Calculate the concentrations of Cd2 +, [Cd (CN) 42] and CN at equilibrium when they dissolve 0, 42mol of

Cd(NO3)2, molarity= 2.50 M
Kf =7.1x10^16

Your post isn't clear to me. I'm confused by the "when THEY dissolve"

What is meant by they?

To calculate the concentrations of Cd2+, [Cd(CN)4]^2-, and CN- at equilibrium, we need to use the equilibrium constant expression and the initial concentrations of the reactants.

First, let's define the given values:
Initial concentration of Cd(NO3)2 = 0.42 mol
Molarity of Cd(NO3)2 = 2.50 M
Kf (formation constant) = 7.1 x 10^16

The balanced chemical equation for the reaction of Cd(NO3)2 with CN- to form [Cd(CN)4]^2- is as follows:
Cd(NO3)2 + 4CN- ⇌ [Cd(CN)4]^2- + 2NO3-

Let's assume that at equilibrium, the concentration of [Cd(CN)4]^2- is x M. Therefore, the concentration of Cd2+ is also x M.

Now, let's set up the equilibrium constant expression:
Kf = [Cd(CN)4]^2- / ([Cd2+][CN-]^4)

Since the stoichiometric coefficient for Cd2+ is 1, and for CN- is 4, and assuming the initial concentration of CN- is 0 M, the concentration of CN- at equilibrium will be 4x M.

Substituting these values into the equilibrium constant expression, we get:
7.1 x 10^16 = x / (x * (4x)^4)

Simplifying the equation:
7.1 x 10^16 = x / (256x^5)

Rearranging the equation:
7.1 x 10^16 * (256x^5) = x

Simplifying further:
1.8176 x 10^21 x^5 = x

This equation is a quintic equation, which does not have a simple analytical solution. However, it can be solved numerically using iterative methods or with the help of advanced mathematical software.

Therefore, the concentrations of Cd2+, [Cd(CN)4]^2-, and CN- at equilibrium when dissolving 0.42 mol of Cd(NO3)2 with a molarity of 2.50 M and a formation constant of 7.1 x 10^16 can be determined by solving the quintic equation numerically.