Question

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 1.040 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

Here is the way to do these.
The reaction is this.
....M^2+ + 4CN^- ==> [M(CN)4]^2-
I..0.170...1.04.......0
C.-0.170..-4*0.170 ...0.170
E...0.....0.36.......0.170
You can see that with a K of a bouat 10^16 that the reaction is essentially complete to the right. Now you start with the reaction as shown line E and calculate what happens to the left; i.e., M(CN)4]^2- going to CN%- and M^2+
I will retype it so you will see it better.
...[M(CN)4]^2- ==> M^2+ + 4CN^-
I..0.170...........0.....0.36
C....-x.............x.....4x
E..0.170x...........x...0.36+4x

Now plug the E line into the formation constant expression and solve for x. Generally you may call 0.170-x = 0.170 and 0.36+4x = 0.36 and x will give you the (M^2+).