The equilibrium constant for the reaction AuBr4- +2Au(s) +2Br- forward and rever

Question

The equilibrium constant for the reaction AuBr4- +2Au(s) +2Br- forward and reverse reaction to 3 AuBr2- 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*10^-4 mol/l of dissolved gold (both ArBr4- and AuBr2-) in 0.400 M hydrobromic acid was at equilibrium in the presence of gold metal. The absorbance was found to be 0.455 in a 1.00 cm cell at 382 nm; it should be noted, however, that AuBr2- is transparent at 382 nm (it does not absorb that wavelength) and did not contribute to the measured absorbance. In a separate experiment, the absorbance of a 8.54*10^-5 M solution containing only AuBr4- (and no AuBr2- in the absence of solid gold metal) in 0.400 M HBr was determined in a 1.00 cm cell to be 0.410 at 382 nm. 1. Calculate the equilibrium concentrations of AuBr4- and AuBr2-. 2. Evaluate the equilibrium constant for the reaction.

Explanation / Answer

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

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