DrBobb22
Calcule the concentrations of Cd^+2 , [Cd(CN)4^-2] and CN^- at equilibrium, when 0.42 moles of Cd(NO3)2 dissolves in 1 L of 2.50 M NaCN
Kf for Cd(CN)4^-2
Help!
Kf is 7.1 x 10^16
If you answered the question I asked I might be able to help.
I will post again ok?
right. See above for my response.
To solve this problem, we need to use the concept of equilibrium calculations and the given equilibrium constant, Kf for Cd(CN)4^-2.
First, let's write the balanced chemical equation for the formation of Cd(CN)4^-2 from Cd^2+ and CN^-:
Cd^2+ + 4 CN^- ⇌ Cd(CN)4^-2
According to the equation, the stoichiometry of Cd^2+ to Cd(CN)4^-2 is 1:1. This means that for every 1 Cd^2+ ion that reacts with 4 CN^- ions, 1 Cd(CN)4^-2 ion will be formed.
Next, let's define the variables:
[X] = concentration of X in moles per liter (M)
Cd^2+ = [Cd^2+] (initial concentration of Cd^2+) - x (change in Cd^2+ concentration)
[Cd(CN)4^-2] = x (concentration of Cd(CN)4^-2 formed)
CN^- = [CN^-] (initial concentration of CN^-) + 4x (change in CN^- concentration)
Given:
Initial concentration of Cd^2+ ([Cd^2+]) = 0.42 moles
Initial concentration of CN^- ([CN^-]) = 2.50 M
We can set up an ICE (Initial, Change, Equilibrium) table to calculate the concentrations at equilibrium:
Cd^2+ + 4 CN^- ⇌ Cd(CN)4^-2
----------------------------------
Initial | 0.42 | 2.50 | 0
Change | -x | -4x | x
Equilibrium| 0.42 - x | 2.50 - 4x | x
Since Kf is the equilibrium constant for the formation of Cd(CN)4^-2, we can write the expression for Kf as follows:
Kf = [Cd(CN)4^-2] / ([Cd^2+] * [CN^-]^4)
Substituting the equilibrium concentrations into the equation:
Kf = x / (0.42 - x) * (2.5 - 4x)^4
Now, we can solve the equation for x by using the given value of Kf. Rearranging the equation and substituting the known values:
Kf = x / (0.42 - x) * (2.5 - 4x)^4
Kf * (0.42 - x) = (2.5 - 4x)^4
Now, we can solve this equation either graphically or iteratively using a numerical method. Once we solve for x, we can substitute its value back into the ICE table to calculate the concentrations of Cd^2+, [Cd(CN)4^-2], and CN^- at equilibrium.
Please note that the specific numerical value of Kf for Cd(CN)4^-2 is not provided in your question, so you need to have that information to obtain the final concentrations.