Question

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.870 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

This is a little hard to show in writing but here goes.
......M^2+ + 4CN^- ==> [M(CN)4]^2-
I..0.150....0.870.......0
C.-0.150...-0.600....+0.150
E.....0.....0.270.....0.150

The first equilibrium shows the formation of the complex on the right. With such a large Kf of over 10^16, it is obvious that essentially ALL of the M^2+ (from the M(NO3)2) reacts to form the complex.

Now that the complex is formed, we reverse the equilibrium (so it appears we are starting with [M(CN)4]^2- to form M^+ and CN^-. In this respect it's just like a Ka or Kb problem in which you're looking for the products. The E line for the first equilibrium is the I line for the reverse equilibrium. So it looks this way.
M^+ + 4CN^- ==> [M(CN)4]^2-
0....0.270......0.150.......I
x......4x........-x.........C
x..0.270+4x....0.150-x......E

Plug the E line into the Kf expression and solve for x = (M^2+).
Post your work if you get stuck and I can help you through it.

Answer 2

Where can I find the Kf expression?

Answer 3

I don't understand, DrBob222.