Eleven worked problems with detailed, step-by-step solutions you can verify in the calculator.
A UV-Vis spectrum of caffeine in water shows A = 0.85 at 272 nm. Given ε = 10,200 L·mol⁻¹·cm⁻¹ and l = 1 cm, find the concentration.
Solution
Identify known values
A = 0.85, ε = 10200, l = 1, c = ?
Rearrange Beer-Lambert law
c = A / (ε × l)
Calculate
c = 0.85 / (10200 × 1) = 8.33 × 10⁻⁵ mol/L
c = 8.33 × 10⁻⁵ M (83.3 μM)
What absorbance would you expect for a 2.5 × 10⁻⁴ M solution of KMnO₄ (ε = 2,455 at 525 nm) in a 1 cm cuvette?
Solution
Identify known values
ε = 2455, l = 1, c = 2.5×10⁻⁴, A = ?
Apply Beer-Lambert law
A = ε × l × c
Calculate
A = 2455 × 1 × 2.5×10⁻⁴ = 0.614
A = 0.614
A 1.0 × 10⁻³ M solution gives A = 0.54 in a 1 cm cell. What is the molar absorptivity?
Solution
Identify known values
A = 0.54, l = 1, c = 1×10⁻³, ε = ?
Rearrange Beer-Lambert law
ε = A / (l × c)
Calculate
ε = 0.54 / (1 × 0.001) = 540 L·mol⁻¹·cm⁻¹
ε = 540 L·mol⁻¹·cm⁻¹
Your lab has a 0.5 cm cuvette instead of 1 cm. If ε = 15,000 and c = 5.0 × 10⁻⁵ M, what absorbance do you expect?
Solution
Identify known values
ε = 15000, l = 0.5, c = 5×10⁻⁵, A = ?
Apply Beer-Lambert law
A = ε × l × c
Calculate
A = 15000 × 0.5 × 5×10⁻⁵ = 0.375
A = 0.375
A solution of methyl orange transmits 18% of incident light at 464 nm. What is the absorbance? If ε = 23,600 L·mol⁻¹·cm⁻¹ and l = 1 cm, what is the concentration?
Solution
Convert percent transmittance
T = 18% = 0.18
Calculate absorbance
A = −log₁₀(T) = −log₁₀(0.18) = 0.745
Find concentration
c = A / (ε × l) = 0.745 / (23600 × 1) = 3.16 × 10⁻⁵ mol/L
A = 0.745, c = 3.16 × 10⁻⁵ M (31.6 μM)
You prepare a standard from a 1.00 × 10⁻³ M stock solution of p-nitrophenol (ε = 18,380 L·mol⁻¹·cm⁻¹ at 400 nm) by diluting it 1:20. What absorbance do you expect in a 1 cm cuvette?
Solution
Calculate diluted concentration
c = 1.00×10⁻³ / 20 = 5.0 × 10⁻⁵ M
Apply Beer-Lambert law
A = ε × l × c
Calculate
A = 18380 × 1 × 5.0×10⁻⁵ = 0.919
A = 0.919
In an enzyme assay, NADH is monitored at 340 nm (ε = 6,220 L·mol⁻¹·cm⁻¹). A patient sample gives A = 0.31 in a 1 cm cuvette. What is the NADH concentration?
Solution
Identify known values
A = 0.31, ε = 6220, l = 1, c = ?
Rearrange Beer-Lambert law
c = A / (ε × l)
Calculate
c = 0.31 / (6220 × 1) = 4.98 × 10⁻⁵ mol/L
c = 4.98 × 10⁻⁵ M (49.8 μM)
Your analyte has ε = 45,000 L·mol⁻¹·cm⁻¹ at 520 nm and c = 1.0 × 10⁻⁵ M. For reliable measurements you need A between 0.1 and 1.0. Will a 1 cm cuvette work? What about a 0.1 cm (1 mm) cuvette?
Solution
Calculate A with 1 cm cuvette
A = 45000 × 1 × 1.0×10⁻⁵ = 0.45 ✓ (within range)
Calculate A with 0.1 cm cuvette
A = 45000 × 0.1 × 1.0×10⁻⁵ = 0.045 ✗ (too low)
Conclusion
Use the 1 cm cuvette. The 0.1 cm cell gives A < 0.1, where detector noise dominates.
1 cm cuvette: A = 0.45 (good). 0.1 cm cuvette: A = 0.045 (unreliable).
You prepare five standards of a dye (0, 10, 20, 30, and 40 μM) and measure absorbance at 490 nm in a 1 cm cuvette: 0, 0.125, 0.250, 0.375, and 0.500. An unknown solution reads A = 0.310. What is the concentration of the unknown?
Solution
Find the slope of A vs. c
Slope = ΔA / Δc = 0.500 / (40×10⁻⁶) = 12,500 M⁻¹ (this equals ε × l)
Solve for unknown concentration
c = A / slope = 0.310 / 12,500 = 2.48 × 10⁻⁵ M
Convert units
c = 24.8 μM. Since ε × l = slope and l = 1, ε = 12,500 L·mol⁻¹·cm⁻¹
c = 2.48 × 10⁻⁵ M (24.8 μM)
You extract DNA and measure A₂₆₀ = 0.35 and A₂₈₀ = 0.18 in a 1 cm cuvette. Using the convention that 1 absorbance unit at 260 nm = 50 μg/mL for dsDNA, what is the DNA concentration? Is the sample pure?
Solution
Apply the dsDNA convention
[DNA] = A₂₆₀ × 50 μg/mL = 0.35 × 50 = 17.5 μg/mL
Assess purity with A₂₆₀/A₂₈₀
Ratio = 0.35 / 0.18 = 1.94. Pure DNA ≈ 1.8; pure RNA ≈ 2.0
Interpret
The ratio of 1.94 is slightly above 1.8, suggesting minor RNA contamination. The 50 μg/mL rule is Beer-Lambert with a = 0.020 (μg/mL)⁻¹cm⁻¹.
[DNA] = 17.5 μg/mL. A₂₆₀/A₂₈₀ = 1.94 (slightly RNA-contaminated).
A solution contains two dyes, P and Q, in a 1 cm cuvette. At 420 nm: ε_P = 12,000 and ε_Q = 3,000. At 560 nm: ε_P = 1,500 and ε_Q = 10,000. You measure A₄₂₀ = 0.570 and A₅₆₀ = 0.360. Find the concentration of each dye.
Solution
Write two simultaneous equations
A₄₂₀ = 12000·c_P + 3000·c_Q = 0.570 A₅₆₀ = 1500·c_P + 10000·c_Q = 0.360
Eliminate c_P (multiply eq 2 by 8, subtract eq 1)
80000·c_Q − 3000·c_Q = 2.880 − 0.570 → 77000·c_Q = 2.310 → c_Q = 3.0 × 10⁻⁵ M
Back-substitute for c_P
12000·c_P = 0.570 − 3000(3.0×10⁻⁵) = 0.480 → c_P = 4.0 × 10⁻⁵ M
c_P = 4.0 × 10⁻⁵ M (40 μM), c_Q = 3.0 × 10⁻⁵ M (30 μM)