For the reaction, `["Ag(CN)"_2]^(-)-> "Ag"^+ + "2CN"^-` the equilibrium constant at 298 K is `4.0 × 10^-10`. The silver ion concentration in a solution which was originally 0.12 M in KCN and 0.04 M in `"AgNO"_3` is?

I got `1.0 xx 10^(-8)"M"`.

The way I see this is that the `"KCN"` and `"AgNO"_3` react to form an initial concentration of `"Ag"("CN")_2^(-)`. The formation of `"Ag"("CN")_2^(-)` is extremely favorable; we know that

`K_f = 1/(4.0 xx 10^(-10)) = 2.5 xx 10^9`.

So, this is first a limiting reactant problem, and then it becomes a common-ion effect problem with leftover excess reactant.

We assume that these solutions are those concentrations after they were mixed, so that the solution volume can be arbitrarily `"1 L"`.

`2"KCN"(aq) + "AgNO"_3(aq) -> "K"["Ag"("CN")_2] (aq) + "KNO"_3(aq)`

I assume you can realize that `"AgNO"_3` is the limiting reactant. We need twice as much `"KCN"`, so we use `"0.08 M"` of the `"KCN"` to convert all `"0.04 M"` of the `"Ag"^(+)` to `"Ag"("CN")_2^(-)` at first, and are left with `"0.04 M KCN"`.

So, our ICE table begins with `"0.04 M Ag"("CN")_2^(-)` and `"0.04 M CN"^(-)`.

`"Ag"("CN")_2^(-)(aq) rightleftharpoons "Ag"^(+)(aq) + 2"CN"^(-)(aq)`

`"I"" ""0.04 M"" "" "" "" ""0.00 M"" "" ""0.04 M"`
`"C"" "-x" "" "" "" "" "+x" "" "" "+2x`
`"E"" "(0.04 - x)"M"" "" "x" M"" "" "(0.04 + 2x)" M"`

This gives an equilibrium expression of:

`K_D -= K_f^(-1)`

`= 4.0 xx 10^(-10) = (["Ag"^(+)]["CN"^(-)]^2)/(["Ag"("CN")_2^(-)])`

`= (x(0.04 + 2x)^2)/(0.04 - x)`

When I use the small x approximation here, I get:

`4.0 xx 10^(-10) = (x(0.04)^2)/(0.04)`

`= 0.04x`

`=> color(blue)(x = ["Ag"^(+)]) = K_D/0.04 = color(blue)(1.0 xx 10^(-8) "M")`