# Question #c0638

May 29, 2014

The procedure to separate a mixture of ions such as Ag⁺, Pb²⁺, and Cu²⁺ from each other is Qualitative Analysis.

THEORY

You use selective precipitation. The ions precipitate in groups, based on differences in their solubility product constants.

1. The Silver Group

This group contains Ag⁺, Pb²⁺, and other ions.

The ions precipitate as the chlorides from hydrochloric acid.

Ag⁺(aq) + Cl⁻(aq) ⇌ AgCl(s); ${K}_{s p}$ = 1.77 × 10⁻¹⁰
Pb²⁺(aq) + 2Cl⁻(aq) ⇌ PbCl₂(s); ${K}_{s p}$ = 1.70 × 10⁻⁵

2. The Copper Group

This group contains Cu²⁺ and other ions. These ions precipitate as the sulfides from hydrochloric acid containing H₂S.

Cu²⁺(aq) + H₂S(g) ⇌ CuS(s) + 2H⁺(aq); ${K}_{s p}$ = 8 × 10⁻³⁷

EXPERIMENT

1. Place 3 mL of your mixture in a small test tube. Add 6 mol/L HCl dropwise until no more precipitate forms. Filter the mixture.

The white precipitate is a mixture of AgCl and PbCl₂. The filtrate is a blue solution of CuCl₂.

2. Pour a small amount of hot water** through the precipitate on the filter paper. Filter the hot mixture.

The solid on the filter paper is AgCl. The filtrate is a solution of PbCl₂.

3. Cool the filtrate from Step 2.

The PbCl₂ crystallizes from the solution.

4. Add dropwise a 1 mol/L solution of Na₂S to the filtrate from Step 1. You have only Cu²⁺ in your solution, so you don't have to use the extremely poisonous H₂S.

The black precipitate is CuS.

Hope this helps.