Complex Ion Formation

Question

Calculate the concentrations of Ag⁺ and Ag(CN)₂^- in a solution prepared by mixing 100.0mL of 5.0x10^-3M AgNO₃ with 100.0mL of 2.00M KCN.

Please help, even if it's just the setup or advice! I don't know how to start it.

Answer 1

The complex ion is the [Ag(CN)2]^- and it is formed as follows:

Ag^+ + 2CN^- <==> [Ag(CN)2]^-

To find the final concentrations of the various species, you will need to know the Kf (which is essentially the same as the Keq for other similar reactions). So, look it up and then use a variation of

Kf = [Ag(CN)2]^-/[Ag^+][CN^-]^2 (see below for the variation)

First find initial concentrations after mixing 100 ml + 100 ml for a total of 200 ml.

Starting concentrations are as follows:

[Ag^+]: (100 ml)(5.0x10^-3 M) = (200 ml)(x M) and x = 2.5x10^-3 M Ag^+

[CN^-]: (100 ml)(2.00 M) = (200 ml)(x M) ane x = 1.00 M CN^-

After the reaction you have:

[Ag^+] = 0

[CN^-] = 1.00 - (2x2.5x10^-3) = 1.00 - 0.005 = 0.995 M

[Ag(CN)2]^- = 2.5x10^-3 M

Up to now, we have looked at the reaction of Ag^+ and CN^- to form the complex. Now, we look at the dissociation of the complex and the equilibrium established and set up an ICE table.

[Ag(CN)2^-] <==> Ag^+ + 2CN^-

2.5x10^-3.............0..........0.995......I...

-x........................+x.........+2x.........C....

2.5x10^-3 - x......x............2x.........E....

Instead of using Kf, you now use 1/Kf = Kd (dissociation constant). So, whatever value you have for Kf, take the reciprocal (1/Kf) and find Kd (this is the aforementioned variation).

1/Kf = Kd = [Ag^+][CN^-]^2 / [Ag(CN)2]^-

Plug in the values from the ICE table and solve for x. That will be [Ag^+]. Then 2.5x10^-3 - x will be the concentration of [Ag(CN)2]^-

Hope this helps