Question

What is \(\left[\mathrm{Br}^{-}\right]\) just as \(\mathrm{AgCl}\) begins to precipitate when \(1.5 \mathrm{M} \mathrm{AgNO}_{3}\) is slowly added to a solution containing $$ 0.050 \mathrm{M} \mathrm{Cl}^{-} \text {and } 0.050 \mathrm{M} \mathrm{Br}^{-} ? $$

Short answer

Answer: The concentration of Br- ions just as AgCl begins to precipitate is 1.41 x 10^{-4} M.

Step-by-step solution

Write the solubility product expressions for AgCl and AgBr

Using Ksp values, we can write the solubility product expressions for AgCl and AgBr as follows: $$ K_{sp}(\mathrm{AgCl}) = [\mathrm{Ag}^+][\mathrm{Cl}^-] $$ and $$ K_{sp}(\mathrm{AgBr}) = [\mathrm{Ag}^+][\mathrm{Br}^-] $$

Find the Ksp values for AgCl and AgBr

Look up the Ksp values for AgCl and AgBr in a table or other reference source. You should find that: $$ K_{sp}(\mathrm{AgCl}) = 1.77 \times 10^{-10} $$ and $$ K_{sp}(\mathrm{AgBr}) = 5.0 \times 10^{-13} $$

Determine the initial concentration of Ag+ ions

Initially, the Ag+ ion concentration is 0, as no AgNO3 has been added yet. As we slowly add 1.5 M AgNO3, the concentration of Ag+ ions will increase. When AgCl starts to precipitate, the ionic product [Ag+][Cl-] will equal Ksp(AgCl). At this point, the concentration of Ag+ ions will be the same in both the AgCl and AgBr solubility product expressions.

Calculate the concentration of Ag+ ions when AgCl starts to precipitate

Using the Ksp value for AgCl and the initial concentration of Cl- ions, we can calculate the concentration of Ag+ ions when AgCl begins to precipitate. $$ [\mathrm{Ag}^+] = \frac{K_{sp}(\mathrm{AgCl})}{[\mathrm{Cl}^-]} $$ Plugging in the values, $$ [\mathrm{Ag}^+] = \frac{1.77 \times 10^{-10}}{0.050} $$ $$ [\mathrm{Ag}^+] = 3.54 \times 10^{-9} \; \mathrm{M} $$

Calculate the concentration of Br- ions when AgCl starts to precipitate

Now that we have the concentration of Ag+ ions when AgCl starts to precipitate, we can use the Ksp value for AgBr and the calculated concentration of Ag+ ions to find the concentration of Br- ions. $$ [\mathrm{Br}^-] = \frac{K_{sp}(\mathrm{AgBr})}{[\mathrm{Ag}^+]} $$ Plugging in the values, $$ [\mathrm{Br}^-] = \frac{5.0 \times 10^{-13}}{3.54 \times 10^{-9}} $$ $$ [\mathrm{Br}^-] = 1.41 \times 10^{-4} \; \mathrm{M} $$

Conclusion

The concentration of Br- ions just as AgCl begins to precipitate is 1.41 x 10^{-4} M.