Verification of the Beer–Lambert law (UV–visible spectrophotometry)

Theory

A slab of thickness dx removes the fraction \(dI/I = -\sigma n\,dx\) of the light passing through it. Integrating gives the Beer–Lambert law

$$ T = \frac{I}{I_0} = 10^{-\varepsilon c l},\qquad A = -\log_{10}T = \varepsilon\,c\,l ,$$

with ε the molar extinction coefficient (L mol⁻¹ cm⁻¹), c in mol/L and l in cm. A plot of A against c at λ_max is a straight line through the origin with slope εl.

Why measure at λ_max?

There the sensitivity is highest and dA/dλ = 0, so small wavelength errors and the finite spectral bandwidth matter least. Try measuring on the steep flank of a band with a wide slit.

Deviations from the law (all simulated)

• Stray light s: \(A_{meas} = -\log_{10}\frac{T+s}{1+s}\). A saturates near \(-\log_{10}s\), so the calibration curves over at high A.
• Polychromatic light: with a bandwidth comparable to the band width, an average of \(10^{-A(\lambda)}\) is measured instead of A(λ), giving negative curvature.
• Chemical: methylene blue dimerises (2M ⇌ D, K ≈ 4000 M⁻¹). The dimer absorbs at 605 nm rather than 664 nm, so A(664) rises more slowly than c.
• Photometric noise is worst at very high and very low A. The relative error in c is smallest near A ≈ 0.43 (T = 37%).

Procedure

1. Choose a solution and scan its spectrum to locate λ_max (KMnO₄ 525 nm, CuSO₄ 810 nm, K₂Cr₂O₇ 350 nm, methylene blue 664 nm).
2. Set λ = λ_max, insert a cuvette of pure solvent and press Blank. Without a blank, the cuvette and solvent absorption (about 0.04) offsets every reading.
3. Prepare a series (1, 2, … 10 mL of stock diluted to 10 mL) and record A for each.
4. Plot A against c and fit within the linear range. Then ε = slope / l.
5. Make an unknown, measure it, and read its concentration from the calibration line.

Reference values

• KMnO₄: ε₅₂₅ ≈ 2.45 × 10³ L mol⁻¹ cm⁻¹ (vibronic peaks at 507, 525, 546, 566 nm)
• K₂Cr₂O₇ (dilute HClO₄): ε₃₅₀ ≈ 3.15 × 10³, ε₂₅₇ ≈ 4.8 × 10³
• CuSO₄(aq): ε₈₁₀ ≈ 12 L mol⁻¹ cm⁻¹ (a d–d band, very weak)
• Methylene blue: ε₆₆₄ ≈ 7.4 × 10⁴ (monomer)

Viva questions

Why is absorbance additive but transmittance is not?

Transmittances of successive layers multiply, so their logarithms add. A is proportional to the number of absorbers in the beam.

Why use a blank?

It cancels reflection at the cuvette windows and absorption by the solvent, so only the solute contributes to A.

What is the colour of KMnO₄ and why?

It absorbs green light (500–560 nm) and transmits red and blue, so it looks purple. The cuvette drawing above uses the transmitted spectrum to set its colour.