DrBobb22
Calcule the concentrations of Cd^+2 , [Cd(CN)4^-2] and CN^- at equilibrium, when 0.42 moles of Cd(NO3)2 dissolves in 1 L of 2.50 M NaCN
Kf for Cd(CN)4^-2 is 7.1x10^16
This is a two part question. First you set up to see how much of the complex is formed, then you run the problem backwards to see how much of the Cd and others components are formed/or remain.
Here is the first part and you assume, with such a huge K value that ALL of the Cd^2+ is used to form the complex. You should confirm these numbers.
...Cd^2+ + 4CN^- ==> [Cd(CN)4]^2-
I..0.42....2.50.......0
C.-0.42...-4*0.42.....0.42
E....0.....0.82.......0.42
Now mentally turn the equilibrium backwards from right to left but keep the Kf the same. Start on the left with the complex at 0.42, CN^- as 0.82 and Cd^2+ as zero; i.e., the I line is the same for the beginning of this part as the E line is for the first part. It looks like this and I'll fill in the others.
I....0.......0.82.......0.42
C...+x........+4x.......-x
E....x.......0.82+4x....0.42-x
Then substitute the E line of the last part into the Kf expression and solve for x, then evaluate the other values. Hope this helps. Post your work if you get confused.