Question

The K_a of $C_6{H}_{5}COOH$ is $6.46\times 10^{-5}$ and K_sp for $C_6{H}_{5}CO{O}^{-}A{g}^{+}$ 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

Answer: 3

${\text{K}}_3=\frac{\left[{\text{C}}_6{\text{H}}_5\text{COOH}\right]\left[{\text{Ag}}^{+}\right]}{\left[{\text{H}}^{+}\right]}=\frac{\text{s}\cdot \text{s}}{\left[{\text{H}}^{+}\right]}=\frac{{\text{s}}^2}{\left[{\text{H}}^{+}\right]}=\frac{{\text{K}}_{\text{sp}}}{{\text{K}}_{\text{a}}}$
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.