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AP Chemistry Help: Beer Lambert Law

Review real example questions for Beer Lambert Law in AP Chemistry.

Question 1 / 10

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A solution has concentration $1.0\times10^{-5}\ \text{M}$ and is measured in a 1.00 cm cuvette at a wavelength where $\varepsilon=3.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$ . What absorbance is expected?

All questions

Question 1

$A=0.30$ (correct answer)
$A=3.0$
$A=0.03$
$A=0.10$
$A=0.003$

Explanation: This question tests the direct application of Beer-Lambert law with the given parameters. Using A = εℓc with c = 1.0×10⁻⁵ M, ℓ = 1.00 cm, and ε = 3.0×10⁴ L mol⁻¹ cm⁻¹: A = (3.0×10⁴)(1.00)(1.0×10⁻⁵) = 0.30. This matches the marked answer exactly. Students might choose option C (0.03) by incorrectly placing the decimal point during multiplication of scientific notation terms. When applying Beer-Lambert law, carefully track powers of 10 throughout your calculation.

Question 2

At 600 nm, a compound has $\varepsilon = 1.0\times 10^4\,\text{M}^{-1}\text{cm}^{-1}$ . A solution of this compound has concentration $c = 2.0\times 10^{-5}\,\text{M}$ and gives absorbance $A = 0.40$ . What path length $\ell$ (in cm) was used?

$0.040\,\text{cm}$
$20\,\text{cm}$
$0.80\,\text{cm}$
$2.0\,\text{cm}$ (correct answer)
$0.20\,\text{cm}$

Explanation: This question tests the application of the Beer-Lambert Law to find path length from absorbance, molar absorptivity, and concentration. The equation A = ε ℓ c rearranges to ℓ = A / (ε c). Inserting the values gives ℓ = 0.40 / (1.0 × $10^4$ M⁻¹ cm⁻¹ × 2.0 × $10^{-5}$ M) = 0.40 / 0.20 = 2.0 cm. This matches choice B. A tempting distractor is 0.20 cm, resulting from inverting the formula incorrectly as ℓ = (ε c) / A, a misconception in algebraic rearrangement. A transferable strategy is to solve for the unknown variable symbolically first, then substitute numbers to minimize errors in Beer-Lambert calculations.

Question 3

A solution is measured at a fixed wavelength where $\varepsilon$ is constant. Which change will decrease the absorbance, assuming the solute remains the same?

Decrease the concentration while keeping path length constant. (correct answer)
Increase the path length while keeping concentration constant.
Increase both concentration and path length by the same factor.
Increase the concentration while keeping path length constant.
Increase the molar absorptivity while keeping concentration constant.

Explanation: This question tests understanding of what factors decrease absorbance according to Beer-Lambert law. Since A = εℓc, absorbance decreases when any of the variables (ε, ℓ, or c) decreases while others remain constant. Option A correctly states that decreasing concentration while keeping path length constant will decrease absorbance. Options B, C, and D would all increase absorbance, while option E is not practically achievable since ε is a molecular property. Students might choose option B by confusing which changes increase versus decrease absorbance. Remember that absorbance is directly proportional to concentration, path length, and molar absorptivity.

Question 4

A solution is measured at a fixed wavelength and follows Beer-Lambert law. Trial 1 uses $\ell=1.0\ \text{cm}$ and $c=0.020\ \text{M}$ and gives $A=0.50$ . In Trial 2, the concentration is doubled and the path length is halved. What absorbance is expected in Trial 2?

$1.00$
$0.25$
$0.50$ (correct answer)
$0.75$
$2.00$

Explanation: This question tests the combined effect of changing concentration and path length in the Beer-Lambert law. Doubling c increases A by 2, but halving ℓ decreases A by 1/2, netting no change: A = 0.50. Alternatively, original ε = 0.50 / (1.0 × 0.020) = 25 M⁻¹cm⁻¹; new A = 25 × 0.5 × 0.040 = 0.50. This matches choice C. A tempting distractor is choice A (1.00), from only considering the concentration doubling and ignoring path halving. Evaluate net effects of multiple changes by multiplying factors in Beer-Lambert applications.

Question 5

Two solutions of the same solute are measured at the same wavelength using identical 1.00 cm cuvettes. Solution 1 has concentration $1.0\times10^{-4}\ \text{M}$ and Solution 2 has concentration $3.0\times10^{-4}\ \text{M}$ . Which statement best compares their absorbances?

Solution 2 has one-third the absorbance of Solution 1.
Solution 2 has three times the absorbance of Solution 1. (correct answer)
Solution 2 has the same absorbance as Solution 1.
Solution 2 has nine times the absorbance of Solution 1.
Solution 2 has one-ninth the absorbance of Solution 1.

Explanation: This question tests understanding of the direct proportionality between concentration and absorbance in the Beer-Lambert law. Since A = εℓc and all other parameters (ε, ℓ, wavelength) remain constant, absorbance is directly proportional to concentration. Solution 2 has concentration 3.0×10⁻⁴ M compared to Solution 1's 1.0×10⁻⁴ M, making it exactly 3 times more concentrated. Therefore, Solution 2 will have exactly 3 times the absorbance of Solution 1. Students might incorrectly choose option D (nine times) by squaring the concentration ratio instead of using the direct proportionality. Remember that in Beer-Lambert law, absorbance scales linearly with concentration when other factors are held constant.

Question 6

A solution with concentration $1.0\times10^{-4}\ \text{M}$ is measured at two different wavelengths in a 1.00 cm cuvette. At $\lambda_1$ , $\varepsilon=2.0\times10^4$ ; at $\lambda_2$ , $\varepsilon=5.0\times10^3$ . Which statement is correct about absorbance?

Absorbance at $\lambda_1$ is twice absorbance at $\lambda_2$ .
Absorbance at $\lambda_2$ is twice absorbance at $\lambda_1$ .
Absorbance at $\lambda_1$ equals absorbance at $\lambda_2$ .
Absorbance at $\lambda_1$ is four times absorbance at $\lambda_2$ . (correct answer)
Absorbance at $\lambda_2$ is four times absorbance at $\lambda_1$ .

Explanation: This question tests understanding of how molar absorptivity affects absorbance at different wavelengths. At λ₁: A₁ = (2.0×10⁴)(1.00)(1.0×10⁻⁴) = 2.0. At λ₂: A₂ = (5.0×10³)(1.00)(1.0×10⁻⁴) = 0.5. The ratio A₁/A₂ = 2.0/0.5 = 4, so absorbance at λ₁ is four times that at λ₂. This occurs because the molar absorptivity at λ₁ is four times larger than at λ₂. Students might choose option A by inverting the ratio. When comparing absorbances at different wavelengths, always calculate both values and determine their ratio carefully.

Question 7

A solution is measured in a 1.00 cm cuvette and has absorbance $A=0.66$ . The molar absorptivity is $\varepsilon=3.3\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$ . What is the concentration?

$5.0\times10^{-4}\ \text{M}$
$1.0\times10^{-4}\ \text{M}$
$5.0\times10^{-5}\ \text{M}$
$2.0\times10^{-4}\ \text{M}$ (correct answer)
$2.0\times10^{-5}\ \text{M}$

Explanation: This question tests the application of Beer-Lambert law to determine concentration from absorbance measurements. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.66/(3.3×10³ × 1.00) = 0.66/(3.3×10³) = 2.0×10⁻⁴ M. This matches the marked answer exactly. Students might choose option B (5.0×10⁻⁴ M) by incorrectly estimating the division result. When working with Beer-Lambert calculations, always perform the division carefully to avoid computational errors.

Question 8

A solution of a colored compound is measured at 400 nm. In a 1.00 cm cuvette, $A=0.10$ for a solution with $c=2.0\times10^{-5}\ \text{M}$ . What is $\varepsilon$ at 400 nm?

$\varepsilon=5.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=1.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=5.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$ (correct answer)
$\varepsilon=2.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=2.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$

Explanation: This question tests the calculation of molar absorptivity from Beer-Lambert law data with very dilute solutions. Using A = εℓc and solving for ε: ε = A/(ℓc) = 0.10/(1.00 × 2.0×10⁻⁵) = 0.10/(2.0×10⁻⁵) = 5.0×10³ L mol⁻¹ cm⁻¹. This matches the marked answer exactly. Students might choose option B (2.0×10⁴) by incorrectly handling the very small concentration value in their calculation. When working with dilute solutions, pay careful attention to powers of 10 in scientific notation.

Question 9

Two solutions of different solutes are measured at the same wavelength using 1.00 cm cuvettes. Solution P has $\varepsilon=1.0\times10^4$ and $c=1.0\times10^{-4}\ \text{M}$ . Solution Q has $\varepsilon=5.0\times10^3$ and $c=2.0\times10^{-4}\ \text{M}$ . Which statement best compares absorbances?

Solution P has half the absorbance of Solution Q.
Solution Q has twice the absorbance of Solution P.
Solution P and Solution Q have the same absorbance. (correct answer)
Solution P has twice the absorbance of Solution Q.
Solution Q has half the absorbance of Solution P.

Explanation: This question tests understanding of how different combinations of ε and c affect absorbance in Beer-Lambert law. For Solution P: A = εℓc = (1.0×10⁴)(1.00)(1.0×10⁻⁴) = 1.0. For Solution Q: A = εℓc = (5.0×10³)(1.00)(2.0×10⁻⁴) = 1.0. Both solutions have the same absorbance because the product εc is identical in both cases. Students might choose option A or B by only comparing individual parameters rather than their combined effect. When comparing absorbances, always calculate the complete Beer-Lambert expression for each solution.

Question 10

A solution has $A=1.20$ in a 2.00 cm cuvette. If the same solution is measured in a 1.00 cm cuvette at the same wavelength, what absorbance is expected?

$A=2.40$
$A=0.60$ (correct answer)
$A=1.20$
$A=0.30$
$A=1.80$

Explanation: This question tests understanding of how path length affects absorbance when moving between different cuvettes. Since A = εℓc and only path length changes (from 2.00 cm to 1.00 cm), the new absorbance is A_new = A_original × (ℓ_new/ℓ_original) = 1.20 × (1.00/2.00) = 0.60. Halving the path length results in halving the absorbance. Students might choose option A (2.40) by incorrectly thinking shorter cuvettes increase absorbance. Remember that absorbance decreases proportionally when path length decreases.