Calculate the exact molar solubility of solid in a aqueous solution of . Assume the complexation mec — Ionic Equilibrium Chemistry Question
Question
Calculate the exact molar solubility of solid $AgBr$ in a $0.40 \text{ M }$ aqueous solution of $NH_3$. Assume the complexation mechanism forms exclusively $[Ag(NH_3)_2]^+$. (Given the formation constant $K_f = 1.0 \times 10^8$ and $K_{sp}$ for $AgBr = 5.0 \times 10^{-13}$).
💡 Solution & Explanation
The dissolution mechanism is coupled with complexation: $AgBr_{(s)} + 2NH_{3(aq)} \rightleftharpoons [Ag(NH_3)_2]^+_{(aq)} + Br^-_{(aq)}$. The overall equilibrium constant is $K_{eq} = K_{sp} \times K_f = (5.0 \times 10^{-13}) \times 10^8 = 5.0 \times 10^{-5}$. Let the solubility be $s$. At equilibrium, $[Complex] = s$, $[Br^-] = s$, and $[NH_3] = 0.40 - 2s$. Thus, $\frac{s^2}{(0.40 - 2s)^2} = 5.0 \times 10^{-5}$. Taking the square root gives $\frac{s}{0.40 - 2s} \approx 7.07 \times 10^{-3}$. Solving for $s$ yields $s \approx 2.8 \times 10^{-3} \text{ M}$.