Question

A 0.130-mole quantity of CoCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Co2 ions at equilibrium? Assume the formation constant of Co(NH3)62 is 5.0× 10^31.

Answer 1

There are a couple of ways to look at this. I'll show you how to do it with a good bit of reasoning. There is another way that I think makes it easier to explain. I can show you that way if you have trouble understanding this.
.........Co^2+ + 4NH3 ==> Co(NH3)4^2+
I......0.130......1.20.......0
C.....-0.130....-4*0.130....0.130
E.......0........0.68.......0.130

The ICE chart above is based on this.
With such a HUGE Kf for the complex ion, essentially ALL of the Co^2+ will be transformed to the complex. Therefore, the equilibrium value for the complex will be essentially 0.130, the Co^2+ at equilibrium will be ALMOST (but not quite) zero (although my ICE chart shows zero), and the NH3 left will be just 1.2 at the start - the amount used. The Kf is
5E31 = [Co(NH3)2^2+/(Co^2+)(NH3)^4
Substitute 0.130 for [Co(NH3)4^2+], substitute 0.68 for [NH3} and substitute x for [Co^2+] and solve for x.

Answer 2

it's actually Co(NH3)6^2. but the whole calculation was correct!!! thank you SO much!

Answer 3

i got it wrong... i got 1.22x10^-32. help!

Answer 4

what is it supposed to be?

Answer 5

what is the answer....

Answer 6

4.74x10^-31 is the answer. NH3 from ice table is 1.2-(6*.130)= [.42] not [.68]

Answer 7

Okay, I know I'm responding quite a few years late, but can someone explain exactly why there is still some cobalt remaining at equilibrium? I mean, in my ICE tables, whenever I have an equilibrium reaction which involves a limiting reagent step (in this case it was the cobalt, in my homework it is actually nickel 2+, but the concept is the same) I always show the reaction going the other way (towards the reactant side), thus, in my reasoning, not only should the cobalt have a very small +x value, but shouldn't the NH3 also have a very small addition of (6 times) the same amount? I just need a bit of an explanation as to why this is not so.