How much could dissolve in of ? Assume that is the only complex formed. Given and . — Ionic Equilibrium Chemistry Question
Question
How much $AgBr$ could dissolve in $1.0 \text{ L}$ of $0.40 \text{ M } NH_3$? Assume that $[Ag(NH_3)_2]^+$ is the only complex formed. Given $K_f([Ag(NH_3)_2]^+) = 1 \times 10^8$ and $K_{sp}(AgBr) = 5 \times 10^{-13}$.
💡 Solution & Explanation
The dissolution reaction is $AgBr_{(s)} + 2NH_{3(aq)} \rightleftharpoons [Ag(NH_3)_2]^+_{(aq)} + Br^-_{(aq)}$. The equilibrium constant $K_{eq} = K_{sp} \times K_f = 5 \times 10^{-13} \times 10^8 = 5 \times 10^{-5}$. Let the solubility be $s$. At equilibrium, $[Complex] = s$, $[Br^-] = s$, and $[NH_3] = 0.40 - 2s$. $K_{eq} = \frac{s^2}{(0.40 - 2s)^2} = 5 \times 10^{-5}$. Taking the square root gives $\frac{s}{0.40 - 2s} = \sqrt{50 \times 10^{-6}} \approx 7.07 \times 10^{-3}$. Solving for $s$ gives $s \approx 2.8 \times 10^{-3} \text{ M}$.