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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

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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

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

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