Calculate the molar solubility of lead (II) chloride (PbCl2) in a 0.010M solution of lead (II) nitrate (Pb(NO3)2). The Ksp of PbCl2 is 1.2×10–5

Question

Calculate the molar solubility of lead (II) chloride (PbCl2) in a 0.010M solution of lead (II) nitrate (Pb(NO3)2). The Ksp of PbCl2 is 1.2×10–5

Answer 1

This is a common ion problem.

PbCl₂(s) ===> Pb²⁺(aq) + 2Cl^-(aq)

Ksp = [Pb²⁺][Cl^-]² = (0.01)(Cl^-)² = 1.2x10^-5

(Cl^-)² = 1.2x10^-7

[Cl^-] = 3.5x10^-4 M

Molar solubility of PbCl₂ in 0.01 M Pb(NO₃)₂ = 1.8x10^-4 M

Answer 2

PbCl₂(s) ==> Pb²⁺(aq) + 2Cl^-(aq)

This is a common ion effect problem. The common ion is Pb²⁺ from the Pb(NO₃)2 and Pb²⁺ from PbCl₂.

Ksp = [Pb²⁺][Cl^-]²

1.2x10^-5 = [0.010][x]²

x² = 1.2x10^-3

x = 3.46x10^-2 M