# The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13.

Question:

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

Solution:

Since $\mathrm{pH}=3.19$.

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=6.46 \times 10^{-4} \mathrm{M}$

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \longleftrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}$

$K_{a} \frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right]}$

$\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{a}}=\frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}}=10$

Let the solubility of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOAg}$ be $x \mathrm{~mol} / \mathrm{L}$

Then,

$\left[\mathrm{Ag}^{+}\right]=x$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right]+\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]=x$

$10\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]+\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]=x$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]=\frac{x}{11}$

$K_{s p}\left[\mathrm{Ag}^{+}\right]\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]$

$2.5 \times 10^{-13}=x\left(\frac{x}{11}\right)$

$x=1.66 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$

Thus, the solubility of silver benzoate in a pH $3.19$ solution is $1.66 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$.

Now, let the solubility of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOAg}$ be $x^{\prime} \mathrm{mol} / \mathrm{L}$.

Then, $\left[\mathrm{Ag}^{+}\right]=x^{\prime} \mathrm{M}$ and $\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=x^{\prime} \mathrm{M} .$

$K_{\text {sp }}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]$

$K_{s p}=\left(x^{\prime}\right)^{2}$

$x^{\prime}=\sqrt{K_{s p}}=\sqrt{2.5 \times 10^{-13}}=5 \times 10^{-7} \mathrm{~mol} / \mathrm{L}$

$\therefore \frac{x}{x^{\prime}}=\frac{1.66 \times 10^{-6}}{5 \times 10^{-7}}=3.32$

Hence, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOAg}$ is approximately $3.317$ times more soluble in a low pH solution.