Silver ion reacts with excess CN- to form a colorless complex ion,Ag(CN)2 , which has a formation constant Kf = 3.0*10^20 .

Question

Calculate the concentration of [Ag] in a solution prepared by mixing equal volumes of 1.8×10−3m NaNO3 and 0.20M NaCN.

When I did the problem which I thought was the correct way i ended up getting 1.56*10^-22M which is wrong, how do you go about doing the problem.

Answer 1

I would do this.
You have equal volumes of 1.8E-3M NaNO3(which I assume is a typo and you meant AgNO3 AND I assume the small m, meaning molality really is meant to be M or molarity) and 0.2M NaCN. So the concns going into the problem are 9E-4 for Ag^+ and 0.1M for NaCN (just 1/2 of the concns quoted in the problem).
............Ag^+ + 2CN^- ==> Ag(CN)2^-2
initial....9E-4.....0.1.......0
change............2*9E-4.....9E-4
equil........x......0.0964....9E-4

K = 3.0E20 = [Ag(CN)2^-]/[Ag^+][(CN)^-]^2 and solve for [Ag^+]
I get double your answer.

Answer 2

You will note I didn't subtract correctly; 0.1-(2*9E-4) = 0.0982 and not 0.964. However, that doesn't change my answer; I punched into the calculator correctly. I still get approximately 3E-22