Question

What is the concentration of Hg^2+ in 500.0 mL of a solution that was originally 0.011 M Hg^2+ and had 73 g of KI added to it?
Hg^2+(aq) + 4I^-(aq)<=> HgI4^2-(aq)

The overall formation constant is 1x10^30

Answer 1

With huge formation constants like this these problems are worked backwards, more or less.
Convert 73g KI in 500 mL to ?M. I get 0.88 but you should confirm that. First, off, make the assumption that with this huge Kf that the problem goes essentially to completion, like this.
........Hg^2+ + 4I^- ==> HgI4^2-
I......0.011....0.88......0
C.....-0.011..-0.044.....0.011
E.......0......0.836.....0.011
================================
NOW, we let the reaction go the other way and work this just as we would a simple ionization problem.
I.......0.......0.836.....0.011
C......+x........+4x.......-x
E.......x......0.836+4x....0.011-x

Substitute the equilibrium values into the expression for the Kf and solve for x = (Hg^2+). Post your work if you get stuck.

Answer 2

I'm messing up somewhere

1x10^30= [.011-x]/[x][.836+4x]^4

4.8x10^29(x)+2.56x10^32(x^5)=.011-x
Factoring out an x and moving all terms
x(4.8x10^29+2.56x10^32(x^4))-.011+x=0
Then because of the relative size I assumed -.011 and x are negligible
x(4.8x10^29+2.56x10^32(x^4))=0
But this gives
x=0
x^4=4.8x10^29/2.56x10^32 = .208

I'm either missing a step or messed up somewhere because I'm certain the concentration is somewhere around the 10^-30 magnitude.

Answer 3

I would assume 0.011-x = 0.011(which you did) and 0.836 + 4x = 0.836 (which you didn't) and not mess with any of that high order stuff.
1E30 = (0.011)/(x)(0.836)^4

Answer 4

Alright, I got it thanks!