Introduction to beer lambert law

of 6
For Any Query, Contact
abrar00adib@gmail.com
ANALYTICAL CHEMISTRY PART I
Lambart Law: This law states that when monochromatic light passes through a transparent
medium, the rate of decrease in intensity with the thickness of the medium is proportional to the
intensity of the light.
This is equivalent to stating that the intensity of the emitted light decreases exponentially as the
thickness of the absorbing medium increases arithmetically, or that any layer of given thickness
of the medium absorbs the same fraction of the light incident upon it.
The law can be expressed as follows-
−
𝑑𝐼
𝑑𝑙
∝ 𝐼
−
𝑑𝐼
𝑑𝑙
= 𝑎𝐼 … … … … … … … … … … … … (𝑖)
Where, I= Intensity of the incident light of wavelength λ
l= Thickness of the medium
a= Proportionality factor
Now, Integrating above equation,
∫ −
𝑑𝐼
𝐼
= 𝑎 ∫ 𝑑𝑙
𝐼
𝐼0
𝑜𝑟, 𝑙𝑜𝑔 𝑒 𝐼 − 𝑙𝑜𝑔 𝑒 𝐼0 = −𝑎𝑙
𝑜𝑟, 𝑙𝑜𝑔 𝑒
𝐼
𝐼0
= −𝑎𝑙
𝑜𝑟,
I
𝐼0
= 𝑒−𝑎𝑙
𝑆𝑜, 𝐼 = 𝐼0 𝑒−𝑎𝑙
………………………………………… (ii)
Beer studied the effect of concentration of the colored constituent in solution upon the light
transmission or absorption. He found the same relationship between transmission and
concentration as Lambert had discovered between transmission and thickness of the layer, i.e.
the intensity of a beam of monochromatic light decreases exponentially as the concentration of
the absorbing substance increases arithmetically.
𝐼 = 𝐼0 𝑒−𝑏𝑐
… … … … … … … … … … … … … … … … (𝑖𝑖𝑖)
From combination of equation (ii) and (iii), we get,
𝐼 = 𝐼0 𝑒−𝑘𝑐𝑙
𝑜𝑟,
𝐼
𝐼0
= 𝑒−𝑘𝑐𝑙
or,

of 6
𝑜𝑟,
𝐼
𝐼0
= 10−0.4343𝑘𝑐𝑙
𝑜𝑟,
𝐼
𝐼0
= 10−𝜀𝑐𝑙
𝑜𝑟, 𝑙𝑜𝑔
𝐼0
𝐼
= 𝜀𝑐𝑙 … … … … … … … … … … … … … … (𝑖𝑣)
𝑨 = 𝜺𝒄𝒍; ∵ 𝐴 = 𝑙𝑜𝑔
𝐼0
𝐼
Here, ε= Molar absorption coefficient
Here, equation (iv) is the fundamental of colorimetry and spectrophotometry, and is often
spoken as the Beer-Lambert Law. The value of ε will depend upon the method of expression of
the concentration.
Again, Since the fraction of radiant energy transmitted decays exponentially with path length,
we can write it in exponential form-
𝑇 =
𝐼
𝐼0
Where, T is called the transmittance, the fraction of radiant energy transmitted.
The percent transmittance is given by,
%𝑇 =
𝐼
𝐼0
× 100 = 100𝑇 … … … … … … … … … … . . . (𝑣)
𝐴 = 𝑙𝑜𝑔
𝐼0
𝐼
𝑜𝑟, 𝐴 = log (
1
𝑇
)
100
%𝑇
) (from equation v)
102
%𝑇
)
𝑜𝑟, 𝐴 = 2log10 − log%T
∴ 𝑨 = 𝟐 − 𝐥𝐨𝐠%𝐓
And, 𝐥𝐨𝐠%𝐓 = 𝟐 − 𝑨
Question: Why do we prefer to express the Beer-Lambert law using absorbance as a measure
of the absorption rather than %T?
Answer: From Beer-Lambert equation, it is found that-
𝑨 = 𝟐 − 𝐥𝐨𝐠%𝐓

of 6
Now, suppose, a solution of copper sulfate (which appears blue because it has an absorption
maximum at 600 nm). Look at the way in which the intensity of the light (radiant power)
changes as it passes through the solution in a 1 cm cuvette. The reduction every 0.2 cm is shown
in the diagram below.
The Law says that the fraction of the light absorbed by each layer of solution is the same. For
this illustration, it will be supposed that this fraction is 0.5 for each 0.2 cm "layer" and will be
calculated the following data:
Path length / cm 0 0.2 0.4 0.6 0.8 1.0
%T 100 50 25 12.5 6.25 3.125
Absorbance 0 0.3 0.6 0.9 1.2 1.5
A = εcl, shows that absorbance depends on the total quantity of the absorbing compound in the
light path through the cuvette. If absorbance is plotted against concentration, a straight line is
obtained passing through the origin (0,0).
The linear relationship between concentration and absorbance is both simple and
straightforward, which is why it is preferred to express the Beer-Lambert law using absorbance
as a measure of the absorption rather than %T.
Limitations to Beer-Lambert's law
The linearity of the Beer-Lambert law is limited by chemical and instrumental factors. Causes of
nonlinearity include:
a) Deviations in absorptivity coefficients at high concentrations (>0.01M) due to
electrostatic interactions between molecules in close proximity

of 6
b) Scattering of light due to particulates in the sample
c) Fluorescence or phosphorescence of the sample
d) Changes in refractive index at high analyte concentration
e) Shifts in chemical equilibria as a function of concentration
f) Non-monochromatic radiation, deviations can be minimized by using a relatively flat
part of the absorption spectrum such as the maximum of an absorption band
g) Stray light
Example: A sample in a 1.0-cm cell transmits 80% light at a certain wavelength. If the
absorptivity of this substance at this wavelenth is 2.0, what is its concentration?
Solution: The percent transmittance is 80%, and so T =0.80.
Also, l= 1 cm and ε= 2 cm-1gm-1L
So,
𝐴 = log (
1
𝑇
) = 𝜀𝑐𝑙
𝑜𝑟. log (
1
0.80
) = 2 × 1 × 𝑐
𝑜𝑟, 0.0969 = 2𝑐
𝑜𝑟, 𝑐 =
0.0969
2
= 0.04846 ≈ 0.05 𝑔/𝐿
Answer: 0.05 g/L.
Example: A solution containing 1.00mg iron (as the thiocyanate complex) in 100mL was
observed to transmit 70.0% of the incident light compared to an appropriate blank.
(a) What is the absorbance of the solution at this wavelength?
(b) What fraction of light would be transmitted by a solution of iron four times as
concentrated?
Solution:
(a) Here, Percent transmittance is 70%, and so T =0.70
𝐴 = log (
1
𝑇
) = 𝑙𝑜𝑔 (
1
0.70
) = log 1.43
∴ 𝐴 = 0.155

of 6
(b) According to Beer’s law, absorbance is linearly related to concentration. If the original
solution has an absorbance of 0.155, a four times more concentrated solution will have
an absorbance four times as much or 4×0.155=0.620.
The transmittance T will be: T =10−0.620 =0.240
Answer: (a) 0.155
(b) 0.240
Example: Amines, RNH2, react with picric acid to form amine picrates, which absorb strongly
at 359nm (ε=1.25×104). An unknown amine (0.1155g) is dissolved in water and diluted to
100mL. A 1-mL aliquot of this is diluted to 250mL for measurement. If this ﬁnal solution
exhibits an absorbance of 0.454 at 359nm using a 1.00-cm cell, what is the formula weight of the
amine?
Solution: Here,
ε=1.25×104 cm-1gm-1L
l = 1 cm
A= 0.454
We know that, A= εcl
𝑜𝑟, 0.454 = 1.25 × 104
× 1 × c
𝑐 = 3.63 × 10−5
𝑚𝑜𝑙/𝐿
(3.63 × 10−5
𝑚𝑜𝑙/𝐿)(0.250 𝐿)
1𝑚𝑙
× 100 𝑚𝐿 = 9.08 × 10−4
𝑚𝑜𝑙
∴
0.1155g
9.08 × 10−4 𝑚𝑜𝑙
= 127.2 𝑔/𝑚𝑜𝑙
Answer: 127.2 g/mol
Example: Chloroaniline in a sample is determined as the amine picrate as described in last
Example. A 0.0265-g sample is reacted with picric acid and diluted to 1L. The solution exhibits
an absorbance of 0.368 in a 1-cm cell. What is the percentage chloroaniline in the sample?
Solution: Here,
ε=1.25×104 cm-1gm-1L
l = 1 cm

of 6
A= 0.386
We know that, A= εcl
𝑜𝑟, 0.386 = 1.25 × 104
× 1 × c
𝑐 = 2.94 × 10−5
𝑚𝑜𝑙/𝐿
(2.94 × 10−5
𝑚𝑜𝑙/𝐿) × (127.6 𝑔/𝑚𝑜𝑙) = 3.75 × 10−3
𝑔 𝑐ℎ𝑙𝑜𝑟𝑜𝑎𝑛𝑖𝑙𝑖𝑛𝑒
3.75 × 10−3
𝑔 𝑐ℎ𝑙𝑜𝑟𝑜𝑎𝑛𝑖𝑙𝑖𝑛𝑒
2.65 × 10−2 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
× 100% = 14.2%
Answer: 14.2%

Introduction to beer lambert law

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Introduction to beer lambert law

Similar to Introduction to beer lambert law (20)

Recently uploaded

Recently uploaded (20)

Introduction to beer lambert law