Question

A 0.140-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)6^2+ is 5.0e31.

Please help, I don't know how to do this problem.

Answer 1

Here is how you do this with a little chicanery. First, we assume the reaction goes to completion to the right. That's not a bad assumption considering the size of the formation constant.
.........Co^2+ + 4NH3 ==> Co(NH3)4^2+
Initial.0.140....1.2........0
change...-0.140.-4*1.2.....0.140
equil.....0.......0.64......0.140

Actually, this isn't at equilibrium since we drove it to completion. Therefore, the numbers above are a little TOO far for equilibrium; therefore, it must move back to the left. How much will it move? We set up another set of conditions but start with where we ended above.
..........Co^2+ + 4NH3 ==> Co(NH3)4^2+
I.........0.......0.64......0.140
C.........+x......+4x.......-x
E.........x.....0.64+4x.....-x

Substitute into K.
K = [Co(NH3)4^2+]/[Co^2+][MH3]^4
5E31 = [0.140-x][x][0.64+4x]^4
Now assume 0.140-x = 0.140 and assume 0.64+4x = 0.64 and solve for x = (Co^2+). It's a very small number.