BEER–LAMBERT LAW●Alternative units of ! in common use are mol 1 dm–3cm–1. These units are con-
venient because the concentration term in the Beer–Lambert equation is now
expressed in mol dm–3and the cell pathlength is in centimetres. The conversion is1 mol–1m^2 = 10 mol–1dm^3 cm–1●For example, 90 mol–1m^2 = 900 mol–1dm^3 cm–1.
Beer–Lambert plots
The Beer–Lambert equation is of the form y=mx, and predicts that a plot of the
absorbance of a compound at a particular wavelength against the concentration of
compound should be a straight line, provided we use the same cell (i.e. same b)
throughout the experiment. The slope of the graph is equal to ! b.
393Example 20.4
Use the spectrum in Exercise 20I to estimate the molar
absorption coefficient of potassium dichromate(VI) at 350 nm.Answer
Figure 20.12 shows that the absorbance of the sample at 350 nm is
approximately 0.78:c= 2.5 10 –4mol dm–3= 0.25 mol m–3; b= 1.0 cm = 0.010 mRearranging the Beer–Lambert equation,! 350 =A 350 /cb= 0.78/(0.25 0.010) = 310 mol–1m^2 (to two significant figs)Therefore,! 350 for potassium dichromate(VI) solution is 310 mol–1m^2.Beer–Lambert law calculations
(i) Chlorophyll ‘a’ shows an absorption maximum at 680 nm with !
1.1 10^4 mol–1m^2.
Estimate the concentration of chlorophyll ‘a’ in a leaf assuming (a) that chlorophyll ‘a’ is
evenly distributed throughout a leaf of thickness 0.010 cm, (b) that the leaf possesses an
absorbance of 1.1, and (c) that all the absorption at this wavelength is entirely due to
chlorophyll ‘a’.
(ii)The molar absorption coefficient of DNA (deoxyribonucleic acid) is about 1000 mol–1m^2
(expressed per mole of base) at 260 nm. What concentration of the compound would be
required to produce a solution of absorbance 0.010 in (a) a 1.0 cm cell and (b) a 4.0 cm
cell?
(iii)A solution of a drug (concentration 0.0020 mol dm–3) was placed in a 1.0 cm cell. The
sample absorbance at 250 nm was 0.40. Calculate the molar absorption coefficient of the
drug at 250 nm.Exercise 20O