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 $pH=3.19$ compared to its solubility in pure water?

• A. $4$
• B. $3.32$
• C. $3.01$
• D. $2.5$

Answer 1

Answer: (B)

$3.32$

Since $pH=3.19$

$\left[H_3{O}^{+}\right]=6.46\times {10}^{-4}M$

$C_6{H}_{5}COOH+H_{2}O\leftrightharpoons C_6{H}_{5}CO{O}^{-}+H_{3}O$

$K_a=\frac{\left[C_6{H}_{5}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}COOH\right]}$

$\frac{\left[C_6{H}_{5}COOH\right]}{\left[C_6{H}_{5}CO{O}^{-}\right]}=\frac{\left[H_3{O}^{+}\right]}{K_a}=\frac{6.46\times {10}^{-4}}{6.46\times {10}^{-5}}=10$

Let the solubility of $C_6{H}_{5}COOAg$ be $xmol/L$

Then,

$\left[A{g}^{+}\right]=x$

$\left[C_6{H}_{5}COOH\right]+\left[C_6{H}_{5}CO{O}^{-}\right]=x$

$10\left[C_6{H}_{5}CO{O}^{-}\right]+\left[C_6{H}_{5}CO{O}^{-}\right]=x$

$\left[C_6{H}_{5}CO{O}^{-}\right]=\frac{x}{11}$

$K_{sp}\left[A{G}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]$

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

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

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

Now, let the solubility of $C_6{H}_{5}COOAg$ be $x^{\prime }M$

Then, $\left[A{g}^{+}\right]=x^{\prime }M$ and $\left[C_6{H}_{5}CO{O}^{-}\right]=x^{\prime }M$

$K_{sp}=\left[A{g}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]$

$K_{sp}=(x^{\prime }{)}^2$

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

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

Hence, $C_6{H}_{5}COOAg$ is approximately $3.32$ times more soluble in low pH solution.