Answer to Question #269906 in Physical Chemistry for kalokalo21

Question

) The solubility product of PbBr2 is 8.9 x 10-6 . Determine the molar solubility, i) in pure water, and, ii) in 0.20 M Pb(NO3)2 solution.

Answer 1

PbBr2(s) <=> Pb2+(aq) + 2 Br-(aq)

Ksp = [Pb2+][Br-]2

(a) [Br-] = 2[Pb2+]

Ksp = [Pb2+] x (2[Pb2+])2

= 4[Pb2+]3 = 8.9 x 10-6

Molar solubility = [Pb2+] = 1.3 x 10-2 M

(b) [Br] = [KBr] = 0.20 M

Ksp = [Pb2+] x (0.20)2 = 8.9 x 10-6

Molar solubility = [Pb2+] = 2.2 x 10-4 M

(c) [Pb2+] = [Pb(NO3)2] = 0.20 M

Ksp = 0.20 x [Br-]2 = 8.9 x 10-6

[Br-] = 6.67 x 10-3 M

Molar solubility = 0.5 x [Br-] = 3.3 x 10-3 M