Question

The ionization constant of benzoic acid is $6.46\times {10}^{-5}$ and Ksp 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?

Answer 1

Since pH = 3.19,
$\left[H_3{O}^{+}\right]=6.46\times {10}^{-4}M{C}_6{H}_{5}COOH+H_{2}O\leftrightarrow C_6{H}_{5}CO{O}^{-}+H_{3}O{K}_a\frac{\left[C_6{H}_{5}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}COOH\right]}\frac{\left[C_6{H}_{5}COOH\right]}{C_6{H}_{5}CO{O}^{-}}=\frac{\left[H_3{O}^{+}\right]}{K_a}=\frac{6.46\times {10}^{-4}}{6.46\times {10}^{-5}}=10$
Let the solubility of $C_6{H}_{5}COOAg$ be x mol/L.
Then,
$\left[A{g}^{+}\right]=x\left[C_6{H}_{5}COOH\right]+\left[C_6{H}_{5}CO{O}^{-}\right]=x10\left[C_6{H}_{5}CO{O}^{-}\right]+\left[C_6{H}_{5}CO{O}^{-}\right]=x$

$11\left[C_6{H}_{5}CO{O}^{-}\right]=x\left[C_6{H}_{5}CO{O}^{-}\right]=\frac{x}{11}{K}_{xp}\left[A{g}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]2.5\times {10}^{-13}=x\left(\frac{x}{11}\right)x=1.66\times {10}^{-6}mol/L$
Thus, the solubility of silver benzoate in a pH 3.19 solution is $1.66\times {10}^{-6}mol/L$.
Now, let the solubility of $C_6{H}_{5}COOAg$ be x' mol/L.
Then, $\left[A{g}^{+}\right]=x^{\prime }Mand\left[C{H}_{3}CO{O}^{-}\right]=x^{\prime }M.K_{sp}=\left[A{g}^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]K_{sp}=(x^{\prime }{)}^2{x}^{\prime }=\sqrt{K_{sp}}=\sqrt{2.5\times {10}^{-13}}=5\times {10}^{-7}mol/L\therefore \frac{x}{x^{\prime }}=\frac{1.66\times {10}^{-6}}{5\times {10}^{-7}}=3.32$
Hence, $C_6{H}_{5}COOAg$ is approximately 3.317 times more soluble in a low pH solution.