Question

The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.500 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.

Answer 1

The reaction will be

Cu2+ + 4NH3 --> Cu(NH3)42+ (Kf = 1.70 × 10¹³)

Now we know that

Kf = [Cu(NH3)4++]/[Cu2+][NH3]^4 = 1.70 × 10¹³

Let us calculate the moles of Cu(NO3)2 and Ammonia solution

however the volume if is not given I assume it to be 10mL for each
10 x 0.04 /1000 Copper nitrate = 0.0004 mole copper nitrate

so concentration = 0.0004 X 1000/ 20 mL = 0.02 Molar
similarly the concentraiton of ammonia solution will be 0.25 Molar

So here the limiting reagent will be copper nitrate so the concentraiton of copper ammonia complex will be 0.02-x
As Kf is very high we can assume completion of reacftion till the reagents are exhausted

So as per stoichiometry

1 mole of copper ion will react with 4 moles of ammonia

so, 0.02 moles will react with 0.08 moles of ammonia to give 0.02 moles of Cu(NH3)42

So at equilibrium

concentraiton of Cu+2 = almost zero

concentration of Cu(NH3)42 = 0.02 moles / L

Concentration of ammonia = 0.25- 0.08 = 0.17 moles / L