How much solid could dissolve in of aqueous solution? Assume that is the only complex formed. Given: — Ionic Equilibrium Chemistry Question
Question
How much solid $AgBr$ could dissolve in $1.0 \text{ L}$ of $0.40 \text{ M } NH_3$ aqueous solution? Assume that $[Ag(NH_3)_2]^+$ is the only complex formed. Given: Formation constant $K_f$ for $[Ag(NH_3)_2]^+ = 1.0 \times 10^8$ and $K_{sp}$ for $AgBr = 5.0 \times 10^{-13}$.
💡 Solution & Explanation
The overall dissolution reaction is $AgBr_{(s)} + 2NH_{3(aq)} \rightleftharpoons [Ag(NH_3)_2]^+_{(aq)} + Br^-_{(aq)}$. The equilibrium constant is $K_{eq} = K_{sp} \times K_f = 5.0 \times 10^{-13} \times 10^8 = 5.0 \times 10^{-5}$. Let solubility be $s$. Equilibrium concentrations: $[NH_3] = 0.40 - 2s$, $[Complex] = s$, $[Br^-] = s$. Thus, $\frac{s^2}{(0.40 - 2s)^2} = 5.0 \times 10^{-5}$. Taking square root: $\frac{s}{0.40 - 2s} = \sqrt{50 \times 10^{-6}} \approx 7.07 \times 10^{-3}$. Solving for $s$ yields $s \approx 2.8 \times 10^{-3} \text{ M}$.