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?
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Answer
Since pH = 3.19,
$[\text{H}_3\text{O}^+]=6.46\times10^{-4}\text{M}$
$\text{C}_6\text{H}_5\text{COOH}+\text{H}_2\text{O}\leftrightarrow\text{C}_6\text{H}_5\text{COO}^-+\text{H}_3\text{O}$
$\text{K}_\text{a}\frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$
$\frac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^-]}=\frac{[\text{H}_3\text{O]}}{\text{K}_\text{a}}=\frac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$
Let the solubility of $C_6H_5COO$Ag be xmol/L.
Then,
$[\text{Ag}^+]=\text{x}$
$[\text{C}_6\text{H}_5\text{COOH}]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$10[\text{C}_6\text{H}_5\text{COO}^-]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$[\text{C}_6\text{H}_5\text{COO}^-]=\frac{\text{x}}{11}$
$\text{K}_\text{sp}[\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-]$
$2.5\times10^{-13}=\text{x}\Big(\frac{\text{x}}{11}\Big)$
$\text{x}1.66\times10^{-6}\text{mol/L}$
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 $C_6H_5COO$ Ag be x'mol/L.
Then,
$[\text{Ag}^+]=\text{x}'\text{M and}[\text{CH}_3\text{COO}^-]=\text{x}'\text{M}.$
$\text{K}_\text{sp}=[\text{Ag}^+][\text{CH}_3\text{COO}^-]$
$\text{K}_\text{sp}=\text{(x}')^2$
$\text{x}'=\sqrt{\text{K}_\text{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\text{mol/L}$
$\therefore\ \frac{\text{x}}{\text{x}'}=\frac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, $C_6H_5COO$Ag is approximately 3.317 times more soluble in a low pH solution.
$[\text{H}_3\text{O}^+]=6.46\times10^{-4}\text{M}$
$\text{C}_6\text{H}_5\text{COOH}+\text{H}_2\text{O}\leftrightarrow\text{C}_6\text{H}_5\text{COO}^-+\text{H}_3\text{O}$
$\text{K}_\text{a}\frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$
$\frac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^-]}=\frac{[\text{H}_3\text{O]}}{\text{K}_\text{a}}=\frac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$
Let the solubility of $C_6H_5COO$Ag be xmol/L.
Then,
$[\text{Ag}^+]=\text{x}$
$[\text{C}_6\text{H}_5\text{COOH}]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$10[\text{C}_6\text{H}_5\text{COO}^-]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$[\text{C}_6\text{H}_5\text{COO}^-]=\frac{\text{x}}{11}$
$\text{K}_\text{sp}[\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-]$
$2.5\times10^{-13}=\text{x}\Big(\frac{\text{x}}{11}\Big)$
$\text{x}1.66\times10^{-6}\text{mol/L}$
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 $C_6H_5COO$ Ag be x'mol/L.
Then,
$[\text{Ag}^+]=\text{x}'\text{M and}[\text{CH}_3\text{COO}^-]=\text{x}'\text{M}.$
$\text{K}_\text{sp}=[\text{Ag}^+][\text{CH}_3\text{COO}^-]$
$\text{K}_\text{sp}=\text{(x}')^2$
$\text{x}'=\sqrt{\text{K}_\text{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\text{mol/L}$
$\therefore\ \frac{\text{x}}{\text{x}'}=\frac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, $C_6H_5COO$Ag is approximately 3.317 times more soluble in a low pH solution.
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