AP Chem 3.13: Beer-Lambert Law

Measuring concentration with light

The more concentrated a colored solution, the darker it looks and the more light it absorbs. A spectrophotometer分光光度计 uses this to measure concentration: it selects one wavelength from a light source, passes it through a cuvette比色皿 holding the solution, and a detector measures how much light gets through.

If the incident light has intensity I₀ and the transmitted light has intensity I, the instrument reports the absorbance吸光度:

A = \log \frac{I_0}{I}

A = 0 means nothing is absorbed; A = 1 means only 10% gets through; A = 2 means only 1% gets through. Absorbance has no units.

The Beer–Lambert law

A = \varepsilon b c

ε is the molar absorptivity摩尔吸光系数, in M⁻¹ cm⁻¹, set by the substance and the wavelength;
b is the path length光程, the distance the light travels through the solution, in cm (a standard cuvette is 1.00 cm);
c is the concentration, in M (mol/L).

At a fixed wavelength and path length, A is directly proportional to c. That is why absorbance can be used to measure concentration.

Good technique

Zero the instrument. First put in a cuvette holding only solvent (a blank空白) and set the absorbance to 0. This subtracts the absorbance of the solvent and the cuvette itself.
Choose the wavelength. Pick λmax, the wavelength of strongest absorption on the absorption spectrum (U3.11): ε is largest there, so the measurement is most sensitive. Standards and samples must be measured at the same wavelength.
Handle the cuvette. Rinse it with the solution to be measured; hold it by the frosted sides; wipe the clear sides clean; keep bubbles out of the light path.
The substance must absorb the light. A visible-light spectrophotometer suits colored substances. In $\ce{2NO(g) + O2(g) -> 2NO2(g)}$ , only $\ce{NO2}$ is colored (red-brown), so the absorbance follows the formation of $\ce{NO2}$ ; colorless reactants and products can’t be measured with visible light.
Keep the absorbance in range. If a sample’s A is above that of the most concentrated standard, dilute it quantitatively, measure again, and multiply by the dilution factor.

Calibration curves

In practice ε is often unknown. Instead, make several standard solutions标准溶液 of known concentration, measure their absorbance at the same wavelength and path length, and plot A against c. The result is a straight line through (or very near) the origin, called a calibration curve标准曲线.

The slope is εb, so a known path length gives ε.
Measure the unknown’s A and use the line to find its c.
Then n = cV gives the amount in moles, and m = nM the mass.

🔧 Math tool card · Slope & Linear RelationshipsSkip it if you already know this

Slope. Choose two points on the line that are far apart, (c₁, A₁) and (c₂, A₂); the slope is (A₂ − A₁) ÷ (c₂ − c₁). Use points on the line, not a data point that sits off it. A has no units and c is in M, so the slope is in M⁻¹.

Intercept. If the instrument was zeroed correctly, the line passes very close to the origin.

From y back to x. To find c from A: c = (A − intercept) ÷ slope; when the intercept is about 0, c = A ÷ slope.

A calibration curve passes through (1.0 × 10⁻⁵ M, 0.20) and (4.0 × 10⁻⁵ M, 0.80). What is its slope?

Try it with a hypothetical dye Y: change the concentration, path length, and wavelength; build a calibration curve to find an unknown concentration; then see what common mistakes do:

Try it: the spectrophotometer and the Beer–Lambert law

Dye Y (hypothetical data): λmax = 620 nm, where ε = 2.00 × 10⁴ M⁻¹ cm⁻¹.

Path length b:

Lab errors: first A, then c

Analyze an error in two steps: does the mistake make the measured A too high or too low? Then, since c = A ÷ slope, the calculated concentration is too high or too low with it.

Situation	Measured A	Calculated concentration
Water droplets left in the cuvette dilute the sample	Too low	Too low
Fingerprints on the clear faces, or bubbles in the light path	Too high	Too high
Instrument not zeroed with a blank	Too high	Too high
Sample measured away from λmax (curve made at λmax)	Too low	Too low
Sample measured in a wider cuvette (longer light path)	Too high	Too high
More solution poured into the cuvette	No change	No change
Standards diluted past the mark when made up	Standards’ A too low, so the slope is too small	Too high

The last row takes one more step: the standards are actually less concentrated than their labels say, so their absorbances are too low and the slope comes out too small. The unknown’s A is fine, but dividing it by a slope that is too small gives a concentration that is too high.

Example 1: From calibration curve to moles

Standard solutions of dye Z are measured at λmax in a 1.00 cm cuvette (hypothetical data):

c (10⁻⁵ M)	1.00	2.00	3.00	4.00
A	0.150	0.300	0.450	0.600

A solid sample is dissolved in water to make 250.0 mL of solution, which gives A = 0.390. Find the molar absorptivity of dye Z, the concentration of the solution, and the moles of dye Z in the sample.

Step-by-step solution

Find the slope of the calibration line. Use two points far apart on the line:
$\begin{aligned} \text{slope} &= \frac{0.600 - 0.150}{(4.00 - 1.00) \times 10^{-5}} \\ &= 1.50 \times 10^{4}\ \text{M}^{-1} \end{aligned}$
Molar absorptivity. A = εbc, so the slope is εb:
$\begin{aligned} \varepsilon &= \frac{\text{slope}}{b} = \frac{1.50 \times 10^{4}}{1.00} \\ &= \boxed{1.50 \times 10^{4}\ \text{M}^{-1}\,\text{cm}^{-1}} \end{aligned}$
Concentration of the sample solution.
$\begin{aligned} c &= \frac{A}{\varepsilon b} = \frac{0.390}{1.50 \times 10^{4}} \\ &= \boxed{2.60 \times 10^{-5}\ \text{M}} \end{aligned}$
Moles. Use the volume of the whole solution (250.0 mL), not the volume in the cuvette:
$\begin{aligned} n &= cV \\ &= (2.60 \times 10^{-5})(0.2500) \\ &= \boxed{6.50 \times 10^{-6}\ \text{mol}} \end{aligned}$

Example 2: When the absorbance is too high

Another solution measured against the calibration curve from Example 1 gives A = 1.80, far above the most concentrated standard (A = 0.600). The student pipets 5.00 mL of it into a 25.00 mL volumetric flask, dilutes to the mark, and measures A = 0.360. Find the concentration of the original solution. What if there were water droplets in the cuvette when the diluted solution was measured?

Step-by-step solution

Why dilute? At A = 1.80 only about 1.6% of the light gets through, so readings are imprecise, and the value is outside the calibration range, where the straight-line relationship isn’t known to hold.
Concentration of the diluted solution.
$\begin{aligned} c_\text{dil} &= \frac{A}{\varepsilon b} = \frac{0.360}{1.50 \times 10^{4}} \\ &= 2.40 \times 10^{-5}\ \text{M} \end{aligned}$
Multiply by the dilution factor. Dilution factor = 25.00 mL ÷ 5.00 mL = 5:
$\begin{aligned} c_\text{orig} &= 5 \times 2.40 \times 10^{-5} \\ &= \boxed{1.20 \times 10^{-4}\ \text{M}} \end{aligned}$
Effect of droplets. The droplets dilute the solution a little further, so the measured A is too low, c_dil is too low, and so is the result after multiplying by 5.

Example 3: Choosing a wavelength when absorptions overlap

Pigments are extracted from 0.500 g of spinach leaves with acetone, and the extract is made up to 25.00 mL. The extract is mostly chlorophyll a, with a little chlorophyll b, and both absorb light. The table gives their molar absorptivities at several wavelengths (illustrative data, in 10⁴ M⁻¹ cm⁻¹):

Wavelength (nm)	430	453	642	662
Chlorophyll a	11.0	4.0	2.5	8.6
Chlorophyll b	6.0	15.0	5.2	1.0

Which wavelength should be used to measure chlorophyll a? At that wavelength the extract gives A = 0.430 in a 1.00 cm cuvette. Ignoring absorption by chlorophyll b, find the mass percent of chlorophyll a in the leaves (M = 893.5 g/mol).

Step-by-step solution

Two conditions for the wavelength. The substance being measured should absorb strongly (sensitive, small relative error), and the interfering substance should absorb weakly (so nearly all of A comes from the substance being measured).
Check each wavelength. At 430 nm chlorophyll a absorbs most, but b absorbs strongly too; at 453 and 642 nm b absorbs more than a; at 662 nm a has ε = 8.6 × 10⁴ while b has only 1.0 × 10⁴, the least interference, so choose it.
Concentration of chlorophyll a in the extract.
$\begin{aligned} c &= \frac{A}{\varepsilon b} = \frac{0.430}{8.6 \times 10^{4}} \\ &= 5.0 \times 10^{-6}\ \text{M} \end{aligned}$
Moles (mol) and mass (g). Use the volume of the whole extract, 25.00 mL:
$\begin{aligned} n &= (5.0 \times 10^{-6})(0.02500) \\ &= 1.25 \times 10^{-7} \\ m &= (1.25 \times 10^{-7})(893.5) \\ &= 1.12 \times 10^{-4} \end{aligned}$
Mass percent.
$\begin{aligned} & \frac{1.12 \times 10^{-4}}{0.500} \times 100\% \\ &= \boxed{0.022\%} \end{aligned}$
Think about error. Chlorophyll b still absorbs a little at 662 nm, so ignoring it makes the measured A slightly too high and the result slightly high; the less b there is, the smaller the error.

Practice: how is the calculated concentration affected?

Fingerprints on the clear faces of the cuvette
Water droplets left in the cuvette
Instrument not zeroed with a blank (the blank absorbs a little)
Sample measured away from λmax (curve made at λmax)
Standards in a 1.00 cm cuvette, sample in a 2.00 cm cuvette
More solution poured into the cuvette (still filling the light path)
Standards diluted past the mark when made up
Bubbles stuck to the inside of the cuvette in the light path
Water accidentally added to the sample before measuring, and not accounted for
Cuvette inserted with a frosted side in the light path
Switching to another clean cuvette of the same size

Common mistakes

Mistake 1: “Higher absorbance means more light gets through.” It’s the reverse. A = log(I₀/I): the more light absorbed, the less gets through, and the larger A is.

Mistake 2: “Any wavelength works just as well.” ε changes with wavelength. It is largest at λmax, where the measurement is most sensitive, and standards and samples must be measured at the same wavelength or they can’t share a calibration curve.

Mistake 3: “More solution in the cuvette gives a higher absorbance.” A = εbc: absorbance depends on the path length and concentration, not the volume. As long as the solution fills the light path, the amount poured in doesn’t matter.

Quick check

At the measurement wavelength, a substance has ε = 1.20 × 10⁴ M⁻¹ cm⁻¹ (hypothetical data). In a 1.00 cm cuvette its solution gives A = 0.360. What is the concentration?

A calibration curve is plotted with absorbance A on the vertical axis and concentration c on the horizontal axis. What does the slope equal?

A spectrophotometer is used to find the concentration of a colored solution. Which piece of information is least useful?

Which reaction is best suited to being followed with a visible-light spectrophotometer?

A student rinses a cuvette with distilled water and then fills it with the sample without rinsing it with the sample first. How is the calculated concentration affected, and why?

An unknown gives A = 1.90, outside the calibration range (the most concentrated standard has A = 0.80). What is the best thing to do?

AP-style free response

A spectrophotometer is used to find the concentration of a blue dye in a sports drink. Standard solutions are measured at λmax in a 1.00 cm cuvette (hypothetical data):

c (10⁻⁶ M)	2.00	4.00	6.00	8.00
A	0.180	0.362	0.538	0.720

(a) Use the slope of the calibration curve to find the molar absorptivity of the dye. Include units.

(b) A sample of the drink has an absorbance of 0.450. Find the concentration of the dye in the drink.

(c) A can holds 355 mL of the drink. Find the amount of dye, in moles, in one can.

(d) When the drink sample was measured, there were fingerprints on the clear faces of the cuvette. Is the calculated concentration too high or too low? Justify your answer.

(e) Solutions of the dye are blue. In which color range is λmax most likely to be? Why is the measurement made at λmax?

Model answer and scoring

(a) Slope = (0.720 − 0.180) ÷ (8.00 × 10⁻⁶ − 2.00 × 10⁻⁶) M = 9.00 × 10⁴ M⁻¹ (a best-fit line giving 8.9 × 10⁴ to 9.1 × 10⁴ is also fine). ε = slope ÷ b = 9.00 × 10⁴ M⁻¹ cm⁻¹.

(b) c = A ÷ slope = 0.450 ÷ (9.00 × 10⁴ M⁻¹) = 5.00 × 10⁻⁶ M.

(c) n = cV = (5.00 × 10⁻⁶ mol/L)(0.355 L) = 1.78 × 10⁻⁶ mol.

(d) Too high. The fingerprints absorb and scatter some of the light, so less reaches the detector; the measured A is too high, and c = A ÷ slope is too high.

(e) In the orange range. The blue color we see is the light not absorbed; the light absorbed is mainly its complement, orange. At λmax, ε is largest, so a given change in concentration gives the largest change in absorbance and the most sensitive measurement.

Scoring: (a) 9.00 × 10⁴ M⁻¹ cm⁻¹ with units; (b) 5.00 × 10⁻⁶ M; (c) 1.78 × 10⁻⁶ mol; (d) too high, because A is too high; (e) orange, with ε largest giving the most sensitive measurement.

Summary

Absorbance A = log(I₀/I); the Beer–Lambert law, A = εbc, makes A proportional to c at a fixed wavelength and path length.
Zero the instrument with a blank and measure at λmax; standards and samples use the same wavelength and path length.
Calibration curve: plot A against c; the slope is εb. Find the unknown’s c from its A, then moles from n = cV.
For errors, first decide whether A is too high or too low: droplets and moving off λmax make it too low; fingerprints, bubbles, skipping the blank, and a longer path make it too high; the volume of solution has no effect. Dilute quantitatively when A is too high.

AP tip: For an error-analysis question, write both steps: how the mistake changes the measured absorbance, then how A = εbc (or c = A ÷ slope) makes the calculated concentration too high or too low. “The volume of solution” or “how full the cuvette is” is usually a distractor.

Beer-Lambert LawBeer-Lambert 定律

Learning goals

Before you start

Measuring concentration with light

The Beer–Lambert law

Good technique

Calibration curves

Lab errors: first A, then c

Example 1: From calibration curve to moles

Example 2: When the absorbance is too high

Example 3: Choosing a wavelength when absorptions overlap

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem