Question

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 $pH3.19$ compared to its solubility in pure water?

Answer 1

Let $s$ M be the solubility of silver benzoate in water.
$\left[A{g}^{+}\right]=\left[PhCO{O}^{-}\right]=sM$

$K_{sp}=\left[PhCO{O}^{-}\right]\left[A{g}^{+}\right]=s\times s=s^2=2.5\times {10}^{-13}$

$s=5.0\times {10}^{-7}M$. This is solubility of silver benzoate in pure water.

$pH=-log\left[H^{+}\right]=3.19$

$\left[H^{+}\right]=6.456\times {10}^{-4}$

Benzoate ions combine with protons to form benzoic acid but hydrogen ion concentration is constant due to buffer solution.

$PhCOOH\rightleftharpoons PhCO{O}^{-}+H^{+}$

$K_a=\frac{\left[PhCO{O}^{-}\right]\left[H^{+}\right]}{\left[PhCOOH\right]}$

$\frac{\left[PhCOOH\right]}{\left[PhCO{O}^{-}\right]}=\frac{\left[H^{+}\right]}{K_a}=\frac{6.456\times {10}^{-4}}{6.46\times {10}^{-5}}=10$
Let y M be the soubiity in the buffer solution.
$y=\left[A{g}^{+}\right]=\left[PhCO{O}^{-}\right]+\left[PhCOOH\right]=\left[PhCO{O}^{-}\right]+10\left[PhCO{O}^{-}\right]=11\left[PhCO{O}^{-}\right]$

$\left[PhCO{O}^{-}\right]=\frac{y}{11}$

$K_{sp}=\left[PhCO{O}^{-}\right]\left[A{g}^{+}\right]$

$2.5\times {10}^{-13}=\frac{y}{11}\times y$

$y=1.66\times {10}^{-6}$

$\frac{y}{s}=\frac{1.66\times {10}^{-6}}{5.0\times {10}^{-7}}=3.32$

Thus, silver benzoate is 3.32 times more soluble in buffer.