Question

A solution is initially 0.10m in mg2+(aq) and 0.10m in fe2+(aq). solid naoh is slowly added. what is the concentration of fe2+ when mg(oh)2 first precipitates? ksp (mg(oh)2) = 6.0 x 10-10; ksp(fe(oh)2) = 7.9 x 10-16

Answer 1

Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻

Ksp = [Mg²⁺] [OH⁻]²

6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²

[OH⁻] = 7.746 x 10⁻⁵ M

when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M

Fe(OH)₂ <—> Fe²⁺ + 2OH⁻

Ksp = [Fe²⁺] [OH⁻]²

7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²

[Fe²⁺] = 1.32 x 10⁻⁷ M

Answer: 1.32 x 10⁻⁷ M