Question

What is the molar solubility of ArBr in 0.001 M NaBr? Ksp for Arbr is 7.7*10^-13

Please explain :)

Answer 1

In a solution containing 0.0010 M NaBr, The following calculation can be made.

AgBr(s) ===== Ag+(aq) + Br-(aq)

[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 supressed.
This means that the amount of Br^- ion produced by dissolving AgBr will be negligible compared to 0.0010 M.
Ksp = [Ag^+][Br^-] = [Ag^+](0.0010 + x) = (x)(0.0100 + x) = (x)(0.0100) = 7.7 x 10^-13
x = (7.7 x 10^-13)/(0.0100) = 7.7 x 10^-11 M = [Ag^+]