A solution is prepared by mixing 50.0 mL of 0.50 M Cu (NO3) 2 with 50.0 mL of 0.50 M Co (NO3) 2. Sodium hydroxide is added to the mixture. Which hydroxide precipitates first and what concentration of hydroxide ions present in solution will accomplish the separation?

Cu (OH) ₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu (OH) ₂ (s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co (OH) ₂ (s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that the concentration of each ion is halved because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

Because Copper requires less concentration of OH⁻ than Cobalt, Cu (OH) ₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M