Question

The K_a of $C_{6}H_{5}COOH$ is $6.46\times 10^{- 5}$ and K_sp for $C_{6}H_{5}COO^{-}Ag^{+}$ is $2.5\times 10^{- 13}$ . How many times the silver benzoate more soluble in a buffer of $pH=3.19$ compared to its solubility in pure water?
Round off your answer to the nearest integer.

Answer 1

$\text{K}_{3} = \frac{\left[\text{C}_{6} \text{H}_{5} \text{COOH}\right] \left[\text{Ag}^{+}\right]}{\left[\text{H}^{+}\right]} = \frac{\text{s} \cdot \text{s}}{\left[\text{H}^{+}\right]} = \frac{\text{s}^{2}}{\left[\text{H}^{+}\right]} = \frac{\text{K}_{\text{sp}}}{\text{K}_{\text{a}}}$
where, s is the solubility of C₆H₅COOAg.
In a buffer of pH = 3.19
$\text{log} \left[\text{H}^{+}\right] = - \text{3.19} = \bar{4} \text{.81}$
$\left[\text{H}^{+}\right] = \text{antilog} \bar{4} \text{.81} = \text{6.46} \times \text{10}^{- 4}$
$\frac{\text{s}^{2}}{\left[\text{H}^{+}\right]} = \frac{\text{K}_{\text{sp}}}{\text{K}_{\text{a}}} \text{or s}^{2} = \frac{\text{K}_{\text{sp}} \times \left[\text{H}^{+}\right]}{\text{K}_{\text{a}}}$
$\text{s} = \sqrt{\frac{\text{2.5} \times \text{10}^{- \text{13}} \times \text{6.46} \times \text{10}^{- 4}}{\text{6.46} \times \text{10}^{- 5}}}$
$\text{s} = \sqrt{\text{2.5} \times \text{10}^{- \text{13}} \times \text{10}}$
s = 1.6 × 10^-6 M (in buffer)
In aqueous solution, solubility of C₆H₅COOAg :
K_sp = [C₆H₅COO^-] [Ag⁺] = s · s = s²
$\text{s} = \sqrt{\text{K}_{\text{sp}}} = \sqrt{\text{2.5} \times \text{10}^{- \text{13}}} = 5 \times \text{10}^{- 7} \text{M}$
$\frac{\left(\text{s}\right)_{\left(\left(\text{C}\right)_{6} \left(\text{H}\right)_{5} \text{COOAg}\right)} \text{in buffer}}{\left(\text{s}\right)_{\left(\left(\text{C}\right)_{6} \left(\text{H}\right)_{5} \text{COOAg}\right)} \text{in aqueous solution}} = \frac{\text{1.6} \times \left(\text{10}\right)^{- 6}}{\text{5.0} \times \left(\text{10}\right)^{- 7}} = \text{3.2}$
C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.