The formation constant of [M(CN)4]^-2 is 7.70 × 10^16, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.840 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

I assume you mean M^2+ ions. Also I assume the 0.150 mol M(NO3)2 solid does not change the volume of the solution.
You do this in two steps.
.....M^2+ + 4CN^- ==> M(CN)4^2-
I..0.150...0.840.........0
C..-0.150..-4*0.150......+0.150
E....0......0.240........0.150

What we have done here is note that the Kf is so huge that essentially all of the metal ion will be used in forming the complex.

Now we work the problem backwards, starting with 0.150M M(CN)4^2- and see how much will dissociate. Althouhg I'm writing the equation backwards it DOES NOT change Kf.
.....M(CN)4^2- ==> M^2+ + 4CN
I....0.150.........0......0.240
C.....-x...........+x.....4x
E...0.150-x.........x.....0.240+4x

Kf = [M(CN)4]^2-/(M^2+)(CN^-)^4
Substitute the E line and solve for x. You can solve this much faster is you assume x is small and can be neglected in relation to 0.150 and 0.240; i.e.,
0.150-x = 0.150 and 0.240+4x = 0.240