Calculate the absorbance of AB in the following Question.

Question

The equilibrium constant for the reaction: AB→←

A+B is 5 x 10^-4 moles/L. The complex AB has maximum absorbance at 370 nm with a molar absorption coefficient of 300. Calculate the

absorbance at 370 nm in a 1 cm cell of solution prepared by mixing 10 ml of 0.01 M solution A

with 10 ml 0.01 M solution B.

Answer 1

To find the absorbance (A), we would use Beers Law: A = ecl

A = absorbance; e = molar absorption coefficient = 300 M^-1cm^-1; c= [AB]; l = path length = 1 cm

So first we must find c, the concentration of AB

AB <===> A + B

Initial [A] = [B] = 10 ml x 1 L / 1000 ml x 0.01 mol/L = 0.0001 mols

AB <===> A + B

0............0.0001.....0.0001.......Initial

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

x........0.0001-x....0.0001-x......Equilibrium

Keq = 5x10^-4 = [A][B] / [AB] = (0.0001-x)² / x assume x is small and ignore it in numerator to simplify the math

5x10^-4 = (0.0001)² / x

x = 2x10^-5 mols AB

[AB] = 2x10^-5 mol / 20 ml x 1000 ml / L = 0.001 M

Absorbance (A) = ecl

A = (300 M^-1cm^-1)(0.001 M)(1 cm)

A = 0.300