Question

Solid potassium hydroxide is slowly added to 175 mL of a 0.0621 M cobalt(II) nitrate solution. The concentration of hydroxide ion required to just initiate precipitation is ...?

Answer 1

Answer:

[OH⁻] = 3,09x10⁻⁷ M

Explanation:

Selective precipitation is a technique of separating ions in an aqueous solution by using a reagent that precipitates one or more of the ions. In this case, you use OH⁻ from potassium hydroxide to precipitate cobalt thus:

Co(OH)₂ ⇄ Co²⁺ + 2 OH⁻ $K_{sp}$ = 5,92x10⁻¹⁵

The $K_{sp}$ definition is:

$K_{sp}$ = [Co²⁺] [OH⁻]²

Knowing  [Co²⁺] is 0,0621M:

5,92x10⁻¹⁵ = 0,0621M ×  [OH⁻]²

Thus, concentration of hydroxide ion required to just initiate precipitation is: [OH⁻] = 3,09x10⁻⁷ M

I hope it helps!