Solid AgNO3 is slowly added to a solution that contains 0.26 M of Cl− and 0.12 M of Br− (assume the volume does not change). What is [Br−] when AgCl(s) starts to precipitate? Ksp of AgCl is 1.8 × 10^−10. Ksp for AgBr is 5.0 × 10^−13.
To determine the concentration of Br− when AgCl(s) starts to precipitate, we need to compare the solubility product constants (Ksp) of AgCl and AgBr.
The balanced chemical equation for the precipitation reaction when AgCl(s) starts to form is:
AgCl(s) ⇌ Ag+(aq) + Cl−(aq)
The equilibrium expression for this reaction is:
Ksp = [Ag+][Cl−]
Given that the Ksp of AgCl is 1.8 × 10^−10, we can write:
1.8 × 10^−10 = [Ag+][Cl−]
Since the Ag+ concentration will be equal to the Cl− concentration when AgCl starts to precipitate, we can substitute [Cl−] with 0.26 M:
1.8 × 10^−10 = [Ag+] × 0.26
Solving for [Ag+], we find:
[Ag+] = 1.8 × 10^−10 / 0.26
[Ag+] ≈ 6.923 × 10^−10 M
Now, to determine the concentration of Br− when AgCl(s) starts to precipitate, we need to compare the solubility product constants of AgBr and AgCl.
The balanced chemical equation for the precipitation reaction when AgBr(s) starts to form is:
AgBr(s) ⇌ Ag+(aq) + Br−(aq)
The equilibrium expression for this reaction is:
Ksp = [Ag+][Br−]
Given that the Ksp of AgBr is 5.0 × 10^−13, we can write:
5.0 × 10^−13 = [Ag+][Br−]
Since we know [Ag+] is approximately 6.923 × 10^−10 M, we can substitute this value into the equation:
5.0 × 10^−13 = (6.923 × 10^−10 M) [Br−]
Solving for [Br−], we find:
[Br−] ≈ (5.0 × 10^−13) / (6.923 × 10^−10 M)
[Br−] ≈ 7.226 × 10^−4 M