The solubility of a sparingly soluble salt derived from a weak acid, generalized strictly as , is me — Ionic Equilibrium Chemistry Question
Question
The solubility of a sparingly soluble salt derived from a weak acid, generalized strictly as $AB$, is meticulously measured at a fixed pH of $3.0$ and found to be $Y \times 10^{-3} \text{ mol L}^{-1}$. If the true solubility product $K_{sp}$ of the salt $AB$ is $2.0 \times 10^{-10}$ and the acid ionization constant $K_a$ of the weak acid $HB$ is $1.0 \times 10^{-8}$, calculate the exact numerical value of $Y$. (Round to two decimal places).
💡 Solution & Explanation
The salt $AB$ dissolves as $AB \rightleftharpoons A^+ + B^-$, with $[A^+] = s$. The weak base anion $B^-$ inevitably hydrolyzes: $B^- + H^+ \rightleftharpoons HB$. Thus, total solubility $s = [A^+] = [B^-] + [HB]$. The acid equilibrium dictates $K_a = \frac{[H^+][B^-]}{[HB]}$, meaning $[HB] = \frac{[H^+][B^-]}{K_a}$. Substituting this into the solubility expression gives: $s = [B^-] + \frac{[H^+][B^-]}{K_a} = [B^-]\left(1 + \frac{[H^+]}{K_a}\right)$. The pH is fixed at $3.0$, so $[H^+] = 10^{-3} \text{ M}$. Then $1 + \frac{[H^+]}{K_a} = 1 + \frac{10^{-3}}{10^{-8}} = 1 + 10^5 \approx 10^5$. Therefore, $[B^-] = s \times 10^{-5}$. The solubility product is $K_{sp} = [A^+][B^-] = s(s \times 10^{-5}) = s^2 \times 10^{-5}$. We are given $K_{sp} = 2.0 \times 10^{-10}$, so $s^2 \times 10^{-5} = 2.0 \times 10^{-10} \Rightarrow s^2 = 2.0 \times 10^{-5} = 20 \times 10^{-6}$. Taking the square root definitively gives $s = \sqrt{20} \times 10^{-3} \approx 4.47 \times 10^{-3} \text{ M}$. Thus, $Y = 4.47$.