Equal volume of a 0.02 M Zn+ solution and a 2.0M NH3 solution are mixed. Kf for [Zn(NH3)4] is 4.1*10^8. If enough sodium oxalate is added to make the solution 0.1 M in oxalate, will ZnC2O4 precipitate. Ksp ZnC2O4= 2.7*10^-8.

Question

I set up an ICE table

which have me:
Kf= (Zn(NH3)4)/ (Zn^2+)(NH3^-)
Kf= (.02-x)/ (1.92+x)(x)
4.1*10^8 = .02/1.92x
x = 2.54*10^-11

but when I used the [] values using x, in the Q equation. I don't get the correct Q for comparing to the K so say whether it will precipitate or not

Answer 1

I might try a little different approach.
Zn^+2 + 4NH3 ==> Zn(NH3)4^+2

First, since the problem states that equal volumes are mixed, that means Zn(II) is 0.02/2 = 0.01 initially and NH3 is 2/2 = 1 M initially.
Since the Kf for the Zn ammine is so LARGE, the final concn of the ammine will be essentially 0.01 M and the final concn of NH3 will be 1 to start less 4*Zn^+2 = 1-0.04 = 0.96 at the end. So plug 0.96 for NH3 (don't forget to raise it to the 4th power), 0.01 for the Zn ammine complex, and the value for Kf, and calculate Zn(II). Then substitute that into Qsp and compare with Ksp.