AP Chemistry 3.13 Beer-Lambert Law Study Notes - New Syllabus Effective fall 2024
AP Chemistry 3.13 Beer-Lambert Law Study Notes- New syllabus
AP Chemistry 3.13 Beer-Lambert Law Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
Explain the amount of light absorbed by a solution of molecules or ions in relationship to the concentration, path length, and molar absorptivity.
Key Concepts:
- Beer-Lambert Law
- Absorbance and Its Dependence on Concentration
The Beer–Lambert Law
The Beer–Lambert Law describes the quantitative relationship between the amount of light absorbed by a solution and three variables: the molar absorptivity (\( \varepsilon \)), path length (\( b \)), and concentration (\( c \)) of the absorbing species. This relationship is fundamental to spectrophotometric analysis.
Equation:
\( \mathrm{A = \varepsilon b c} \)
- \( \mathrm{A} \): Absorbance (unitless)
- \( \mathrm{\varepsilon} \): Molar absorptivity (L·mol⁻¹·cm⁻¹) — indicates how strongly a species absorbs light of a given wavelength
- \( \mathrm{b} \): Path length (cm) — distance light travels through the solution
- \( \mathrm{c} \): Concentration (mol·L⁻¹)
Key Properties:
- 1. Proportionality of Absorbance: Absorbance (\( \mathrm{A} \)) increases linearly with both the path length and concentration, assuming constant wavelength and molar absorptivity.
- 2. Molar Absorptivity (\( \varepsilon \)): A unique constant for each compound at a specific wavelength; it measures how effectively the species absorbs light.
- 3. Path Length (\( b \)): The distance the light beam passes through the solution typically 1 cm in a standard cuvette. Longer path → more absorbing species encountered → higher absorbance.
- 4. Concentration (\( c \)): Higher solute concentration means more absorbing particles in the light path, resulting in greater absorbance.
Example:
The absorbance of a solution at 520 nm is measured as 0.500. The molar absorptivity (\( \varepsilon \)) is \( \mathrm{1.25 \times 10^3\, L·mol^{-1}·cm^{-1}} \) and the path length (\( b \)) is 1.00 cm. Calculate the concentration of the solution.
▶️ Answer/Explanation
Step 1: Use the equation \( \mathrm{A = \varepsilon b c} \)
\( \mathrm{0.500 = (1.25 \times 10^3)(1.00)(c)} \)
Step 2: Solve for \( \mathrm{c} \): \( \mathrm{c = \dfrac{0.500}{1.25 \times 10^3} = 4.00 \times 10^{-4}\, mol·L^{-1}} \)
Final Answer: The concentration of the solution is \( \mathrm{4.00 \times 10^{-4}\, M.} \)
Absorbance and Its Dependence on Concentration
In most spectrophotometric experiments, the path length (\( b \)) and wavelength of light (\( \lambda \)) are kept constant. Under these conditions, absorbance is directly proportional to the concentration (\( c \)) of the absorbing species in the solution.
Equation (Simplified):
\( \mathrm{A = k c} \)
where \( \mathrm{k = \varepsilon b} \) is a constant for a fixed wavelength and cuvette path length.
Key Properties:
- 1. Linear Relationship: Because \( \mathrm{b} \) and \( \mathrm{\varepsilon} \) are constants in an experiment, absorbance varies linearly with concentration. This linearity enables accurate calibration curves for quantitative analysis.
- 2. Wavelength Selection: The spectrophotometer is set to the wavelength of maximum absorbance (\( \lambda_{max} \)) for the analyte.
- At \( \lambda_{max} \), the molar absorptivity (\( \varepsilon \)) is largest.
- This ensures maximum sensitivity and precision — small changes in concentration produce large changes in absorbance.
- 3. Practical Application: By measuring absorbance at \( \lambda_{max} \), a calibration curve of \( \mathrm{A} \) vs. \( \mathrm{c} \) is constructed from standard solutions. The concentration of an unknown sample can then be determined from its measured absorbance using this linear relationship.
Key Idea: When wavelength and path length are constant, absorbance is directly proportional to concentration. This makes spectrophotometry a reliable, precise, and widely used technique in chemical analysis for determining unknown concentrations of absorbing species.
Example:
In a spectrophotometric analysis of a red dye, the absorbance at 520 nm is found to be 0.300 for a solution of concentration \( \mathrm{2.0 \times 10^{-5}\, M} \). What absorbance would be expected for a \( \mathrm{5.0 \times 10^{-5}\, M} \) solution under identical conditions?
▶️ Answer/Explanation
Step 1: Since \( \mathrm{A \propto c} \), \( \mathrm{\dfrac{A_1}{A_2} = \dfrac{c_1}{c_2}} \)
Step 2: Substitute: \( \mathrm{\dfrac{0.300}{A_2} = \dfrac{2.0 \times 10^{-5}}{5.0 \times 10^{-5}}} \)
Step 3: Solve for \( \mathrm{A_2} \): \( \mathrm{A_2 = 0.300 \times \dfrac{5.0}{2.0} = 0.750.} \)
Final Answer: The expected absorbance is \( \mathrm{0.750.} \)