Beer-Lambert Law: Study Guide (Chemistry 3.13)

What the law says

Shine light of one wavelength through a coloured solution and some of it is absorbed. In Topics 3.11 and 3.12 you saw why: a molecule or ion absorbs a photon when the photon’s energy matches an electronic transition. Topic 3.13 asks a quantitative question: how much light is absorbed, and what controls it?

The answer is the Beer-Lambert law (often just called Beer’s law):

A = εbc

Symbol	Name	Usual unit	What it describes
A	absorbance	none	how much light the sample absorbs (bigger A, less light gets through)
ε	molar absorptivity	M⁻¹ cm⁻¹	how strongly this species absorbs light of this wavelength
b	path length	cm	the distance the light travels through the solution (the inside width of the cuvette)
c	concentration	M (mol L⁻¹)	the concentration of the absorbing species

Check the units: M⁻¹ cm⁻¹ × cm × M leaves nothing, so A has no unit. If your units do not cancel, your setup is wrong.

The law is a direct proportion in each variable. Double the path length and A doubles. Double the concentration and A doubles. Switch to a species (or a wavelength) with twice the molar absorptivity and A doubles.

Why b and c matter: counting particles in the beam

Picture the beam as a narrow column of light passing through the cuvette. Each absorbing particle in that column has a chance of absorbing a photon. What matters is how many absorbing particles the light meets on its way through.

• A longer path (larger b) puts more particles in the column, because the column is longer.
• A higher concentration (larger c) puts more particles in each centimetre of the column.

So b and c both scale the number of absorbing particles in the light path, and A follows that number. This gives a useful check: a 2.0 × 10⁻⁴ M solution in a 1.0 cm cuvette and a 1.0 × 10⁻⁴ M solution of the same species in a 2.0 cm cuvette have the same product bc (2.0 × 10⁻⁴ M cm). The light meets the same number of absorbing particles, so the absorbances are equal.

Notice what does not matter: the volume of solution in the cuvette. Pouring in more solution makes the liquid deeper, but the beam still crosses the same width at the same concentration.

Molar absorptivity and the choice of wavelength

ε is a property of a particular species at a particular wavelength. A dye that looks purple absorbs some wavelengths strongly and others hardly at all, so its ε is large at some wavelengths and small at others. ε does not change when you change the concentration; it is a constant for that species, solvent and wavelength.

An absorption spectrum plots A against wavelength for one solution. Figure 1 shows the spectrum of a 5.00 × 10⁻⁵ M solution of an invented dye, Dye V, which we use again in Worked example 2.

Because b and c are fixed for this one solution, the shape of the spectrum is the shape of ε against wavelength. At 560 nm, ε = 0.559 ÷ (1.00 cm × 5.00 × 10⁻⁵ M) ≈ 1.12 × 10⁴ M⁻¹ cm⁻¹. At 500 nm (A ≈ 0.18) it is only about 3.6 × 10³ M⁻¹ cm⁻¹, roughly a third as large.

Chemists set the spectrophotometer to the wavelength of maximum absorbance, λmax, for two reasons:

1. Sensitivity. The largest ε gives the biggest change in A for a given change in concentration. Small differences between solutions show up clearly.
2. Stability. The top of the peak is nearly flat, so if the instrument’s wavelength setting drifts by a nanometre or two, A hardly changes. On the steep side of the peak, the same drift would change A noticeably.

If another substance in the mixture also absorbs at λmax, you may need a different wavelength where only your species absorbs, so that the reading belongs to one species alone.

How a spectrophotometer measures absorbance

Before measuring any sample, you fill a cuvette with the solvent only (the blank) and set the instrument to A = 0. This cancels light lost to the cuvette walls and the solvent, so later readings measure only the absorbing species.

Background (not needed for calculations in this course): the instrument actually measures the fraction of light transmitted, T = I / I₀, and converts it with A = −log₁₀ T. That is why %T falls as concentration rises, but only A is proportional to concentration.

Calibration curves

In most experiments you use the same cuvette (fixed b) and the same wavelength (fixed ε). Then εb is one constant, and A is proportional to c alone. You can use this in two ways:

• One standard. Measure a solution of known concentration, calculate ε, then use it for the unknown (Worked example 1).
• Several standards (better). Measure a series of known concentrations, plot A against c and draw the best-fit straight line. Its gradient is εb, and it should pass through (or very close to) the origin. Read the unknown’s concentration from the line (Worked example 2).

A calibration curve averages out small errors in individual standards, and it shows you whether the readings really are linear. Keep the unknown’s absorbance inside the range of your standards. Very high absorbances, when hardly any light reaches the detector, are less reliable, and the line can bend at high concentration. If the unknown reads too high, dilute it by a known factor and multiply back afterwards.

Worked example 1: one standard, one unknown

Question. A standard solution of an invented blue dye, Dye B, has concentration 2.50 × 10⁻⁵ M. In a 1.00 cm cuvette at the dye’s λmax, its absorbance is 0.380. A drink sample measured in the same way has A = 0.252. Find ε and the concentration of Dye B in the drink.

1. Rearrange A = εbc for ε: ε = A ÷ (bc) = 0.380 ÷ (1.00 cm × 2.50 × 10⁻⁵ M) = 1.52 × 10⁴ M⁻¹ cm⁻¹.
2. Rearrange for c: c = A ÷ (εb) = 0.252 ÷ (1.52 × 10⁴ M⁻¹ cm⁻¹ × 1.00 cm) = 1.658 × 10⁻⁵ M.

Answer. c = 1.66 × 10⁻⁵ M (3 significant figures, matching the data).

Check by proportion. Same cuvette and wavelength, so c is proportional to A: 2.50 × 10⁻⁵ M × (0.252 ÷ 0.380) = 1.66 × 10⁻⁵ M. The drink’s absorbance is about two thirds of the standard’s, and so is its concentration.

Worked example 2: calibration curve and a diluted sample

Question. A student makes five standards of Dye V and measures them at 560 nm in a 1.00 cm cuvette, after zeroing with a water blank:

c (× 10⁻⁵ M)	1.00	2.00	3.00	4.00	5.00
A	0.112	0.221	0.338	0.446	0.559

A sample was too concentrated to read, so she diluted 10.00 mL of it to 100.0 mL with water. The diluted solution has A = 0.284. Find ε and the concentration of Dye V in the original sample.

1. Gradient. The best-fit line passes through the origin and, for example, through (5.00 × 10⁻⁵ M, 0.559). Gradient = 0.559 ÷ 5.00 × 10⁻⁵ M ≈ 1.12 × 10⁴ M⁻¹. (A least-squares fit of all five points gives the same value to 3 significant figures.)
2. Molar absorptivity. Gradient = εb, so ε = 1.12 × 10⁴ M⁻¹ ÷ 1.00 cm = 1.12 × 10⁴ M⁻¹ cm⁻¹.
3. Diluted solution. c = A ÷ gradient = 0.284 ÷ 1.12 × 10⁴ M⁻¹ = 2.54 × 10⁻⁵ M. This lies inside the range of the standards, so the reading is trustworthy.
4. Undo the dilution. The sample was diluted from 10.00 mL to 100.0 mL, a factor of 10.0. Original c = 2.54 × 10⁻⁵ M × 10.0 = 2.54 × 10⁻⁴ M.

Check. By the law, the undiluted sample would read A ≈ 2.84, far above the highest standard (0.559). That is why the dilution was needed.

Extension. In a 0.500 cm cuvette, the same diluted solution would give A = 0.284 ÷ 2 = 0.142, because the light meets half as many absorbing particles.

Worked example 3: predicting the effect of an error

Every error question has the same chain: what happens to the measured A? Then, because c = A ÷ (εb), what happens to the calculated c?

Question. Using Dye B from Worked example 1 (ε = 1.52 × 10⁴ M⁻¹ cm⁻¹, b = 1.00 cm), predict the effect of each error on the calculated concentration of the drink.

(a) A fingerprint on the cuvette during the drink measurement only. The smudge absorbs and scatters some light, so less reaches the detector and the measured A is too high. If it added 0.015 to the reading, A = 0.267 and c = 0.267 ÷ (1.52 × 10⁴ × 1.00) = 1.76 × 10⁻⁵ M, about 6% too high.

(b) Water droplets left in the cuvette from rinsing, before the drink is added. The drink is slightly diluted, so there are fewer absorbing particles in the beam: A is too low, and the calculated c is too low.

(c) Wavelength set away from λmax for both the standard and the drink. ε is smaller, so both absorbances are smaller, but ε is found from the standard at that same wavelength. The answer is not biased, but it is less precise, because small reading errors are now a larger fraction of each A.

Error	Measured A	Calculated c
Smudge, scratch or air bubble in the beam (unknown only)	too high	too high
Cloudy sample (suspended solid scatters light)	too high	too high
Cuvette wet with water, unknown diluted	too low	too low
Unknown read without zeroing on the blank, though the standards were zeroed (solvent and cuvette absorb a little)	too high	too high
Unknown measured in a cuvette with a longer path than the standards	too high	too high

Always name the direction of the error in A first, then link it to c through the equation. “It would be inaccurate” earns nothing on its own.

Common misconceptions

• “Absorbance has a unit.” It does not. ε carries the unit M⁻¹ cm⁻¹ so that εbc has none.
• “A more concentrated solution has a larger ε.” ε depends on the species and the wavelength only. Concentration changes A, not ε.
• “More solution in the cuvette means a larger absorbance.” The beam crosses the cuvette sideways. Depth of liquid does not change b.
• “Choose the wavelength where most light gets through, so the detector gets a strong signal.” That wavelength gives small, insensitive absorbances. Choose λmax.
• “Percent transmittance is proportional to concentration.” Only absorbance is. %T falls as c rises, but not in a straight line.
• “Fingerprints make the solution seem weaker.” They block light, so the solution seems stronger (A and c too high).
• Forgetting the dilution factor. The calibration line gives the concentration of the solution in the cuvette. Multiply back to get the original sample.
• “Colourless means no absorbance.” A colourless species may still absorb ultraviolet light; it simply does not absorb visible light.

Where this leads

This topic completes Unit 3. It builds on photon absorption from Topic 3.12, Properties of Photons and on molarity and dilution from Topic 3.7. Unit 4 starts with Topic 4.1, Introduction for Reactions. In Unit 5 (kinetics) you will use absorbance readings taken over time to follow how a concentration changes during a reaction. Try the practice questions now, then use the revision notes and the checklist to consolidate.