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The Beer–Lambert law connects how much light a solution absorbs to how concentrated it is. It’s the basis of colorimetry, UV-visible spectroscopy, and countless routine lab measurements. These seventeen problems cover every common question type. For the theory, see the Beer–Lambert law, colorimetry and UV-vis spectroscopy.
Key equations
- A = εcl
- A = absorbance (no units)
- ε = molar absorptivity (dm³ mol⁻¹ cm⁻¹)
- c = concentration (mol dm⁻³)
- l = path length (cm), usually 1.00 cm
- Transmittance: T = I ÷ I₀ (often given as %T = T × 100)
- A = −log₁₀ T, or A = 2 − log₁₀(%T)
Part A: Absorbance and transmittance
1. A solution transmits 25% of the incident light. What is its absorbance?
Answer: T = 0.25. A = −log(0.25) = 0.602
2. A solution has an absorbance of 1.00. What percentage of light is transmitted?
Answer: T = 10⁻¹·⁰⁰ = 0.100, so 10% transmitted (90% absorbed).
3. If the absorbance doubles from 0.30 to 0.60, what happens to the transmittance?
Answer: T goes from 10⁻⁰·³⁰ = 0.501 (50.1%) to 10⁻⁰·⁶⁰ = 0.251 (25.1%). Doubling the absorbance squares the transmittance (0.501² ≈ 0.251); it doesn’t halve it. Absorbance is logarithmic.
Part B: Direct calculations
4. A solution of a dye has ε = 12 000 dm³ mol⁻¹ cm⁻¹ at 520 nm. In a 1.00 cm cell, its absorbance is 0.480. Find its concentration.
Answer: c = A ÷ (εl) = 0.480 ÷ (12 000 × 1.00) = 4.00 × 10⁻⁵ mol dm⁻³
5. A 2.50 × 10⁻⁴ mol dm⁻³ solution has an absorbance of 0.625 in a 1.00 cm cell. Calculate ε.
Answer: ε = A ÷ (cl) = 0.625 ÷ (2.50 × 10⁻⁴ × 1.00) = 2500 dm³ mol⁻¹ cm⁻¹
6. The solution in question 5 is measured in a 0.500 cm cell. Predict its absorbance.
Answer: A is proportional to l, so A = 0.625 × 0.500 = 0.313
7. A copper(II) solution has ε = 12.0 dm³ mol⁻¹ cm⁻¹ at 800 nm (a weakly absorbing ion). What concentration gives A = 0.600 in a 1.00 cm cell?
Answer: c = 0.600 ÷ 12.0 = 0.0500 mol dm⁻³. Transition metal ions often have small ε values, so relatively concentrated solutions are needed, or a ligand (such as ammonia) is added to form a more intensely coloured complex.
Part C: Calibration graphs
8. Standards of potassium manganate(VII) give these absorbances at 525 nm:
| Concentration (× 10⁻⁴ mol dm⁻³) | 0.0 | 1.0 | 2.0 | 3.0 | 4.0 |
|---|---|---|---|---|---|
| Absorbance | 0.000 | 0.235 | 0.470 | 0.705 | 0.940 |
An unknown gives A = 0.587. Find its concentration.
Answer: The graph is a straight line through the origin with gradient 0.235 per 1.0 × 10⁻⁴ mol dm⁻³ (i.e. 2350 dm³ mol⁻¹ with l = 1.00 cm). c = 0.587 ÷ 2350 = 2.50 × 10⁻⁴ mol dm⁻³ See calibration curves.
9. Why is it better to use a calibration graph than a single value of ε from a data book?
Answer: A calibration graph uses the actual instrument, cells, wavelength setting and conditions, so it corrects for differences between instruments, stray light and slight wavelength errors. It also shows whether the response is linear over the range used.
10. The unknown in a calibration experiment gives A = 1.35, but the highest standard only had A = 0.94. What should the analyst do?
Answer: Don’t extrapolate. Dilute the unknown by a known factor so its absorbance falls within the calibration range, measure again, then multiply by the dilution factor.
Part D: Dilutions and real samples
11. 5.00 cm³ of a food colouring is diluted to 250.0 cm³. The diluted solution has A = 0.420 at 630 nm, where ε = 1.05 × 10⁵ dm³ mol⁻¹ cm⁻¹ (l = 1.00 cm). Find the concentration of dye in the original colouring.
Answer: Diluted: c = 0.420 ÷ 1.05 × 10⁵ = 4.00 × 10⁻⁶ mol dm⁻³ Dilution factor = 250.0 ÷ 5.00 = 50 Original: 4.00 × 10⁻⁶ × 50 = 2.00 × 10⁻⁴ mol dm⁻³
12. An iron tablet is dissolved and made up to 100.0 cm³. 10.00 cm³ of this is treated to form a red iron(II) complex and made up to 100.0 cm³. The calibration graph shows the final solution contains 1.20 × 10⁻⁴ mol dm⁻³ of iron. Find the mass of iron in the tablet (Fe = 55.85).
Answer: Moles in final 100.0 cm³ = 1.20 × 10⁻⁴ × 0.1000 = 1.20 × 10⁻⁵ mol. This all came from the 10.00 cm³ portion. Moles in the whole 100.0 cm³ tablet solution = 1.20 × 10⁻⁵ × (100.0 ÷ 10.00) = 1.20 × 10⁻⁴ mol Mass = 1.20 × 10⁻⁴ × 55.85 = 6.70 × 10⁻³ g = 6.70 mg
Part E: Wavelength, mixtures and limitations
13. Why is the wavelength of maximum absorbance (λmax) usually chosen for measurements?
Answer: At λmax, absorbance changes most for a given change in concentration, giving the greatest sensitivity. The absorbance curve is also flattest at its peak, so small errors in the wavelength setting have the least effect on the reading.
14. A mixture contains two dyes, X and Y. At 450 nm only X absorbs; at 600 nm both absorb. Outline how both concentrations could be found.
Answer: Absorbances are additive. Measure A at 450 nm to find [X] directly. At 600 nm, A(total) = ε(X,600)[X]l + ε(Y,600)[Y]l. Knowing [X] and both ε values (from standards), solve for [Y].
15. Give three reasons why a plot of A against c might curve at high concentrations.
Answer: Any three of:
- At high concentrations, molecules interact with each other, changing their absorbing properties.
- Chemical changes with concentration, such as association, dissociation or changing equilibria.
- Stray light reaching the detector becomes significant when very little light passes through the sample.
- The light isn’t perfectly monochromatic.
- Very high absorbances (above about 1.5–2) mean so little light is transmitted that the measurement becomes imprecise.
Part F: Units and connections
16. A solution contains 2.00 mg dm⁻³ of a dye with molar mass 400 g mol⁻¹ and ε = 5.00 × 10⁴ dm³ mol⁻¹ cm⁻¹. Predict its absorbance in a 1.00 cm cell.
Answer: First convert to mol dm⁻³: 2.00 × 10⁻³ g dm⁻³ ÷ 400 g mol⁻¹ = 5.00 × 10⁻⁶ mol dm⁻³. A = 5.00 × 10⁴ × 5.00 × 10⁻⁶ × 1.00 = 0.250
17. A student measures a coloured solution in a plastic cuvette that has a scratch on one clear face. Readings vary each time the cuvette is reinserted. Explain, and suggest a fix.
Answer: The scratch scatters light, reducing the light reaching the detector by an amount that depends on how the cuvette is positioned, so the apparent absorbance changes. Use an unscratched cuvette, always insert it the same way round, handle it only by the frosted sides, and wipe the clear faces before each reading.
Common mistakes
- Using %T directly as if it were A.
- Forgetting the path length when it isn’t 1 cm.
- Forgetting to multiply by the dilution factor.
- Extrapolating beyond the calibration range.
- Mixing up units of concentration (mol dm⁻³ vs mg dm⁻³); convert using molar mass. See molarity explained.
- Not zeroing the instrument with a blank (the solvent and reagents without the analyte).
Key takeaways
- A = εcl; absorbance is logarithmically related to transmittance.
- Absorbance is proportional to both concentration and path length.
- Calibration graphs are more reliable than single literature ε values.
- Dilute samples into the calibration range and multiply back up.
- Measure at λmax, zero with a blank, and remember the law works best at low absorbances.
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