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 .

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.

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.

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

To determine the concentration of [Ag] in the solution, we need to use equilibrium calculations. The reaction between silver ions (Ag+) and cyanide ions (CN-) to form the complex ion Ag(CN)2 can be represented as follows:

Ag+ + 2CN- ⇌ Ag(CN)2

Given that the formation constant (Kf) is 3.0 x 10^20, it corresponds to the equilibrium constant (K_eq) for the above reaction.

The equilibrium constant expression is given by:

K_eq = [Ag(CN)2] / ([Ag+] * [CN-]^2)

Since we are mixing equal volumes of NaNO3 and NaCN, we can assume that the total volume remains constant. Consequently, the final concentration of NaNO3 and NaCN will be halved.

Let's assign "x" as the initial concentration of Ag+ ions (which is equal to the concentration of Ag+ in the final solution). Since the initial concentration of NaCN is 0.20 M and it becomes 0.10 M in the final solution, the concentration of CN- ions can be written as (0.10 - x).

Using the equilibrium expression, we can write:

K_eq = [Ag(CN)2] / (x * [(0.10 - x)]^2)

As mentioned earlier, the formation constant (Kf) is equal to K_eq, so we have:

3.0 x 10^20 = [Ag(CN)2] / (x * [(0.10 - x)]^2)

Now, to solve for [Ag(CN)2], we need to find the value of x, the concentration of Ag+.

Now, let's solve the equation to find the concentration of [Ag]:

3.0 x 10^20 = [Ag(CN)2] / (x * [(0.10 - x)]^2)

Now, let's plug in the known values and solve for x:

3.0 x 10^20 = (x) / (x * [(0.10 - x)]^2)

Rearranging the equation:

3.0 x 10^20 = 1 / [(0.10 - x)]^2

Taking the reciprocal of both sides and simplifying:

[(0.10 - x)]^2 = 1 / (3.0 x 10^20)

Taking the square root of both sides:

0.10 - x = √(1 / (3.0 x 10^20))

Now, to solve for x:

x = 0.10 - √(1 / (3.0 x 10^20))

Calculate this value and you will have the concentration of [Ag] in the solution.