CoCl2 is soluble; therefore, it has no Ksp.
The easiest way to do these is to note that with a Kf so extremely high you know the solution at equilibrium will be far to the right; therefore, we set up an ICE chart and make the reaction go 100% to the right. That is an ok assumption because of the huge Kf for the complex ion.
.......Co^2+ + 6NH3 ==> [Co(NH3)6]^2+
I.....0.170....1.20........0
C....-0.170...-1.02.......0.170
E........0.....0.18.......0.170
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Then you turn the thing around and set up and ICE chart in reverse. Completed it looks like this and you start with 0.170 of the complex and zero of Co^2+and 0.18 for NH3 (just the E line of the first equilibrium).
........Co^2+ + 6NH3 --> [Co(NH3)6]^2+
I........0.......0.18......0.170
C........x.......+6x........-x
E........x.....0.18+6x.....0.170-x
Now set up the Kf =
[{Co(NH3)6}^2+]/(Co^2+)(NH3)^6
Substitute the E line into this Kf expression and solve for x = (Co^2+).
Making the assumption that 0.170-x = 0.170 and (0.18+6x)^6 = (0.18)^6 makes it easier to work and doesn't affect the answer because x is very small.
A 0.170-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 (Kf) of Co(NH3)62 is 5.0× 1031 M–6.
I don't really know how to start. At first I thought about setting up an ICE chart and getting it to equal to the Keq =Kf*Ksp, but then I can't find the Ksp for CoCl2.
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