Result

The Beer-Lambert law

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Statement

Let a collimated, monochromatic beam of irradiance \(I_0\) pass through a homogeneous, non-scattering, non-luminescent sample of uniform thickness \(l\) containing a single absorbing species at molar concentration \(c\), whose molecules attenuate the beam independently of one another. Then the transmitted irradiance falls exponentially with \(cl\), so that the transmittance \(T=I/I_0\) and the decadic absorbance \(A=-\log_{10}T\) satisfy \(A=\varepsilon c l\), where the molar absorptivity \(\varepsilon\) depends on the species and the wavelength but not on \(c\) or \(l\). For several independent absorbers the absorbances add: \(A=l\sum_i \varepsilon_i c_i\).

Why it matters

Kinetics needs a concentration read out repeatedly, quickly, and without disturbing the reaction mixture. Titrating an aliquot destroys the sample and takes minutes; an absorbance reading takes milliseconds, is non-invasive, and can be logged continuously. The Beer-Lambert law is what licenses reading a detector voltage as a concentration, and it is therefore the measurement layer underneath almost every experimental rate constant in this unit: the exponential decays fitted in integrated rate laws, the initial-rate and isolation experiments that establish orders in the rate law and reaction order, and the temperature series that yields \(E_a\) in the Arrhenius equation are, in the laboratory, almost always absorbance-time traces.

The law also earns its place because of the specific way it fails. Its derivation makes exactly one physical demand — that absorbers act independently — and one instrumental demand — that the light really is monochromatic and really is all the light reaching the detector. A calibration line that bends is therefore not noise: it is evidence about aggregation, about a concentration-dependent equilibrium, or about the spectrometer. Knowing which of the two demands has broken is what separates a usable kinetic run from a discarded one, and the same reasoning supports the calibration-curve and standard-addition protocols of calibration and standard addition.

Hypotheses

The illuminating light is monochromatic, or its band is narrow enough that \(\varepsilon\) is effectively constant across it.Because \(\varepsilon\) varies with wavelength — that variation is precisely what an absorption spectrum plots — a finite band makes the detector average transmittances, not absorbances. Step 8 shows this always biases the measured absorbance downwards, so a wide slit on a sharp band curves the calibration line even for a perfectly ideal solution.

Absorbers act independently: each molecule presents the same cross-section \(\sigma\) whatever its neighbours are doing.Dropping it removes the linearity in \(c\), not merely its accuracy. If the absorber dimerises, the dimer is a different chromophore with its own \(\varepsilon\), and the monomer fraction itself depends on \(c\); the apparent \(\varepsilon\) then drifts with concentration.

Absorption is the only process removing light from the beam, and the sample is optically homogeneous over the beam cross-section.Scattering by turbid samples, fluorescence re-emitted into the detector's acceptance cone, and photochemical bleaching during the measurement each break the bookkeeping of Step 3. A suspension of cells or crystallites also violates homogeneity: the absorber is concentrated in particles, some photons miss them entirely, and the measured absorbance flattens below \(\varepsilon c l\).

The detector sees only light that has traversed the full path \(l\), and responds linearly to irradiance.Stray light of relative magnitude \(s\) that bypasses the sample imposes a ceiling \(A_{\max}=\log_{10}\!\left((1+s)/s\right)\) on any measurable absorbance, no matter how concentrated the sample; this is an instrument property, not a chemical one, and it is why absorbances above about \(2\) are distrusted on a routine spectrometer.

Proof

The argument is a photon-accounting one: work out the probability that a photon survives a slab of thickness \(\mathrm{d}x\), then compound those probabilities along the path. Nothing quantum-mechanical is needed beyond the existence of a fixed absorption cross-section per molecule; the base-\(10\) logarithm enters only at the very end, as a unit convention.

\[ T \equiv \frac{I}{I_0}, \qquad A \equiv -\log_{10} T = \log_{10}\frac{I_0}{I} \]

Definitions. \(I\) is the irradiance (radiant power per unit area, which at a single wavelength is proportional to the photon flux per unit area per unit time used below) leaving the sample, \(I_0\) that entering it, both already corrected against a blank cell of pure solvent so that reflection at the windows and solvent absorption cancel. Absorbance is dimensionless; transmittance is often quoted as a percentage. A

\[ \text{fraction removed in }\mathrm{d}x \;=\; \sigma N\,\mathrm{d}x, \qquad N = \text{absorbers per unit volume} \]

Assign every absorbing molecule an absorption cross-section \(\sigma\) at the working wavelength: the effective target area it presents to a photon. In a slab of thickness \(\mathrm{d}x\) and unit face area there are \(N\,\mathrm{d}x\) molecules, blocking a total area \(\sigma N\,\mathrm{d}x\). Two molecules can only shadow one another over an area of order \((\sigma N\,\mathrm{d}x)^2\), which vanishes faster than \(\mathrm{d}x\), so in the differential limit the independence hypothesis is exact rather than approximate. B

\[ \mathrm{d}I = -\,\sigma N I\,\mathrm{d}x \]

The photons removed from the beam in the slab are the incident ones times the blocked fraction. The minus sign records that absorbed photons never return: no scattering back into the beam, no fluorescence emitted along it, no stimulated emission from a populated excited state — the third hypothesis. A

\[ \int_{I_0}^{I}\frac{\mathrm{d}I}{I} = -\sigma N\!\int_0^l \mathrm{d}x \quad\Longrightarrow\quad \ln\frac{I_0}{I} = \sigma N l \]

Separate variables and integrate along the path. Taking \(\sigma\) and \(N\) outside the integral is where uniformity is used: \(N\) must not vary along \(x\), which is why a settling suspension or an unstirred cell with a concentration gradient does not obey the law even when everything else is ideal. A

\[ N = \frac{N_{\mathrm{A}}\,c}{10^{3}\,\text{cm}^{3}\,\text{L}^{-1}} \qquad (c\ \text{in mol L}^{-1},\ N\ \text{in cm}^{-3}) \]

Convert number density to the chemist's variable using the Avogadro constant \(N_{\mathrm{A}}=6.022\times10^{23}\,\text{mol}^{-1}\); the factor \(10^{3}\) is only the litre-to-cubic-centimetre conversion, carried explicitly because it survives into the definition of \(\varepsilon\). A

\[ A = \frac{1}{\ln 10}\,\ln\frac{I_0}{I} = \underbrace{\frac{\sigma N_{\mathrm{A}}}{10^{3}\ln 10}}_{\textstyle \varepsilon}\; c\, l \;=\; \varepsilon c l \]

Rewriting the natural logarithm of Step 4 as a base-\(10\) one divides by \(\ln 10 = 2.303\); everything constant is then collected into a single symbol \(\varepsilon\), the molar absorptivity, with units \(\text{L mol}^{-1}\text{cm}^{-1}\) when \(c\) is in \(\text{mol L}^{-1}\) and \(l\) in \(\text{cm}\). Note what \(\varepsilon\) is not: it is not a fundamental constant but a repackaging of the molecular cross-section, \(\sigma = (10^{3}\ln 10/N_{\mathrm{A}})\,\varepsilon = 3.82\times10^{-21}\,\varepsilon\ \text{cm}^2\). B

\[ \sigma N \longrightarrow \sum_i \sigma_i N_i \quad\Longrightarrow\quad A = l\sum_i \varepsilon_i c_i \]

If several non-interacting species absorb at the same wavelength, the blocked area in Step 2 is the sum of their contributions, and every later step goes through unchanged. Absorbances are therefore additive while transmittances multiply — the basis of multi-component analysis, and the reason two cells in series behave as one cell of the summed absorbance. A

\[ T_{\text{meas}} = \int w(\lambda)\,10^{-\varepsilon(\lambda)cl}\,\mathrm{d}\lambda \;\ge\; 10^{-\left(\int w(\lambda)\varepsilon(\lambda)\,\mathrm{d}\lambda\right) cl} \quad\Longrightarrow\quad A_{\text{meas}} \le \langle\varepsilon\rangle c l \]

Real light has a finite bandwidth with normalised weight \(w(\lambda)\). The detector integrates transmittances, so the measured absorbance is the logarithm of a mean, not the mean of logarithms. Since \(x\mapsto 10^{-x}\) is convex, Jensen's inequality makes the measured absorbance a strict underestimate whenever \(\varepsilon\) varies across the band, with equality only for genuinely constant \(\varepsilon\). The deviation is therefore always negative and always grows with \(cl\) — a theorem, not an empirical tendency. C

Result

\[ T=\frac{I}{I_0}=10^{-\varepsilon c l}, \qquad A=-\log_{10}T=\varepsilon c l, \qquad \varepsilon=\frac{\sigma N_{\mathrm{A}}}{10^{3}\ln 10} \]

Reading. Transmittance decays exponentially in concentration and path length; absorbance, its negative base-\(10\) logarithm, is linear in both and additive over independent absorbers. The proportionality constant \(\varepsilon\) is a molecular cross-section in disguise, wavelength-specific and independent of \(c\) and \(l\).

Scope. Holds for a homogeneous, non-scattering, non-emitting sample of independent absorbers in monochromatic light, measured against a matched blank on a linear detector with negligible stray light. Every departure from linearity traces back to one of those clauses, and Step 8 guarantees that the finite-bandwidth departure is always downwards.

Corollaries & converses

Concentration and path length are interchangeable. Only the product \(cl\) appears, so halving the concentration and doubling the cuvette length leaves \(A\) unchanged; a \(10\,\text{cm}\) cell recovers a trace analyte that is invisible at \(1\,\text{cm}\).
Optimal working absorbance. If the photometric uncertainty \(\Delta T\) is constant, then \(\Delta c/c = \Delta T/\left(T\ln(1/T)\right)\), which is minimised at \(T=1/e=0.368\), i.e. \(A=\log_{10}e=0.4343\). This is why the routine working window is quoted as roughly \(0.1\le A\le 1.0\): relative precision degrades sharply outside it, at low \(A\) because the signal is small and at high \(A\) because \(T\) is.
Isosbestic points. For a clean conversion \(\mathrm{A}\to\mathrm{B}\) with \(c_{\mathrm{A}}+c_{\mathrm{B}}=c_0\) fixed, additivity gives \(A(\lambda)=l\left[\varepsilon_{\mathrm{B}}c_0+(\varepsilon_{\mathrm{A}}-\varepsilon_{\mathrm{B}})c_{\mathrm{A}}\right]\), so at any wavelength where \(\varepsilon_{\mathrm{A}}=\varepsilon_{\mathrm{B}}\) the absorbance is constant in time. A sharp isosbestic point in a kinetic run is strong evidence that exactly two absorbing species are interconverting with no accumulating intermediate — a direct experimental test of the mechanisms treated under the steady-state approximation.
Rate constants do not need \(\varepsilon\). Since \(c\propto A-A_\infty\) with the same constant of proportionality at all times, a first-order fit of \(\ln(A_t-A_\infty)\) against \(t\) returns \(k\) with \(\varepsilon\) and \(l\) cancelling. Only rate constants of order other than one, and absolute concentrations, require \(\varepsilon\) to be known.
Converse. Linearity of \(A\) in \(c\) over a range is evidence for, not proof of, ideality: a monomer-dimer system with \(\varepsilon_{\text{dimer}}=2\varepsilon_{\text{monomer}}\) at the monitored wavelength is perfectly linear while being chemically far from a single independent species.

Fails without

Monochromatic light dropped (finite slit width): with two equally weighted wavelengths at which \(\varepsilon=2.00\times10^{4}\) and \(1.00\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\), a \(5.00\times10^{-5}\,\text{mol L}^{-1}\) solution in a \(1.00\,\text{cm}\) cell reads \(A=0.682\) instead of the mean-\(\varepsilon\) value \(0.750\), and the apparent \(\varepsilon\) drifts from \(1.50\times10^{4}\) at infinite dilution to \(1.36\times10^{4}\) here. Worked Example 2 carries this through.
Independence dropped (aggregation or a concentration-dependent equilibrium): a dye that dimerises, \(2\mathrm{M}\rightleftharpoons \mathrm{D}\), has a monomer fraction that itself falls as \(c\) rises, so the apparent \(\varepsilon\) at the monomer band decreases with concentration and the calibration line bends towards the axis. Methylene blue is the standard classroom case; the diagnostic is that the shape of the spectrum, not just its size, changes with dilution.
Stray light (detector linearity dropped): a relative stray-light level \(s=1.0\times10^{-3}\) reads \(A=1.959\) for a true \(A=2.000\), and can never report more than \(A_{\max}=\log_{10}(1.001/0.001)=3.00\) however black the sample. The measured curve saturates at that ceiling, which is easily mistaken for chemical saturation.
Homogeneity dropped (particulate samples): when the absorber is packaged into cells, chloroplasts or crystallites rather than dissolved, photons that miss a particle escape unattenuated while those that hit one are absorbed almost completely. The measured absorbance flattens below \(\varepsilon c l\) — the sieve effect — and the apparent \(\varepsilon\) depends on particle size, not only on the chromophore.
Non-absorptive losses (scattering or fluorescence): turbidity adds an apparent absorbance that rises steeply toward the blue and is not proportional to analyte concentration at all, while a fluorescent analyte returns photons into the detector's cone and biases \(A\) low. Both survive blank subtraction, because neither is present in the blank.

Common errors

“Transmittance is proportional to concentration.” Transmittance is exponential in \(c\); it is absorbance that is linear. Averaging replicate \(T\) values and then taking a logarithm is not the same as averaging their absorbances, and the difference is systematic, by Step 8's convexity.
“\(\varepsilon\) is a constant of the substance.” It is a constant of the substance at one wavelength, in one solvent, at one temperature and one protonation state. Applying an \(\varepsilon\) quoted at \(\lambda_{\max}\) to an absorbance read on the side of the band is one of the commonest sources of a wrong concentration.
Mixing napierian and decadic quantities. Literature values expressed as a cross-section \(\sigma\), or as a napierian coefficient, differ from \(\varepsilon\) by the factor \(\ln 10=2.303\) (and by the litre-to-\(\text{cm}^3\) factor). Dropping it is a \(130\) per cent error.
Working outside \(0.1\le A\le 1.0\) without dilution. At \(A=2.5\) fewer than \(1\) photon in \(300\) reaches the detector, stray light dominates, and the reading is biased low; the corollary above quantifies the precision loss on both sides of the window.
Using \(I_0\) from an empty beam rather than a blank cuvette. Window reflections and solvent absorption then appear as analyte, giving a non-zero intercept in the calibration line that is often wrongly attributed to contamination.
Reading a kinetic trace as \(\ln A_t\) instead of \(\ln(A_t-A_\infty)\). Whenever the product or the solvent absorbs at the monitoring wavelength, \(A_\infty\ne 0\) and the uncorrected plot curves, faking a change of order.

Discussion

The three names attached to this law record three separate discoveries. Pierre Bouguer, in his 1729 Essai d'optique sur la gradation de la lumière, established that equal thicknesses of an absorbing medium remove equal fractions of light — the geometric progression that is the content of Step 4. Johann Heinrich Lambert restated and popularised the exponential form in his Photometria of 1760. August Beer's contribution in 1852 was the second variable: that concentration enters the exponent in exactly the same way as path length, which is precisely the statement that dissolved absorbers act independently. It is worth noticing that the modern derivation runs in the opposite direction to the history — independence is now the assumption and the exponential is the conclusion.

Modern terminology has largely retired the older names. IUPAC deprecates “extinction coefficient” for \(\varepsilon\), because extinction and attenuation properly name the total loss from the beam — absorption and scattering together — whereas \(\varepsilon\) as used here counts absorption alone; the recommended name is molar absorption coefficient, with molar attenuation coefficient reserved for the case where scattering genuinely contributes. That is exactly the distinction the turbidity failure mode above makes expensive to blur. Absorbance without qualification means the decadic quantity \(\log_{10}(I_0/I)\); the napierian absorbance \(\ln(I_0/I)\) appears in atmospheric and laser physics, where cross-sections rather than molar quantities are natural.

The cross-section \(\sigma\) is where this law meets quantum mechanics. For an electronic transition, \(\sigma\) at the band maximum is fixed by the square of the transition dipole moment between the two states, and the band area \(\int\varepsilon(\tilde\nu)\,\mathrm{d}\tilde\nu\) is proportional to the Einstein coefficient for stimulated absorption and hence to the oscillator strength. This has a practical consequence for kinetics: the integrated area is the robust quantity, while the peak height depends on how the oscillator strength is spread over vibrational structure, so a solvent change that broadens a band lowers \(\varepsilon_{\max}\) without any change in the underlying transition probability. It is also the reason a strongly allowed transition tops out near \(\varepsilon\sim10^{5}\,\text{L mol}^{-1}\text{cm}^{-1}\): by Step 6 that corresponds to \(\sigma\approx 4\times10^{-16}\,\text{cm}^2\), a few square ångströms, which is within an order of magnitude of the projected area of the conjugated system itself. An effective absorption cross-section cannot indefinitely exceed the molecule presenting it.

For kinetic work the practical upshot is that the spectrometer measures \(cl\varepsilon\) and nothing else, so the experimental design question is always which wavelength maximises the change in absorbance over the run, not which maximises absorbance itself. Monitoring at \(\lambda_{\max}\) of the reactant is standard, but monitoring at an isosbestic point deliberately — where absorbance is invariant — is the standard check that the run is a clean two-species conversion before any rate law is fitted. Where \(\varepsilon\) is unknown or the matrix interferes, the calibration and standard-addition procedures of calibration and standard addition replace a single \(\varepsilon\) with a fitted slope, and the propagation of the resulting uncertainty into \(k\) follows error propagation.

Common misconceptions. That the law is exact and its deviations are experimental sloppiness — Step 8 shows one deviation to be mathematically inevitable for any real slit width. That a bent calibration curve invalidates the law — it localises which hypothesis has failed, and the direction of the bend discriminates between them: bandwidth and stray light bend it downwards at high \(A\), whereas a dimerisation with a more strongly absorbing dimer can bend it upwards. And that absorbance is a property of a sample — it is a property of a sample and a cuvette, since \(l\) is in the product; concentration is the sample property.

Worked examples

Example 1. A dye is hydrolysed by hydroxide held in large excess, so the reaction is pseudo-first-order in the dye. Absorbance is logged at the dye's \(\lambda_{\max}\) in a \(1.00\,\text{cm}\) cell: \(A_0=0.842\) at \(t=0\), \(A=0.418\) at \(t=120\,\text{s}\), and \(A_\infty=0.020\) after ten half-lives. The molar absorptivity at that wavelength is \(\varepsilon=8.5\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\). Find the pseudo-first-order rate constant, the half-life, the initial dye concentration and the initial rate.

\[ A_t - A_\infty = \varepsilon l\,c_t \quad\Longrightarrow\quad \frac{c_t}{c_0} = \frac{A_t-A_\infty}{A_0-A_\infty} \]

The residual absorbance \(A_\infty\) is contributed by the product and the solvent, which do not decay; subtracting it isolates the dye's contribution, and additivity (Step 7) guarantees the subtraction is legitimate. The ratio is what the rate law needs, and \(\varepsilon\) and \(l\) cancel from it. A

\[ c_t=c_0e^{-kt} \quad\Longrightarrow\quad k=\frac{1}{t}\ln\!\frac{A_0-A_\infty}{A_t-A_\infty} \]

Insert the first-order integrated rate law and solve for \(k\) symbolically before any number is substituted, so that the whole calculation reduces to one logarithm of a ratio of absorbance differences. A

\[ k=\frac{1}{120\,\text{s}}\ln\frac{0.842-0.020}{0.418-0.020}=\frac{\ln(0.822/0.398)}{120\,\text{s}}=\frac{0.7254}{120\,\text{s}}=6.04\times10^{-3}\,\text{s}^{-1} \]

Substitute. The ratio \(0.822/0.398=2.065\) is a little above \(2\), so \(120\,\text{s}\) is a little more than one half-life — a useful order-of-magnitude check before trusting the arithmetic. A

\[ t_{1/2}=\frac{\ln 2}{k}=\frac{0.6931}{6.04\times10^{-3}\,\text{s}^{-1}}=115\,\text{s} \]

The first-order half-life is concentration-independent, so it can be quoted without knowing \(c_0\) — and it confirms the reading of Step 3. A

\[ c_0=\frac{A_0-A_\infty}{\varepsilon l}=\frac{0.822}{8.5\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\times1.00\,\text{cm}}=9.67\times10^{-6}\,\text{mol L}^{-1} \]

Only now is \(\varepsilon\) needed: converting a relative measurement into an absolute concentration is the one operation that cannot avoid it. The initial absorbance sits inside the \(0.1\)-\(1.0\) window, so the reading is in the precise regime. A

\[ \text{rate}_0 = k c_0 = 6.04\times10^{-3}\,\text{s}^{-1}\times 9.67\times10^{-6}\,\text{mol L}^{-1}=5.85\times10^{-8}\,\text{mol L}^{-1}\text{s}^{-1} \]

The initial rate follows from the pseudo-first-order rate law. Equivalently \(\text{rate}=-(\varepsilon l)^{-1}\mathrm{d}A/\mathrm{d}t\): the spectrometer measures the slope of an absorbance trace, and \(\varepsilon l\) is the only thing needed to convert it. A

\[ k=6.04\times10^{-3}\,\text{s}^{-1},\quad t_{1/2}=115\,\text{s},\quad c_0=9.67\times10^{-6}\,\text{mol L}^{-1},\quad \text{rate}_0=5.85\times10^{-8}\,\text{mol L}^{-1}\text{s}^{-1} \]

Reading. Absorbance differences alone fix \(k\) and \(t_{1/2}\); \(\varepsilon\) is required only for the absolute concentration and hence the absolute rate. This is why a spectrophotometric rate constant is often far more reliable than the concentration it was derived from.

Scope. Valid because the excess hydroxide makes the decay exponential and because \(A_0\) lies in the linear window; had \(A_0\) been near \(2.5\), the early points would have been biased low by stray light and the fitted \(k\) too small.

Example 2. A monochromator passes two wavelengths of equal intensity at which the analyte has \(\varepsilon_1=2.00\times10^{4}\) and \(\varepsilon_2=1.00\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\). In a \(1.00\,\text{cm}\) cell, what absorbance is measured for \(c=5.00\times10^{-5}\,\text{mol L}^{-1}\), and for half that concentration? Is the calibration line straight?

\[ T_{\text{meas}}=\tfrac12\left(10^{-\varepsilon_1cl}+10^{-\varepsilon_2cl}\right), \qquad A_{\text{meas}}=-\log_{10}T_{\text{meas}} \]

The detector sums photons, so it is transmittances — not absorbances — that are averaged over the band (Step 8 with a two-point weight \(w\)). Setting this up correctly is the whole content of the example. B

\[ A_1=\varepsilon_1cl=1.000,\qquad A_2=\varepsilon_2cl=0.500 \]

Evaluate the two monochromatic absorbances the sample would show if each wavelength were used alone: \(2.00\times10^{4}\times5.00\times10^{-5}\times1.00\) and half of it. A

\[ T_{\text{meas}}=\tfrac12\left(10^{-1.000}+10^{-0.500}\right)=\tfrac12(0.1000+0.3162)=0.2081 \]

Substitute. The weaker-absorbing wavelength dominates the transmitted light — \(0.3162\) against \(0.1000\) — which is the mechanism of the bias: the beam that reaches the detector is progressively enriched in the wavelengths the sample absorbs least. B

\[ A_{\text{meas}}=-\log_{10}(0.2081)=0.682 \quad \text{versus} \quad \langle\varepsilon\rangle cl = 1.50\times10^{4}\times5.00\times10^{-5}\times1.00=0.750 \]

The measured value falls \(0.068\) below the mean-\(\varepsilon\) prediction, a \(9.1\) per cent negative deviation, exactly as the convexity inequality of Step 8 requires. A

\[ c'=2.50\times10^{-5}: \quad T'=\tfrac12(10^{-0.500}+10^{-0.250})=\tfrac12(0.3162+0.5623)=0.4393,\quad A'=0.357 \]

Repeat at half the concentration. Apparent absorptivities: \(A_{\text{meas}}/(cl)=1.36\times10^{4}\) and \(A'/(c'l)=1.43\times10^{4}\), against the infinite-dilution limit \(1.50\times10^{4}\). A

\[ \frac{A_{\text{meas}}}{A'}=\frac{0.682}{0.357}=1.908 \ne 2 \]

Doubling the concentration multiplies the absorbance by \(1.908\), not \(2\): the calibration line is genuinely curved, concave towards the concentration axis, and the curvature worsens as \(cl\) grows. A

\[ A_{\text{meas}}=0.682\ \ (\text{ideal }0.750), \qquad A'_{\text{meas}}=0.357, \qquad A_{\text{meas}}/A'_{\text{meas}}=1.908 \]

Reading. A two-fold spread in \(\varepsilon\) across the passband costs nearly ten per cent at \(A\approx0.7\) and destroys linearity, with no chemical fault whatsoever in the sample.

Scope. The fix is instrumental: narrow the slit, or move to a flatter part of the spectrum. Measuring at \(\lambda_{\max}\) is preferred partly for signal but mainly because \(\mathrm{d}\varepsilon/\mathrm{d}\lambda=0\) there, so the spread of \(\varepsilon\) across the passband is second order in the bandwidth rather than first.

Problems

A solution transmits \(27.5\) per cent of the incident light in a \(1.00\,\text{cm}\) cell at a wavelength where \(\varepsilon=1.20\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\). Find the absorbance and the concentration.
Solution
\(T=0.275\), so \(A=-\log_{10}(0.275)=0.561\). Rearranging the Result, \(c=A/(\varepsilon l)=0.561/(1.20\times10^{4}\times1.00)=4.67\times10^{-5}\,\text{mol L}^{-1}\). The absorbance sits inside the \(0.1\)-\(1.0\) precision window, so no dilution is needed. Note that the analogous calculation with \(T\) treated as linear in \(c\) would give a wholly different answer — the logarithm is not optional.
A mixture of X and Y is measured in a \(1.00\,\text{cm}\) cell at two wavelengths, giving \(A_1=0.780\) and \(A_2=0.510\). The molar absorptivities, in \(\text{L mol}^{-1}\text{cm}^{-1}\), are \(\varepsilon_{X1}=1.20\times10^{4}\), \(\varepsilon_{Y1}=2.00\times10^{3}\), \(\varepsilon_{X2}=1.50\times10^{3}\), \(\varepsilon_{Y2}=8.00\times10^{3}\). Find both concentrations.
Solution
Additivity (Step 7) gives two simultaneous equations, \(A_1=l(\varepsilon_{X1}c_X+\varepsilon_{Y1}c_Y)\) and \(A_2=l(\varepsilon_{X2}c_X+\varepsilon_{Y2}c_Y)\). With \(l=1.00\,\text{cm}\), Cramer's rule gives determinant \(D=\varepsilon_{X1}\varepsilon_{Y2}-\varepsilon_{Y1}\varepsilon_{X2}=1.20\times10^{4}\times8.00\times10^{3}-2.00\times10^{3}\times1.50\times10^{3}=9.60\times10^{7}-3.00\times10^{6}=9.30\times10^{7}\).
\(c_X=(A_1\varepsilon_{Y2}-A_2\varepsilon_{Y1})/D=(0.780\times8.00\times10^{3}-0.510\times2.00\times10^{3})/9.30\times10^{7}=(6240-1020)/9.30\times10^{7}=5.61\times10^{-5}\,\text{mol L}^{-1}\).
\(c_Y=(A_2\varepsilon_{X1}-A_1\varepsilon_{X2})/D=(0.510\times1.20\times10^{4}-0.780\times1.50\times10^{3})/9.30\times10^{7}=(6120-1170)/9.30\times10^{7}=5.32\times10^{-5}\,\text{mol L}^{-1}\).
Check: \(A_1=1.20\times10^{4}\times5.61\times10^{-5}+2.00\times10^{3}\times5.32\times10^{-5}=0.673+0.106=0.779\approx0.780\). The determinant is small compared with \(\varepsilon_{X1}\varepsilon_{Y2}\) alone only if the two wavelengths discriminate poorly between X and Y; a near-singular \(D\) is what makes multi-component analysis ill-conditioned.
A chromophore has \(\varepsilon=4.6\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\) at its band maximum. Find its absorption cross-section in \(\text{cm}^2\) and in square ångströms, and comment.
Solution
From Step 6, \(\sigma=\varepsilon\,10^{3}\ln 10/N_{\mathrm{A}}\). Numerically \(10^{3}\times2.303/(6.022\times10^{23})=3.82\times10^{-21}\,\text{cm}^2\) per unit of \(\varepsilon\), so \(\sigma=3.82\times10^{-21}\times4.6\times10^{4}=1.76\times10^{-16}\,\text{cm}^2\). Since \(1\,\text{Å}^2=10^{-16}\,\text{cm}^2\), this is \(1.8\,\text{Å}^2\).
Comment: \(\sigma\) is an effective target area set by the transition dipole, not a geometric one, and here it is roughly a tenth of the projected area of a benzene ring (of order \(30\,\text{Å}^2\)). So even an intensely allowed transition intercepts only a modest fraction of the photons that geometrically strike the molecule — but the two areas are within an order of magnitude of each other, and that is what bounds \(\varepsilon\) in practice: \(\varepsilon\sim10^{5}\) already implies \(\sigma\approx3.8\,\text{Å}^2\), and values far beyond it would demand a cross-section exceeding the chromophore itself. Single chromophores are not observed above about \(10^{5}\,\text{L mol}^{-1}\text{cm}^{-1}\).
A reaction is followed at a wavelength where only the reactant absorbs, with \(l=1.00\,\text{cm}\) and \(\varepsilon=1.10\times10^{4}\,\text{L mol}^{-1}\text{cm}^{-1}\). The absorbances are \(0.900,\ 0.546,\ 0.331,\ 0.201\) at \(t=0,\ 100,\ 200,\ 300\,\text{s}\). Establish the order, find the rate constant and the initial concentration, and state which of those two answers would change if \(\varepsilon\) were wrong by ten per cent.
Solution
Since only the reactant absorbs, \(A_\infty=0\) and \(A\propto c\). Test first order by taking logarithms: \(\ln A = -0.1054,\ -0.6051,\ -1.1056,\ -1.6045\). The successive differences are \(-0.4998,\ -0.5005,\ -0.4988\) over equal \(100\,\text{s}\) intervals — constant to within rounding, so the decay is exponential and the reaction is first order in the absorbing reactant.
\(k=0.500/100\,\text{s}=5.00\times10^{-3}\,\text{s}^{-1}\), and \(t_{1/2}=\ln 2/k=139\,\text{s}\). The initial concentration is \(c_0=A_0/(\varepsilon l)=0.900/(1.10\times10^{4}\times1.00)=8.18\times10^{-5}\,\text{mol L}^{-1}\).
A ten per cent error in \(\varepsilon\) changes \(c_0\) by ten per cent but leaves \(k\) untouched, because \(k\) is extracted from ratios of absorbances in which \(\varepsilon l\) cancels. Had the reaction been second order, \(1/A\) rather than \(\ln A\) would be linear in \(t\), and the fitted second-order rate constant \(k_2=\text{slope}\times\varepsilon l\) would then inherit the full ten per cent error.
A spectrometer passes a fraction \(s\) of \(I_0\) to the detector as stray light that never traverses the sample, so that \(T_{\text{meas}}=(10^{-A}+s)/(1+s)\) where \(A\) is the true absorbance. For \(s=1.0\times10^{-3}\), find the measured absorbance when \(A=1.000\) and when \(A=2.000\), and the largest absorbance the instrument can ever report. Comment on the direction of the resulting calibration error.
Solution
At \(A=1.000\): \(T_{\text{meas}}=(0.1000+0.0010)/1.0010=0.10090\), so \(A_{\text{meas}}=-\log_{10}(0.10090)=0.9961\) — low by \(0.004\), about \(0.4\) per cent.
At \(A=2.000\): \(T_{\text{meas}}=(0.01000+0.00100)/1.0010=0.010989\), so \(A_{\text{meas}}=1.959\) — low by \(0.041\), about \(2.1\) per cent.
As \(A\to\infty\), \(10^{-A}\to0\) and \(T_{\text{meas}}\to s/(1+s)=9.99\times10^{-4}\), giving \(A_{\max}=\log_{10}(1.001/0.001)=3.00\). No sample, however opaque, can read above \(3.00\) on this instrument.
The error is always negative and grows rapidly with \(A\), so the calibration line bends downward at high concentration and eventually saturates — the same qualitative signature as chemical aggregation. The two are distinguished by their behaviour on dilution into a longer cell: stray light is a property of the instrument at a given \(A\), so a sample rediluted tenfold and measured in a \(10\,\text{cm}\) cell (identical \(cl\), identical \(A\)) reproduces the same bias, whereas aggregation depends on \(c\) alone and disappears on dilution.