Chemistry, Third edition

394 20 · LIGHT AND SPECTROSCOPY

Example 20.5

An example of a Beer–Lambert plot is shown in Fig. 20.34. This

shows the aborbance of 4-nitrophenol (C 6 H 4 OHNO 2 ) in ethanol

solution at a wavelength of 312 nm. The cell pathlength was 1.0

cm. Calculate the molar absorption coefficient of 4-nitrophenol

at 312 nm.

Answer

The slope of the graph is

0.115 – 0.000

————————— = 9.60 mol–1m–3=!b

12.0 10 –3– 0.00

Since b 1.0 cm 0.010 m,

9.6 mol–1m^3

!= —————— = 9.6 10 2 mol–1m^2

0.010 m

Therefore,! 312 for 4-nitrophenol is 9.6  102 mol–1m^2.

0.12

0.08

0.04

0

Absorbance

2 4 6 8

(Concentration/mol m–3)/10–3

0 10 12

Fig. 20.34Beer–Lambert plot for 4-nitrophenol in ethanol

solution at 312 nm.

Beer–Lambert law with two

components

The molar absorption coefficient of 2-nitrophenol

at 312 nm is 1.2  102 mol–1m^2. A solution

contains 2-nitrophenol and 4-nitrophenol, both at

a concentration of 1.0  10 –3mol m–3. What is

the absorbance of the mixture at 312 nm in a

1.0 cm cell? Assume that the two phenols do

not react with each other. (Hint: The mixture

absorbance is the sum of the individual

absorbances.)

Exercise 20P