Question

Solid ammonium cyanide is slowly added to 125 mL of a 0.0541 M nickel(II) acetate solution. The concentration of cyanide ion required to just initiate precipitation is ...?

Answer 1

Answer:

[CN⁻] = 2,35x10⁻¹¹ 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 CN⁻ from ammonium cyanide to precipitate nickel(II)thus:

Ni(CN)₂ ⇄ Ni²⁺ + 2 CN⁻ $K_{sp}$ = 3,0x10⁻²³

The $K_{sp}$ definition is:

$K_{sp}$ = [Ni²⁺] [CN⁻]²

Knowing  [Ni²⁺] is 0,0541M:

3,0x10⁻²³ = 0,0541M ×  [CN⁻]²

Thus, concentration of cyanide ion required to just initiate precipitation is: [CN⁻] = 2,35x10⁻¹¹ M

I hope it helps!