The Ksp for lead iodide PbI2 is 1.4×10^-8 Calculate the solubility of lead iodide in each of the following

Question

A. Water

B. 0.10M Pb (NO3) 2

C. 0.010M NaI

Answer 1

PbI2 <==> Pb²⁺ + 2I^-

Ksp = 1.4x10^-8 = [Pb²⁺][I^-]² = (x)(2x)²

4x³ = 1.4x10^-8

x = 1.52x10^-3 M = solubility in water

For B, repeat above calculations substituting 0.1 for [Pb²⁺] and solve for [I^-].

For C, repeat above calculations substituting 0.01 for [I^-]

These are known as "common ion effects" problems.