Calculate the concentrations of Cd ⁺ ², [Cd (CN)₄-²] and CN at equilibrium when they dissolve mole of 0.42 Cd (in the ₃)²⁺ 2.50 M. Given: Kf = 7, 1 x 10 ^16
2 answers
Your post makes no sense to me as written. I understand what is being asked but I don't understand how the solution was made.
Let's try this... Adding 0.42 mole of Cd^+2 salt into a 2.50M solution of Cyanide salt. The concentration of Cd(CN)4^2- with a Kf = 7.1 x 10^16 will be 0.42mole/L. (100% conversion) Now, the Kd for the Cd-complex will be (7.1 x 10^16)^-1 = 1.4 x 10^-17.
Cd(CN)4^2- <=> Cd^+2 + 4CN^-
Ci: 0.42M 0 0
∆C: -x +x +4x
Ceq: 0.42-x x 4x
~0.42M
Kd = [Cd^+2][CN^-]^4/[Cd(CN)4^2-]
1.4E-17 = (x)(4x)^4/(0.42)
= 256x^5/0.42
Solving for x = [(0.42)(1.4E-14) / (256)]^1/5 = 1.2E-4
[Cd(CN)4^2-] = 0.42M - x ~ (0.42)M
[Cd^+2] = x = (1.2E-4)M
[CN^-] = 4x = 4(1.2E-4)M = (4.8E-4)M
Cd(CN)4^2- <=> Cd^+2 + 4CN^-
Ci: 0.42M 0 0
∆C: -x +x +4x
Ceq: 0.42-x x 4x
~0.42M
Kd = [Cd^+2][CN^-]^4/[Cd(CN)4^2-]
1.4E-17 = (x)(4x)^4/(0.42)
= 256x^5/0.42
Solving for x = [(0.42)(1.4E-14) / (256)]^1/5 = 1.2E-4
[Cd(CN)4^2-] = 0.42M - x ~ (0.42)M
[Cd^+2] = x = (1.2E-4)M
[CN^-] = 4x = 4(1.2E-4)M = (4.8E-4)M