How to calculate concentrations at equilibrium

Question

A solution is formed by mixing 50.0 mL of 11.2 M NaX with 50.0 mL of 2.0 10-3 M CuNO3? Assume that Cu+  forms complex ions with  as follows:

Cu+(aq) + X -(aq)<--> CuX(aq) K1 = 1.0 102

CuX(aq) + X -(aq) <--> CuX2-(aq) K2 = 1.0 104

CuX2-(aq) + X -(aq) <--> CuX32-(aq) K3 = 1.0 103

The overall reaction is the following.

Cu+(aq) + 3 X -(aq) <--> CuX32-(aq) K = 1.0 109

Calculate the following concentrations at equilibrium.

a) CuX32- = ?

b) CuX2- = ?

c)Cu+=?

if i could get a hint or a head start on how to go about answering this quesiton that would be great

Answer 1

NaX + CuNO₃ ==> CuX + NaNO₃

Overall:

Cu⁺(aq) + 3X^-(aq) <==> [CuX₃^2-]

Cu⁺(aq) + X ^-(aq)<--> CuX(aq) K1 = 1.0x10²

CuX(aq) + X ^-(aq) <--> CuX₂^-(aq) K2 = 1.0x10⁴

CuX₂^-(aq) + X ^-(aq) <--> CuX₃^2-(aq) K3 = 1.0x10³

a) K = [CuX₃^2-]/[CuX₂^-][X^-] = K1 x K2 x K3 = 10² x 10⁴ x 10³ = 1x10⁹

b) K = [CuX₂^-]/[CuX][X^-] = K1 x K2 = 10² x 10⁴ = 1x10⁷

c) K = [CuX]/[Cu⁺][X^-] = 1x10²

Concentration of X^-: 0.05 L X- x 11.2 mol/L = 0.56 moles X- in 0.1 L = 0.056 M

Concentration of Cu⁺: 0.05 L Cu⁺ x 2.0x10^-3 mol/L = 1x10^-4 mol/0.1 L = 1x10^-3 M

Use these concentrations and the above equilibrium constants to solve for concentrations