A saturated solution of silver benzoate, , has pH of 8.6. for benzoic acid is . The value of for sil — Ionic Equilibrium Chemistry Question
Question
A saturated solution of silver benzoate, $\text{AgOCOC}_6\text{H}_5$, has pH of 8.6. $K_a$ for benzoic acid is $5.0 \times 10^{-5}$. The value of $K_{sp}$ for silver benzoate is ($\log 2 = 0.3$)
💡 Solution & Explanation
Since $\text{pH} = 8.6$, $\text{pOH} = 5.4 \implies [\text{OH}^-] = 10^{-5.4} = 10^{0.6} \times 10^{-6} \approx 4 \times 10^{-6}\text{ M}$. The benzoate ion ($\text{B}^-$) hydrolyzes: $\text{B}^- + \text{H}_2\text{O} \rightleftharpoons \text{HB} + \text{OH}^-$. The hydrolysis constant $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{5.0 \times 10^{-5}} = 2.0 \times 10^{-10}$. Using $K_h = \frac{[\text{OH}^-]^2}{[\text{B}^-]}$, we find $[\text{B}^-] = \frac{(4 \times 10^{-6})^2}{2.0 \times 10^{-10}} = \frac{16 \times 10^{-12}}{2.0 \times 10^{-10}} = 0.08\text{ M}$. Let the solubility be $S$. Then $[\text{Ag}^+] = S$. Total dissolved benzoate $= [\text{B}^-] + [\text{HB}] = 0.08 + 4 \times 10^{-6} \approx 0.08\text{ M}$, so $S \approx 0.08\text{ M}$. Thus, $K_{sp} = [\text{Ag}^+][\text{B}^-] = 0.08 \times 0.08 = 6.4 \times 10^{-3}$. Therefore, correct answer is B.