What is the concentration of Hg 2+ in 500.0 ml of a solution that was originally M Hg 2+ and 0.78 M I-?

Question

The overall formation constant for HgI42- is 1.0 x 10 ^30.

1.0 x 10^30 = HgI42-/(Hg2+)(I-)^4

Answer 1

You didn't provide the original (initial) concentration of Hg²⁺ so I will assume it to be 0.01 M. If you meant a different concentration, then replace that for the 0.01 and continue as described.

Set up an ICE table

Hg²⁺ + 4I^- ==> HgI₄^2-

0.01...0.78........0..........Initial

-0.01....-0.04....+0.01..........Change

0.........0.74......0.01...........Equilibrium

Now, let the reaction proceed the other direction and set up another ICE table:

0.......0.74............0.01...........Initial

+x....+4x.............-x...............Change

x........0.74+4x..........0.01-x........Equilibrium

Kf = [HgI₄^2-] / [Hg²⁺][I^-]⁴

1x10³⁰ = 0.01-x / (x)(0.74+4x)

Because x is going to be such a small number, I think we can simplify as follows:

1x10³⁰ = 0.01 / (x)(0.74)⁴

1x10³⁰ = 0.01 / 0.3x

3x10²⁹x = 0.01

x = 3.3x10^-32