Question

calculate the solubility in g/l of AgBr in (a) pure water and (b) 0.0066 M NaBr. (in scientific notation)

Answer 1

Equilibrium: AgBr(s) <=> Ag^+(aq) + Br^-(aq)

In pure water all of the Ag^+(aq) and Br^-(aq) is present because of dissolved solid AgBr. The concentrations of the two species is identical in this situation, so [Ag^+] = [Br^-]. If we let x = [Ag^+], then we can calculate its concentration.

Ksp = [Ag^+][Br^-] = 7.7 x 10^-13
x^2 = 7.7 x 10^-13
x = sq.root(7.7 x 10^-13) = 8.77 x 10^-7 M in pure water.
molar mass AgBr = 187.77 g/mol
solubility = (8.77 x 10^-7mol/L)(187.77 g/mol) = 1.65 x 10^-4 g/L

In a solution containing 0.0060 M NaBr, The following calculation can be made.

[Ag^+] = that dissolved from solid AgBr.
[Br^-] = that from NaBr + that from dissolved from AgBr.
As the [Br^-] is increased by adding NaBr, the solubility of AgBr will be suppressed.
This means that the amount of Br^- ion produced by dissolving AgBr will be negligible compared to 0.0060 M.
Ksp = [Ag^+][Br^-] = [Ag^+](0.0060 + x) = (x)(0.0600 + x) = (x)(0.0600) = 7.7 x 10^-13
x = (7.7 x 10^-13)/(0.0600) = 1.28 x 10^-11 M = [Ag^+]
Using the example above you calculate the solubility in g/L