What are the concentrations of Cu^2+, NH3, and Cu(NO3)4^2+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a .400 M solution of aqueous ammonia? Assume that the reaction goes to completion and forms Cu(NH3)4^2+.

Answer 1

I think you made a typo with that Cu(NO3)4^2. I assume you meant [Cu(NH3)4]^2. 18.8 g Cu(NO3)2 = ?M. That's mols = grams/molar mass. Let's call that about 0.1 but you should go through and confirm that.

........Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2
I....... 0.1....0.400.......0
C.......-0.1...-0.400......0.1
E.........0.......0........0.1

We make the ICE table above. You will need to look up Kf for [Cu(NH3)4]^2+ and you will find it is a huge number which means the reaction will go essentially to completion. Now you turn the reaction around and work the problem as IF it were a weak acid (it isn't of course) problem like this.
.........[Cu(NH3)4]^2 ==> Cu^2+ + 4NH3
I.........0.1..............0.......0
C..........-x..............x.......x
E.........0.1-x............x.......x

Now substitute the E line into Kf expression for the complex and solve for x, then evaluate 0.1-x.