Question

Calculate the molar solubility of Pbl2

Answer 1

Explanation:

PbI2(s) ==> Pb2+(aq) + 2I-(aq)

Ksp = 1.4x10-8 = [Pb2+][I-]2

(a) In pure water:

Let x = [Pb2+]; then [I-] = 2x

Ksp = 1.4x10-8 = (x)(2x)2 = 4x3

x3 = 3.5x10-9

x = 1.52x10-3 M

(b) In 0.50 L containing 15.0 g FeI3

This is a common ion problem. The solubility will be LESS than that in pure water.

[FeI3] = 15.0 g x 1 mol/437 g /0.5 L = 0.0686 M assuming it is soluble (?)

[I-] = 0.0686 x 3 = 0.206 M

1.4x10-8 = (x)(0.206)2

0.0424x = 1.4x10-8

x = 3.3x10-7 M (compared to 1.5x10-3 M in pure water)