Beer-Lambert Law
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AP Chemistry › Beer-Lambert Law
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?
$A=0.003$
$A=0.10$
$A=3.0$
$A=0.30$
$A=0.03$
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.
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?
$20,\text{cm}$
$0.20,\text{cm}$
$2.0,\text{cm}$
$0.80,\text{cm}$
$0.040,\text{cm}$
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.
A solution absorbs light at 430 nm with molar absorptivity $\varepsilon = 150,\text{M}^{-1}\text{cm}^{-1}$. A student uses a cuvette with $\ell = 2.00,\text{cm}$ and measures $A = 0.60$. What is the concentration of the solution?
$0.0010,\text{M}$
$0.20,\text{M}$
$0.0040,\text{M}$
$0.0080,\text{M}$
$0.0020,\text{M}$
Explanation
This question tests the application of the Beer-Lambert Law to calculate the concentration of a solution from absorbance, molar absorptivity, and path length. The Beer-Lambert Law states A = ε ℓ c, so concentration c = A / (ε ℓ). Substituting the values gives c = 0.60 / (150 M⁻¹ cm⁻¹ × 2.00 cm) = 0.60 / 300 = 0.0020 M. This matches choice A. A tempting distractor is 0.0040 M, which results from omitting the path length and calculating c = A / ε = 0.60 / 150 = 0.0040 M, stemming from the misconception that path length does not affect absorbance in the equation. A transferable strategy is to memorize the full Beer-Lambert equation and check that all variables are included before performing calculations.
A solution is measured at a fixed wavelength where $\varepsilon$ is constant. Which change will decrease the absorbance, assuming the solute remains the same?
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.
Increase the path length while keeping concentration constant.
Decrease the concentration while keeping path length 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.
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?
$2.00$
$0.25$
$1.00$
$0.75$
$0.50$
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.
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 nine times the absorbance of Solution 1.
Solution 2 has one-ninth the absorbance of Solution 1.
Solution 2 has the same absorbance as Solution 1.
Solution 2 has three times 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.
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 $\lambda_2$ is four times absorbance at $\lambda_1$.
Absorbance at $\lambda_1$ is four times absorbance at $\lambda_2$.
Absorbance at $\lambda_1$ equals absorbance at $\lambda_2$.
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.
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?
$2.0\times10^{-4}\ \text{M}$
$5.0\times10^{-4}\ \text{M}$
$5.0\times10^{-5}\ \text{M}$
$1.0\times10^{-4}\ \text{M}$
$2.0\times10^{-5}\ \text{M}$
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.
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?
$\varepsilon=2.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=5.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=2.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=1.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$
$\varepsilon=5.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$
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.
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 twice the absorbance of Solution Q.
Solution Q has half the absorbance of Solution P.
Solution Q has twice the absorbance of Solution P.
Solution P has half the absorbance of Solution Q.
Solution P and Solution Q have the same absorbance.
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.