Question

Calculate the equilibrium molar concentration of free Zn²⁺(aq) in a solution that contains 0.021 mol of Zn²⁺ per liter to which 0.498 mole of NH₃ is added. K_f for Zn(NH₃)₄²⁺ is 2.9x10⁹. Show the equilibrium equation.

Answer 1

Zn2+ (aq)   + 4 NH3 (aq)   <------------> Zn(NH3)42+

1                        4                                           1

0.021                0.498

here limiting reagent is Zn2+

remaining NH3 = 0.498 - (4 x 0.021) = 0.414 M

[ Zn(NH3)42+ ] = 0.021 M

Kf= [ Zn(NH3)42+ ] / [Zn2+] [NH3]^4

2.9 x 10^9 = 0.021 / [Zn2+] (0.414)^4

[Zn2+] = 2.46 x 10^-10 M