The solubility of a salt of weak acid () at is . Given that the value of solubility product of () is — Ionic Equilibrium Chemistry Question
Question
The solubility of a salt of weak acid ($AB$) at $pH = 3$ is $Y \times 10^{-3} \text{ mol L}^{-1}$. Given that the value of solubility product of $AB$ ($K_{sp}$) is $2 \times 10^{-10}$ and the value of ionization constant of $HB$ ($K_a$) is $1 \times 10^{-8}$. The value of $Y$ is (rounded to two decimal places):
💡 Solution & Explanation
The dissolution is $AB \rightleftharpoons A^+ + B^-$, with $[A^+] = S$. The weak base anion hydrolyzes: $B^- + H^+ \rightleftharpoons HB$. Total solubility $S = [B^-] + [HB]$. From $K_a$, $[HB] = [H^+][B^-]/K_a$. Substituting gives $S = [B^-](1 + [H^+]/K_a)$. At $pH = 3$, $[H^+] = 10^{-3} \text{ M}$. $1 + [H^+]/K_a = 1 + 10^{-3}/10^{-8} \approx 10^5$. Thus, $[B^-] = S \times 10^{-5}$. $K_{sp} = [A^+][B^-] = S \times (S \times 10^{-5}) = S^2 \times 10^{-5}$. We are given $K_{sp} = 2 \times 10^{-10}$, so $S^2 \times 10^{-5} = 2 \times 10^{-10} \Rightarrow S^2 = 2 \times 10^{-5} = 20 \times 10^{-6}$. $S = \sqrt{20} \times 10^{-3} \approx 4.47 \times 10^{-3} \text{ M}$. Therefore, $Y = 4.47$.