Question

The tetramminecopper(II) ion, Cu(NH3)4 2+, has a formation constant Kf = 1.1 x 10^13. What is the minimum concentration of free ammonia in solution required to ensure that at least 99.9% of the dissolved copper(II) ion is found in the form of its ammonia complex? (A) 9 x 10^–14 M (B) 9 x 10^–11 M (C) 8 x 10^–4 M (D) 3 x 10^–3 M.

The answer is D. Please show how you get to the solution

Answer 1

The correct answer is option D, that is, 3 × 10⁻³ M.

The following is the equation that represent the equilibrium for the formation of tetramminecopper (II) ion:

Kf = [Cu(NH₃)₄)] / [Cu] [NH₃]₄ ----------- Equation (i)

It is known that 99.9 % of copper is in the form of ammonia complex and 0.01 % of it is found in its free state, the reaction will be:

[Cu(NH₃)₄] / [Cu] = 0.999 / 0.001

Substituting the equation (i) we get:

1.1 × 10¹³ = 0.999 / (0.001) (NH₃)₄

After solving for ammonia we get:

[NH₃] = (999 / 1.1 × 10¹³)1/4

[NH₃] = 3.08 × 10⁻³ M

Answer 2

The balanced formation reaction is
Cu2+ + 4 NH3 <---> Cu(NH3)4 2+
The formation constant is
Kf = [Cu(NH3)4 2+] / [Cu2+] [NH3]^4
Assume the reactants are at 1 M for Cu2+ and 4 M for NH3

Cu2+ + 4 NH3 <---> Cu(NH3)4 2+
Initial 1 M 4 M x
Change -0.999 -0.999*4 0.999
Equilibrium 0.001 0.004 x - 0.999

Kf = (x - 0.999) / (0.001)(0.004)^4 = 1.1x10^13
Solve for x and you'd get
D