A 0.170-mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2 ions at equilibrium? Assume the formation constant* of Ni(NH3)62 is 5.5 × 108.

Answer 1

You do these problems by working it twice, once forward and once in reverse like this. Since the formation constant is such a large number you assume this reaction goes to completion.
.......Ni^2+ + 6NH3 ==> Ni(NH3)6^2+
I....0.00131....1.20.....0
C...-0.00131...-0.00787...0.00131
E........0......1.20.... 0.00131
The 0.00131 is mols NiCl2/1L = 0.00131M.

Then use the same equation but work right to left. The E line from above becomes the I line for this.
.......Ni^2+ + 6NH3 ==> Ni(NH3)6^2+
I......0........1.20....0.00131
C......x........6x.......-x
E......x.......1.20+6x..0.00131-x

Then substitute the E line into the Kf expression and solve for x = (Ni^2+).