Chemistry: AP Chemistry: Spectroscopy and Beer-Lambert Law Worksheet

Question 1

State the Beer-Lambert law and identify what each variable represents.

Accepted Answer

The Beer-Lambert law is A = εbc. A is absorbance, ε is molar absorptivity in L mol^-1 cm^-1, b is path length in cm, and c is concentration in mol/L.

Question 2

A solution has a molar absorptivity of 1.50 x 10^4 L mol^-1 cm^-1 at 520 nm. If the path length is 1.00 cm and the concentration is 2.00 x 10^-5 M, calculate the absorbance.

Accepted Answer

Using A = εbc, A = (1.50 x 10^4 L mol^-1 cm^-1)(1.00 cm)(2.00 x 10^-5 mol/L) = 0.300. The absorbance is 0.300.

Question 3

A sample in a 1.00 cm cuvette has an absorbance of 0.742 at its wavelength of maximum absorbance. The molar absorptivity is 3.71 x 10^3 L mol^-1 cm^-1. Calculate the concentration of the sample.

Accepted Answer

Rearranging A = εbc gives c = A/(εb). The concentration is c = 0.742/[(3.71 x 10^3 L mol^-1 cm^-1)(1.00 cm)] = 2.00 x 10^-4 M.

Question 4

A calibration curve for a colored ion gives the equation A = 2450c + 0.012, where c is in mol/L. An unknown solution has an absorbance of 0.502. Calculate the concentration of the unknown.

Accepted Answer

Substitute the absorbance into the calibration equation: 0.502 = 2450c + 0.012. Then c = (0.502 - 0.012)/2450 = 2.00 x 10^-4 M, so the unknown concentration is 2.00 x 10^-4 M.

Question 5

Explain why the wavelength of maximum absorbance, λmax, is usually chosen for quantitative spectrophotometric analysis.

Accepted Answer

The wavelength of maximum absorbance is chosen because the analyte absorbs most strongly there, giving better sensitivity. Small concentration changes produce larger absorbance changes at λmax.

Question 6

A solution transmits 25.0 percent of the incident light at a certain wavelength. Calculate its absorbance.

Accepted Answer

Percent transmittance must be converted to transmittance, so T = 0.250. Absorbance is A = -log(T) = -log(0.250) = 0.602.

Question 7

Two solutions of the same absorbing species are measured in the same 1.00 cm cuvette at the same wavelength. Solution 1 has concentration 1.20 x 10^-4 M and absorbance 0.360. Solution 2 has concentration 2.40 x 10^-4 M. Predict the absorbance of solution 2.

Accepted Answer

Because absorbance is directly proportional to concentration when ε and b are constant, doubling the concentration doubles the absorbance. The predicted absorbance is 0.720.

Question 8

A student measures the absorbance of a solution and obtains A = 1.95. Explain why the student might dilute the solution before making a final concentration determination.

Accepted Answer

An absorbance of 1.95 is relatively high and may be outside the most reliable linear range of the instrument. Diluting the solution can bring the absorbance into a range where the Beer-Lambert law and the calibration curve are more dependable.

Question 9

A 10.00 mL sample of an unknown solution is diluted to 50.00 mL. The diluted solution has a concentration of 3.60 x 10^-5 M based on spectrophotometric data. Calculate the concentration of the original unknown solution.

Accepted Answer

Using M1V1 = M2V2, M1 = (M2V2)/V1 = (3.60 x 10^-5 M)(50.00 mL)/(10.00 mL) = 1.80 x 10^-4 M. The original concentration was 1.80 x 10^-4 M.

Question 10

An absorption spectrum shows a strong peak at 620 nm. What color of visible light is most strongly absorbed, and what color might the solution appear to the eye?

Accepted Answer

Light near 620 nm is orange-red light, so the solution most strongly absorbs orange-red wavelengths. The solution may appear blue-green because transmitted and reflected light is often close to the complementary color of the absorbed light.

Question 11

A blank cuvette is not used to zero a spectrophotometer before measuring samples. Describe one likely effect on the measured absorbance values and explain why this matters.

Accepted Answer

The measured absorbance values may be too high because absorbance from the solvent, cuvette, or background light loss is included with the analyte signal. This matters because the calculated concentration would likely be overestimated.

Question 12

The following calibration data are collected for an absorbing species in a 1.00 cm cuvette: 0.00 x 10^-5 M gives A = 0.000, 1.00 x 10^-5 M gives A = 0.115, 2.00 x 10^-5 M gives A = 0.230, and 3.00 x 10^-5 M gives A = 0.345. Determine the molar absorptivity ε.

Accepted Answer

The slope of A versus concentration equals εb. Using the data, slope = 0.115/(1.00 x 10^-5 M) = 1.15 x 10^4 L mol^-1. Since b = 1.00 cm, ε = 1.15 x 10^4 L mol^-1 cm^-1.

Chemistry: AP Chemistry: Spectroscopy and Beer-Lambert Law

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