Chemistry question

Question

Determine the equilibrium concentrations of both Cu(NH3)4 2+ and Cu2+ in a solution made initially to be 0.010 M Cu(NO3)2 and 0.100 M NH3.

The Kf for Cu(NH3)4 2+ is 1.0 x 10^-12

Answer 1

Cu²⁺(aq) + 4NH₃(aq) ==> Cu(NH₃)₄²⁺(aq) .. Kf = 1.0x10^-12 ??? I'll assume it to be 1.0x10¹²

I've assumed the Kf to be 10¹² instead of 10^-12 because pretty much all Kf values are large. That being said...

Because it takes 4 mols NH₃ to react with each 1 mol Cu²⁺, it will take 4x0.01 = 0.04 M NH3 to be used up. This is shown in the table below:

Cu²⁺(aq) + 4NH₃(aq) ==> Cu(NH₃)₄²⁺(aq)

0.01...........0.100..................0............Initial

0..............0.06.....................0.01........After rxn

Since the Kf is so large, the equilibrium will lie far to the right and so after complete formation of Cu(NH₃)₄²⁺,

it will dissociate into Ca2+ and NH3 until equilibrium is reached. This is shown in the following table...

Cu²⁺(aq) + 4NH₃(aq) ==> Cu(NH₃)₄²⁺(aq)

0.................0.06..................0.01........Initial

+x..............+4x......................-x..........Change

x............0.06+x..................0.01-x......Equilibrium

Kf = [Cu(NH₃)₄²⁺] / [Cu²⁺][NH₃]⁴ = (0.01-x) / (x)(0.06+x)⁴

Simplified, we have...

1x10¹² = 0.01 / (x)(0.06)⁴

1x10¹² = 0.01 / 0.000013x

x = ~7.7x10^-10 M

Final concentrations would be as follows:

[Cu²⁺] = 7.7x10^-10 M

[Cu(NH₃)₄²⁺] = 0.01 M