A 0.140-mole quantity of CoCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Co^2+ ions at equilibrium? Assume the formation constant of Co(NH3)6^2+ is 5.0× 10^31.

Answer 1

The way these complex ion problems are solved is almost like working it backwards but that is what we do. You see such a large K (5E31 is huge) that it makes it almost impossible to solve the complex equations. So we FORCE the equation to go to the right ASSUMING that all of it goes to completion. Then we make it go backwards (because that's what our calculations can do) and solve for the small amount that IT DID NOT GO. The idea here is that it's easier to solve for a small amount than a large amount so we make it work as we can calculate.
.........Co^2+ + 6NH3 ==> Co(NH3)6^2+
I.......0.140M...1.20M......0
change.-0.140...-0.84......+0.140
equil....0......0.36.......0.140
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That's the forcing part. We just acted as if all if it reacted completely because K is so large. Now we make it go backwards and calculate the x part.
I.........0........0.36......0.140
C.........x........6x...........-x
E.........x........6x........0.140-x

K = [Co(NH3)6]^2+/(Co^2+)(NH3)^6
Substitute from the ICE chart (the lower one) and solve for x = (Co^2+).