Question

Multistep problem that I'm confused on the entire thing:

A) The addition of 5E-3 total moles of Zn2+ to a 1.0L solution of NaCN gives a solution of the complex ion [Zn(CN)₄^-2](Kf=4.2E19). What is the concentration of uncomplexed Zn²⁺ ions if the concentration of cyanide ions in the final solution is 0.5M?

B) ZnCO₃ is sparaingly soluble salt with a Ksp=1.0E-7. The addition of CN- (aq) to ZnCO₃(s) yields the complex ion [Zn(Cn)₄^2-] (aq) with the K_f mentioned before in Part A. Write the reaction.

C) What will be the Keq for this overall reaction?

D) What is the solubility of ZnCO₃ (s) in a solution of NaCN (aq) where the equilibrium concentration of cyanide in the solution is 0.2 M? Use the calculated Keq from Part C.

Answer 1

Kf is formation of complex

Ksp is dissolution constant.

A) Zn2+ + 4NaCN --------> [Zn(CN)4]2-

concentration of Zn2+ = 5 x 10^-3 moles / 1L = 5 x 10^-3

     Zn2+      +      4NaCN --------> [Zn(CN)4]2-
5 x 10^-3        4(5 x 10^-3)                 0

-x                          0.5                     5 x 10^-3

Feed the values,

Kf = 4.2 x 10^-3 = 5 x 10^-3 / [x](0.5)^4

[x] = 1.9 x 10^-21

Zn2+ uncomplexed in solution = 5 x 10^-3 - 1.9 x 10^-21 = 5 x 10^-3 M

B) The chemical equation would be,

ZnCO3(s) + 4CN-(aq) -------------> [Zn(CN)4]2-(aq) + CO32-(aq)

C) Keq = Kf x Ksp

            = 4.2 x 10^19 x 1 x 10^-7

            = 4.2 x 10^12

D) Keq. = [Zn(CN)4]2-/[Zn(CO3)][CN-]^4

4.2 x 10^12 = 5 x 10^-3/(ZnCO3)(0.2)^4

[Zn(CO3)] = 7.44 x 10^-20 is the solubility of ZnCO3(s)