Complex ion formation

Question

Calculate the concentration of free cadmium ion, [Cd2+], in a solution that contains 0.20 M Cd(NO3)2 in 2.0 M NaCN. Cadmium ion forms the complex ion, Cd(CN)42- for which Kf is 6.0 x 10^18.

For the this problem, I know that the overall reaction will turn out as:

Cd^2+ + 4CN- ➡️ {Cd(CN)4}^-2

In the solutions I’ve seen for this problem, they flipped the equation and used the dissociation constant (kd) instead of kf. I think this is because its assumed that all of Cd has converted to complex ion. So how do you know that Cd is the limiting reactant? Usually when determining the limiting reactant moles are used and not molarity. Thanks!

Answer 1

Yes, it is assumed that all of the Cd²⁺ is in the form of the complex ion Cd(CN)₄^2-.

You know (or assume) that Cd²⁺ is limiting because of the large value of the K_f suggesting that the reaction Cd²⁺ + 4CN^- <==> Cd(CN)₄^2- lies very, very far to the right. We can now use this to set up a ICF table.

0.2M....2.0M.............0........Initial

-0.2M..-4x0.2M.......+0.2M...Change

0.........1.2M............0.2M.....Final (this is not equilibrium because we have to consider dissociation of complex)

Now, using these calculated values, we can consider the dissociation of the complex ion, as follows:

Cd(CN)₄^2- <====> Cd²⁺ + 4CN^-

0.2 M.....................0 M..........1.2 M.......Initial

-x...........................+x............+4x...........Change

0.2-x........................x............1.2+4x.......Equilibrium

Since we are looking at the dissociation, the constant will be the K_d (the inverse of the K_f) = 1 ÷ 6x10¹⁸

Kd = 1 / 6x10¹⁸ = [Cd²⁺][CN^-]⁴ / [Cd(CN)₄^2-

1/6x10¹⁸ = (x)(1.2+4x)⁴ / 0.2-x ... here we can assume that x is very, very small relative to 1.2 and 0.2 and we can ignore it so as to greatly simplify the calculation. Doing so gives us 1/6x10¹⁸ = (x)(1.2)⁴ / 0.2 = 2.07x/0.2

Solving for x:

x =0.2/(2.07)(6x10¹⁸) = 1.6x10^-20 M [Cd²⁺]