Answer:
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 <em>the concentration of each ion is halved </em>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
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
