If a 0.1 M CuSO₄ solution has A = 0.5 at 635 nm in a 1 cm cuvette, what is ε?

A = εbc → ε = A/(bc) = 0.5/(1 × 0.1) = 5 L/(mol·cm).

If you double the concentration, what happens to the absorbance?

A = εbc is linear: doubling c doubles A (within Beer's Law range).

An unknown solution has A = 0.35 at 520 nm. From the calibration curve, ε = 3.5 L/(mol·cm) at this wavelength with b = 1 cm. Find the concentration.

C = A/(εb) = 0.35/(3.5 × 1) = 0.1 mol/L.

Pro 🔒~18 min

Beer's Law Lab

Absorbance, transmittance, and spectrophotometry

How it works

Beer-Lambert Law states that absorbance is directly proportional to both the concentration of the absorbing species and the path length of light through the solution. A = εbc, where ε is the molar absorptivity (a constant for each substance at a given wavelength), b is the path length in cm, and c is the molar concentration. Transmittance T = I/I₀ is the fraction of light that passes through. A = -log₁₀(T). A calibration curve (A vs c) at fixed wavelength and path length gives a straight line through the origin. The slope is εb. To find an unknown concentration, measure its absorbance and read from the calibration curve. Deviations from Beer's Law occur at high concentrations (>0.01 M for most species) due to intermolecular interactions.

Upgrade to Pro to access this experiment

Step-by-step

Select a solution type and adjust concentration with the slider.
Observe the solution color change in the cuvette — darker = more concentrated.
The spectrophotometer beam passes through and measures absorbance.
Try different path lengths and wavelengths.
The graph plots absorbance vs concentration in real time.

Key formulas

$A = \varepsilon \cdot b \cdot c$ Beer-Lambert Law: A = absorbance, ε = molar absorptivity (L/(mol·cm)), b = path length (cm), c = concentration (mol/L)
$A = -\log_{10}(T) = -\log_{10}\left(\frac{I}{I_0}\right)$ Absorbance from transmittance: T = I/I₀ = fraction of light transmitted

Frequently asked questions

If a 0.1 M CuSO₄ solution has A = 0.5 at 635 nm in a 1 cm cuvette, what is ε?: A = εbc → ε = A/(bc) = 0.5/(1 × 0.1) = 5 L/(mol·cm).
If you double the concentration, what happens to the absorbance?: A = εbc is linear: doubling c doubles A (within Beer's Law range).
An unknown solution has A = 0.35 at 520 nm. From the calibration curve, ε = 3.5 L/(mol·cm) at this wavelength with b = 1 cm. Find the concentration.: C = A/(εb) = 0.35/(3.5 × 1) = 0.1 mol/L.