Chemical Equilibrium?

Question

I'm having trouble on this question. Any help would be appreciated.

The equilibrium constant for the reaction
AuBr4- + 2Au(s) + 2Br- <--> 3AuBr2-
can be determined by preparing a solution of AuBr4- and AuBr2- in contact with a piece of gold metal. The absorbance of the solution due to AuBr4- is measured at 382 nm. In one experiment, a solution containing a total of 6.41 x10-4 mol/L of dissolved gold (both AuBr4-, AuBr2-) in .4 M of hydrobromic acid was at equilibrium in the presence of gold metal. The total absorbance was found to be .445 in a 1.0 cm cell at 382 nm; however, AuBr2- does not absorb at 382 nm.
In a seperate experiment, the absorbance of 8.54x10-5 M solultion containing only AuBr4- (no AuBr2-) in .4 M HBr was determined in a 1.0 cm cell to be .41 at 382 nm.

1. Calculate the equilibrium concentrations of AuBr4- and AuBr2-.
2. Evaluate the equilibrium constant for the reaction.

Answer 1

Assuming that the absorbance of AuBr4- is linear, then the concentration of the AuBr4- in your first solution is:

(8.54 x 10^-5)(.445/.41) = 9.27 x 10^-5

The AuBr2- concentration is then 6.41 x 10^-4 - 9.27 x 10^-5 or 5.48 x 10^-4.

The equilibrium expression is
K = [AuBr2-]^3 / ([Br-]^2 * [AuBr4-]

Substitute the concentrations to get the value for K.

K = (5.48 x 10^-4)^3 / ((0.4)^2 *9.27 x 10^-5)
K = 1.65 x 10^-10 / 1.48 x 10^-5
K = 1.11 x 10^-5