left\{\frac{2.5 \times 1 0^{- 13} \times 7.106 \times 1 0^{- 4}}{6.46 \times 1 0^{- 5}}\right\}^{\frac{1}{2}}$
$=\left(2.75 \times 1 0^{- 12}\right)^{\frac{1}{2}}$
$=1.658\times 10^{- 6} \, M$
So the ratio of $\frac{S ′}{S}=\frac{1.658 \times 1 0^{- 6}}{5.0 \times 1 0^{- 7}}=3.32$
Silver benzoate is $3.32$ times more soluble in buffer of $pH = 3.19$ than in pure water.