Calculate the molar solubility of PbI2 (Ksp = 1.4x10^-8)in (a) Pure water (b) On 0.50 L of solution that contains 15.0 g of FeI3 .

Question

Please answer in details with all steps. 

Answer 1

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)