Posted by kitor on .
The formation constant of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.170mole quantity of M(NO3)2 is added to a liter of 1.51 M NaCN solution. What is the concentration of M2 ions at equilibrium?

chemistry 
DrBob222,
The best way to work these is in a two step process. The first step is to assume the complex forms completely (and with a huge formation constant that is a good assumption) with the aid of an ICE chart. The second step is to start with those materials and see how much they will dissociate using a second ICE chart. As follows:
.......M^+2 + 6CN^ ==> M(CN)6^4
I....0.170...1.51........0
C...0.170..6*0.170.....0.170
E....0.......0.49........0.170
=================================
......0......0.49........0.170....I
......x.....+6x...........x......C
......x....0.49+6x.......0.170x..E
Kf = 2.50E17 = [M(CN)6^4]/(M^2+)(CN^)^6
Substitute the last E line into Kf expression and solve for x.