Equal volumes of a 0.020 M Zn²⁺(aq)solution and a 2.0 M NH₃(aq)solution are mixed.K_f for [Zn(NH₃)₄]²⁺ is 4.1 × 10⁸.If enough sodium oxalate is added to make the solution 0.10 M in oxalate,will ZnC₂O₄(s)precipitate? What is Q_sp? K_sp ZnC₂O₄ = 2.7 × 10^-8

Question

Quizplus · Accepted Answer

The first step is to write the balanced chemical equation for the formation of Zn(NH3)4^2+ and ZnC2O4 precipitation with balanced charges: 
ZnS + 4NH3 + H2C2O4 → [Zn(NH3)4]2+ + H2S + (NH4)2C2O4 
[Zn(NH3)4]2+ + C2O4^2- → ZnC2O4(s) + 4NH3 
The value of Kf = 4.1 × 10^18 for [Zn(NH3)4]2+ and Ksp = 2.7 × 10^-18 for ZnC2O4(s). 
At equilibrium, the reaction quotient Qsp is: 
Qsp = [Zn2+][C2O4^2-] 
The initial concentration of Zn2+ and C2O4^2- are both 0.010 M. 
If enough H2C2O4 is added to make the solution 0.10 M in oxalate, the concentration of the C2O4^2- ion will be 0.10 M, but the concentration of Zn2+ ion will not change because H2C2O4 is not a source of Zn2+. 
Now solve for Qsp: 
Qsp = [Zn2+][C2O4^2-] = (0.010)(0.10) = 1.0 × 10^-3 
Since Qsp > Ksp, the reaction will proceed to the left to reduce the concentration of products and reach equilibrium. This means that the precipitation of ZnC2O4(s) will not occur, so the answer is (C) no, Qsp = 2.9 × 10^-12.

Equal Volumes of a 0