An amount of of is added to one litre of water. Next, the crystals of are added until 75% of the is — Ionic Equilibrium Chemistry Question
Question
An amount of $0.10\text{ moles}$ of $\text{AgCl(s)}$ is added to one litre of water. Next, the crystals of $\text{NaBr}$ are added until 75% of the $\text{AgCl}$ is converted to $\text{AgBr(s)}$, the less soluble silver halide. What is $\text{Br}^-$ at this point? $K_{sp}$ of $\text{AgCl} = 2 \times 10^{-10}$ and $K_{sp}$ of $\text{AgBr} = 4 \times 10^{-13}$.
💡 Solution & Explanation
The substitution reaction is $\text{AgCl(s)} + \text{Br}^- \rightleftharpoons \text{AgBr(s)} + \text{Cl}^-$. The equilibrium constant is $K = \frac{[\text{Cl}^-]}{[\text{Br}^-]} = \frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgBr})} = \frac{2 \times 10^{-10}}{4 \times 10^{-13}} = 500$. We convert 75% of the initial $0.10\text{ moles}$ of AgCl, which produces exactly $0.075\text{ moles}$ of $\text{Cl}^-$ in the $1\text{ L}$ solution. Hence, $[\text{Cl}^-] = 0.075\text{ M}$. Since both solids are present, the equilibrium ratio dictates $[\text{Br}^-] = \frac{[\text{Cl}^-]}{500} = \frac{0.075}{500} = 1.5 \times 10^{-4}\text{ M}$. Therefore, correct answer is C.