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?
(C) Given $pH = 3.19$,then $[H_{3}O^{+}] = 10^{-3.19} \approx 6.46 \times 10^{-4} \, M$.
For the equilibrium: $C_{6}H_{5}COOH + H_{2}O \longleftrightarrow C_{6}H_{5}COO^{-} + H_{3}O^{+}$.
The ratio of concentrations is $\frac{[C_{6}H_{5}COOH]}{[C_{6}H_{5}COO^{-}]} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}} = 10$.
Let the solubility of $C_{6}H_{5}COOAg$ in the buffer be $x \, mol/L$.
Then $[Ag^{+}] = x$ and $[C_{6}H_{5}COOH] + [C_{6}H_{5}COO^{-}] = x$.
Substituting the ratio: $10[C_{6}H_{5}COO^{-}] + [C_{6}H_{5}COO^{-}] = x$,so $[C_{6}H_{5}COO^{-}] = \frac{x}{11}$.
Using $K_{sp} = [Ag^{+}][C_{6}H_{5}COO^{-}] = x \times \frac{x}{11} = 2.5 \times 10^{-13}$.
$x^{2} = 27.5 \times 10^{-13} = 2.75 \times 10^{-12}$,so $x \approx 1.66 \times 10^{-6} \, mol/L$.
In pure water,let solubility be $x^{\prime}$. $K_{sp} = (x^{\prime})^{2} = 2.5 \times 10^{-13}$.
$x^{\prime} = \sqrt{2.5 \times 10^{-13}} \approx 5.0 \times 10^{-7} \, mol/L$.
The ratio of solubility is $\frac{x}{x^{\prime}} = \frac{1.66 \times 10^{-6}}{5.0 \times 10^{-7}} = 3.32$.
For the equilibrium: $C_{6}H_{5}COOH + H_{2}O \longleftrightarrow C_{6}H_{5}COO^{-} + H_{3}O^{+}$.
The ratio of concentrations is $\frac{[C_{6}H_{5}COOH]}{[C_{6}H_{5}COO^{-}]} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}} = 10$.
Let the solubility of $C_{6}H_{5}COOAg$ in the buffer be $x \, mol/L$.
Then $[Ag^{+}] = x$ and $[C_{6}H_{5}COOH] + [C_{6}H_{5}COO^{-}] = x$.
Substituting the ratio: $10[C_{6}H_{5}COO^{-}] + [C_{6}H_{5}COO^{-}] = x$,so $[C_{6}H_{5}COO^{-}] = \frac{x}{11}$.
Using $K_{sp} = [Ag^{+}][C_{6}H_{5}COO^{-}] = x \times \frac{x}{11} = 2.5 \times 10^{-13}$.
$x^{2} = 27.5 \times 10^{-13} = 2.75 \times 10^{-12}$,so $x \approx 1.66 \times 10^{-6} \, mol/L$.
In pure water,let solubility be $x^{\prime}$. $K_{sp} = (x^{\prime})^{2} = 2.5 \times 10^{-13}$.
$x^{\prime} = \sqrt{2.5 \times 10^{-13}} \approx 5.0 \times 10^{-7} \, mol/L$.
The ratio of solubility is $\frac{x}{x^{\prime}} = \frac{1.66 \times 10^{-6}}{5.0 \times 10^{-7}} = 3.32$.
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