A volume of 500 ml of solution, 250 ml of solution and 250 ml of solution are mixed. The final conce — Ionic Equilibrium Chemistry Question
Question
A volume of 500 ml of $0.01 \text{ M} - \text{AgNO}_3$ solution, 250 ml of $0.02 \text{ M} - \text{NaCl}$ solution and 250 ml of $0.02 \text{ M} - \text{NaBr}$ solution are mixed. The final concentration of bromide ion in the solution is ($K_{sp}$ of AgCl and AgBr are $10^{-10}$ and $5 \times 10^{-13}$, respectively.)
💡 Solution & Explanation
The mixture contains $5 \text{ mmol}$ of $\text{Ag}^+$, $5 \text{ mmol}$ of $\text{Cl}^-$, and $5 \text{ mmol}$ of $\text{Br}^-$ in 1 L. Because AgBr is significantly less soluble than AgCl, the $\text{Ag}^+$ will preferentially precipitate the $\text{Br}^-$. The 5 mmol of $\text{Ag}^+$ precipitates almost exactly 5 mmol of $\text{Br}^-$, leaving the $\text{Cl}^-$ largely unprecipitated ($[\text{Cl}^-] \approx 0.005 \text{ M}$). At equilibrium where both could theoretically precipitate, the ratio of ions in solution must obey $\frac{[\text{Cl}^-]}{[\text{Br}^-]} = \frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgBr})} = \frac{10^{-10}}{5 \times 10^{-13}} = 200$. Therefore, $[\text{Br}^-] = \frac{[\text{Cl}^-]}{200} = \frac{0.005}{200} = 2.5 \times 10^{-5} \text{ M}$. Therefore, correct answer is D.