Complex Ion Equilibria

Question

Determine the concentration of the species present at equilibrium after 0.25 mol of NaCl is dissolved in 1.50 L of a solution that contains 7.5 × 10–3 M Pb2+. (Kf = 2.5 × 1015 for [PbCl4]2–)

Answer 1

Pb²⁺(aq) + 4Cl^-(aq) ==> Pb(Cl₄)^2-

moles Pb2+ = 1.50 L x 7.5x10-3 mol/L = 0.01125 moles Pb2+

moles Cl- = 0.25 mol Cl-

Setting up an ice table....

Pb²⁺(aq) + 4Cl^-(aq) ==> Pb(Cl₄)^2-

0.01125..........0.25................0..........Initial

-x.................-4x....................+x.......Change

0.01125-x......0.25-4x...........x..........Equilibrium

Kf = 2.5x10¹⁵ = [Pb(Cl₄)^2-] / [Pb²⁺][Cl^-]⁴

Solve for x which will be the moles of Pb(Cl4)2-. Moles of Pb2+ will be 0.01125-x and moles of Cl- will be 0.25-4x. Divide all by 1.50 L to find the final concentration of each species.