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Spectrophotometry: the Beer–Lambert law

A colored solution absorbs part of the light that crosses it, and the more concentrated it is, the more it absorbs. That is the basis of spectrophotometry, the most used method to measure concentrations in the laboratory. This topic explains the Beer–Lambert law, transmittance, calibration curves and how a mixture of several species is analyzed.

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Transmittance and absorbance

If an intensity $I_{0}$ arrives and $I$ leaves, the transmittance is $T = I / I_{0}$ (or %T = 100·T). Each thin layer of solution absorbs the same fraction of the light reaching it, so the intensity falls exponentially and what grows in proportion to concentration is the logarithm, the absorbance:

A = - lo g_{10} T = 2 - lo g_{10} (% T) A = ε b c

The light crosses the cell: it leaves attenuated, and the attenuation grows with the path length and with the concentration.

$ε$ is the molar absorptivity (L·mol⁻¹·cm⁻¹), specific to each species and each wavelength; $b$ is the path length (almost always 1 cm) and $c$ the molar concentration. A = 1 lets 10% of the light through and A = 2 lets 1%: that is why it works best with absorbances between 0.1 and 1, where the relative reading error is small.

Calibration curve

In practice $ε$ is rarely known in advance: standards of known concentration are prepared, their absorbance is measured at the wavelength of maximum absorption and A is plotted against c. The slope is $ε b$ ; the concentration of a sample is read on the line. With a single standard, $ε = A_{s} / (b c_{s})$ .

Illustrative calibration curve with five standards and a sample read on the line.

The law holds for dilute solutions and single-wavelength light. It deviates at high concentrations (molecules interact, the refractive index changes), with stray light or if the species reacts or associates on dilution. That is why samples are diluted to fall within the range of the standards.

Mixtures: additivity

If the sample has several species that do not react with each other, their absorbances add up. Measuring at as many wavelengths as species (preferably near each one's maximum) leaves a linear system with the concentrations as unknowns:

A (λ_{j}) = b i \sum ε_{ij} c_{i} \Rightarrow (ε_{11} ε_{12} ε_{21} ε_{22}) (c_{1} c_{2}) = \frac{1}{b} (A_{1} A_{2})

The system is well conditioned when each wavelength «sees» mostly one species; if the spectra are very similar, a small reading error changes the concentrations a lot. With more wavelengths than species it is solved by least squares, which averages the errors, and the residual tells how well the readings fit.

Absorption bands (schematic) of Co²⁺ and Ni²⁺: at 510 nm mostly cobalt absorbs and at 656 nm mostly nickel.

Common mistakes

Using %T instead of A in $A = ε b c$ , or taking $A = - lo g (% T)$ without converting to a fraction.
Mixing units: ε in L·mol⁻¹·cm⁻¹ requires b in cm and c in mol/L.
Extrapolating the calibration curve beyond the last standard.
Using the ε of one wavelength for readings taken at another.
Forgetting to subtract the blank (the cell with only the solvent).

Worked examples

Example 1From %T to concentration

A sample lets 45% of the light through in a 1 cm cell. The species has ε = 2455 L·mol⁻¹·cm⁻¹ at that wavelength. Find A and the concentration.

Solution

$T = 0.45$ → $A = - lo g 0.45 = 2 - lo g 45 = 0.3468$ .
$c = A / (ε b) = 0.3468/ (2455 \cdot 1) = 1.413 \times 1 0^{- 4}$ M = 141.3 µM.

A = 0.3468

c = 1.413 \times 1 0^{- 4}

Solve it in Simuveliq

In Chemistry → Spectrophotometry (Beer–Lambert), the Beer's law: A = ε b c card with the reading in %T = 45, ε = 2455, b = 1 cm and c blank (the button sets it): it gives c = 1.413×10⁻⁴ M and both steps (A = −log T, then c = A/εb). The Absorbance ↔ transmittance card does just the conversion.

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Example 2Mixture of two species

A sample with Co²⁺ and Ni²⁺ gives A = 0.446 at 510 nm and A = 0.326 at 656 nm (b = 1 cm). The ε are: Co²⁺ 36 400 and 1240; Ni²⁺ 5520 and 17 500 (at 510 and 656 nm). Find the concentrations.

Solution

System: $36400 c_{C o} + 5520 c_{N i} = 0.446$ and $1240 c_{C o} + 17500 c_{N i} = 0.326$ .
Determinant: $36400 \cdot 17500 - 5520 \cdot 1240 = 6.3016 \times 1 0^{8}$ . By Cramer: $c_{C o} = (0.446 \cdot 17500 - 5520 \cdot 0.326) /6.3016 \times 1 0^{8} = 9.530 \times 1 0^{- 6}$ M.
$c_{N i} = (36400 \cdot 0.326 - 1240 \cdot 0.446) /6.3016 \times 1 0^{8} = 1.795 \times 1 0^{- 5}$ M.
Check at 510 nm: Co contributes $36400 \cdot 9.53 \times 1 0^{- 6} = 0.347$ and Ni $0.099$ ; they add up to 0.446.

c_{C o} = 9.530

µM and

c_{N i} = 17.95

µM.

Solve it in Simuveliq

In the Mixture of several species at several wavelengths card, «Or load an example» → «Co²⁺ and Ni²⁺ at 510 and 656 nm (2 species)». The Result gives both concentrations; the Setup writes both equations, and the bar chart shows each species' contribution to each absorbance.

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Example 3Readings in %T

A mixture of dichromate and permanganate gives 35.2 %T at 440 nm and 21.4 %T at 545 nm (b = 1 cm). ε (L·mol⁻¹·cm⁻¹): Cr₂O₇²⁻ 369 and 11; MnO₄⁻ 95.5 and 2350. Find the concentrations in mM.

Solution

First to absorbance: $A_{440} = 2 - lo g 35.2 = 0.4535$ and $A_{545} = 2 - lo g 21.4 = 0.6696$ .
System: $369 c_{1} + 95.5 c_{2} = 0.4535$ and $11 c_{1} + 2350 c_{2} = 0.6696$ , with determinant 866 100.
$c_{1} = 1.157 \times 1 0^{- 3}$ M = 1.157 mM of dichromate and $c_{2} = 2.795 \times 1 0^{- 4}$ M = 0.2795 mM of permanganate. The ε are small (hundreds, not tens of thousands), so millimolar concentrations are needed to read useful absorbances.

Cr₂O₇²⁻ 1.157 mM and MnO₄⁻ 0.2795 mM.

Solve it in Simuveliq

Example «Dichromate and permanganate, readings in %T»: in step 1 the instrument gives %T and the concentrations are in mM; the tool converts each reading to A before solving.

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Example 4ε from standards

Standards of 1.0×10⁻⁴ M of A and of B separately give, at 450 and 620 nm: A 0.980 and 0.120; B 0.150 and 0.850. A mixture gives 0.620 and 0.510. Find the concentrations.

Solution

Each standard gives its ε: $ε = A_{s} / (b c_{s})$ , that is, A: 9800 and 1200; B: 1500 and 8500.
System: $9800 c_{A} + 1500 c_{B} = 0.620$ and $1200 c_{A} + 8500 c_{B} = 0.510$ → $c_{A} = 5.528 \times 1 0^{- 5}$ M and $c_{B} = 5.220 \times 1 0^{- 5}$ M.

c_{A} = 55.28

µM and

c_{B} = 52.20

µM.

Solve it in Simuveliq

Example «ε from 1.0×10⁻⁴ M standards»: in step 2 each species has «I get it from a standard of that species alone», and in step 3 the absorbances of the standards are typed instead of ε.

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Example 5More wavelengths than species

Two species P and Q are measured at 4 wavelengths (420, 480, 540 and 600 nm), with A = 0.312, 0.318, 0.381 and 0.450. Find their concentrations.

Solution

Four equations and two unknowns: there is no exact solution, because the readings have errors. Least squares finds the $c$ that minimize the sum of the squared differences between measured and computed A; one solves $(E^{T} E) c = E^{T} A / b$ .
Result: $c_{P} = 30.11$ µM and $c_{Q} = 39.88$ µM. The recomputed A differ from the measured ones by less than 0.003 (root-mean-square error 0.00157): the readings are consistent.

c_{P} = 30.11

µM and

c_{Q} = 39.88

µM (least-squares fit).

Solve it in Simuveliq

Example «More wavelengths than species (least squares)»: the result says so («4 wavelengths for 2 unknowns: least-squares fit») and the table compares, for each λ, the measured A with A = b Σ εc and their difference.

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Spectrophotometry: the Beer–Lambert law

Transmittance and absorbance

Calibration curve

Mixtures: additivity

Common mistakes

Worked examples

Example 1From %T to concentration

Solution

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Example 2Mixture of two species

Solution

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Example 3Readings in %T

Solution

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Example 4ε from standards

Solution

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Example 5More wavelengths than species

Solution

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