Question

Calculate the molar solubility of PbBr₂ in a solution that contains 0.65M NaBr. Assume the NaBr dissociates 100%.

PbBr_2(aq) <->Pb²⁺_(aq)   + 2Br^-_(aq)                     Ksp = 4.6 x 10^-6                  .

Answer 1

Concentration of NaBr = 0.65M

NaBr-------------- Na+ Br-

[Br-]= 0.65M

PbBr2(aq) ----------------- Pb +2    + 2 Br-

                               s mole/L     ( 2s + 0.65) mol/L

Ksp = [Pb+2]]Br-]^2

4.6x10^-6 = s x(2s+0.65)^2

4.6x10^-6 = Sx(4S2 + 0.4225 + 2.6S)

4.6x10^-6 = 4S3 + 0.4225S + 2.6S2

4.6x10^-6 = 0.4225 S     ( S3 and S2 values are neglected)

S= 4.6x10^-6/0.4225

S= 10.887 x10^-6

molar solubility of PbBr2 = 1.09x10^-5M