3.13 Beer-Lambert Law

1. AThe volume of liquid in the cuvette
   A student who thinks the amount of solution in the cuvette sets the absorbance picks this. The b in A = εbc is the distance across the cuvette along the beam; the volume of solution does not appear in the relationship.
2. BThe mass of the compound dissolved
   A student who thinks absorbance depends on the amount of solute rather than its concentration picks this. A = εbc uses the molar concentration, which is found from A, ε, and b without knowing the mass dissolved.
3. CThe brightness of the lamp used
   A student who thinks a brighter light makes a solution absorb more picks this. Absorbance compares the light reaching the detector with and without the sample, so it does not depend on the brightness of the source.
4. DThe path length of the cuvette Correct
   A = εbc, so c = A/(εb). With the absorbance and the molar absorptivity known, the only other quantity needed is the path length, the distance the light travels through the solution.

Working A = εbc, so c = A/(εb). A and ε are known, so the only other quantity needed is the path length, b.

CED 3.13.A.1 · Read this in Fix

1. AZeroing the instrument with a cuvette of distilled water
   A student who thinks zeroing with a blank subtracts part of each sample's absorbance picks this. Zeroing with the solvent sets zero for everything except the dye, so later readings show the dye's absorbance; it is a correct step for the standards and the drink alike.
2. BFilling the cuvette only three-quarters full with the drink
   A student who thinks a cuvette holding more solution gives a greater absorbance picks this. The path length is the distance across the cuvette along the beam; as long as the beam passes through the drink, the depth of liquid does not change the absorbance.
3. CUsing light of 620 nm, which is orange-red, to measure a blue dye
   A student who thinks a solution absorbs light of its own color picks this, expecting a blue dye to absorb little orange light. A blue dye looks blue because it transmits blue light; it absorbs mainly orange and red light, so 620 nm is a suitable wavelength.
4. DLeaving water from the rinse in the cuvette with the drink Correct
   Water left in the cuvette dilutes the drink, so the concentration of dye in the light path, and therefore the absorbance, is lower than for the undiluted drink. The calibration graph then gives a concentration that is too low.

CED 3.13.A.2 · Read this in Fix

1. ACuvette 1
   A student who thinks absorbance depends only on concentration picks this, the most concentrated solution (4 particles in the narrow beam region). Its path is half as long as that of cuvette 2, so the beam meets only 4 particles there compared with 6 in cuvette 2.
2. BCuvette 2 Correct
   The absorbance is proportional to the number of absorbing particles in the light path, which depends on both the path length and the concentration. The beam meets 6 particles in this cuvette, more than in any other: it is wider than cuvettes 1 and 3 and more concentrated than cuvette 4.
3. CCuvette 3
   A student who thinks a cuvette holding more solution gives a greater absorbance picks this, the cuvette with the most solution and the most particles in total. Only the 3 particles inside the beam absorb its light; the solution above the beam does not affect the absorbance.
4. DCuvette 4
   A student who thinks absorbance measures the light that passes through the sample picks the cuvette that lets the most light through, which has only 2 particles in the beam. Absorbance measures the light absorbed, so this cuvette has the smallest absorbance, not the greatest.

Working Absorbance is proportional to the number of absorbing particles in the light path (A = εbc with ε fixed). Particles inside the beam: cuvette 1, 4; cuvette 2, 6; cuvette 3, 3; cuvette 4, 2. The greatest absorbance is in cuvette 2, the wide cuvette with 6 particles in the beam; the solution above the beam does not absorb the beam's light.

CED 3.13.A.1 · Read this in Fix

1. A6.0 × 10⁻⁵ M Correct
   The graph was made with a 1.00 cm path; doubling the path length doubles the absorbance, so in a 1.00 cm cuvette the diluted solution would have A = 0.24, which the graph gives as 1.2 × 10⁻⁵ M. The sample was diluted from 5.00 mL to 25.0 mL, a factor of 5, so the original concentration is 6.0 × 10⁻⁵ M.
2. B1.2 × 10⁻⁴ M
   A student who thinks absorbance depends only on concentration reads the graph at 0.48 without correcting for the path length: 2.4 × 10⁻⁵ M, × 5 = 1.2 × 10⁻⁴ M. The 2.00 cm path doubles the absorbance, so the 1.00 cm value is 0.24.
3. C2.4 × 10⁻⁴ M
   A student who thinks absorbance measures the light passing through the sample reasons that the longer path lets less light through and so lowers the absorbance, and doubles the reading to 0.96: 4.8 × 10⁻⁵ M, × 5 = 2.4 × 10⁻⁴ M. A longer path increases the absorbance, so the reading must be halved.
4. D1.2 × 10⁻⁵ M
   A student who thinks dilution does not change concentration stops at the concentration of the diluted solution, 1.2 × 10⁻⁵ M. The original sample was diluted fivefold (5.00 mL to 25.0 mL), so its concentration is five times greater.

Working Slope of the graph = 0.20 per 1.0 × 10⁻⁵ M (path 1.00 cm). In the 2.00 cm cuvette the absorbance is twice what it would be in 1.00 cm, so the 1.00 cm absorbance of the diluted solution is 0.48/2 = 0.24, giving c = 0.24/(0.20 per 10⁻⁵ M) = 1.2 × 10⁻⁵ M. Dilution factor = 25.0 mL/5.00 mL = 5, so the original concentration = 5 × 1.2 × 10⁻⁵ M = 6.0 × 10⁻⁵ M. Distractors: ignoring the path length gives 5 × 2.4 × 10⁻⁵ = 1.2 × 10⁻⁴ M; doubling instead of halving the absorbance gives 5 × 4.8 × 10⁻⁵ = 2.4 × 10⁻⁴ M; ignoring the dilution gives 1.2 × 10⁻⁵ M.

CED 3.13.A.2 · Read this in Fix

1. A1.50 × 10⁴ M⁻¹ cm⁻¹
   A student who thinks absorbance depends only on concentration leaves out the path length: 0.300/(2.00 × 10⁻⁵ M) = 1.50 × 10⁴. The path is 0.500 cm, half the usual 1 cm, so ε is twice this value.
2. B1.00 × 10⁷ M⁻¹ cm⁻¹
   A student who thinks absorbance depends on the amount of dye uses the moles in the cuvette, (2.00 × 10⁻⁵ M)(3.00 × 10⁻³ L) = 6.00 × 10⁻⁸ mol, in place of the concentration. A = εbc uses molar concentration; the amount of solution in the cuvette does not matter.
3. C3.00 × 10⁴ M⁻¹ cm⁻¹ Correct
   A = εbc, so ε = A/(bc) = 0.300/[(0.500 cm)(2.00 × 10⁻⁵ M)] = 3.00 × 10⁴ M⁻¹ cm⁻¹. The volume of solution in the cuvette does not enter the calculation.
4. D5.00 × 10³ M⁻¹ cm⁻¹
   A student who thinks b measures the amount of solution in the cuvette uses 3.00 (the volume in mL) as the path length: 0.300/[(3.00)(2.00 × 10⁻⁵ M)] = 5.00 × 10³. The path length is the 0.500 cm distance the light travels through the solution.

Working ε = A/(bc) = 0.300/[(0.500 cm)(2.00 × 10⁻⁵ M)] = 3.00 × 10⁴ M⁻¹ cm⁻¹. Distractors: leaving out b gives 0.300/(2.00 × 10⁻⁵) = 1.50 × 10⁴; using moles of dye, (2.00 × 10⁻⁵ M)(3.00 × 10⁻³ L) = 6.00 × 10⁻⁸ mol, in place of concentration gives 1.00 × 10⁷; using the 3.00 mL volume as the path length gives 5.00 × 10³.

CED 3.13.A.1 · Read this in Fix

1. AA change in concentration changes the absorbance most at 620 nm Correct
   The dye absorbs most strongly at 620 nm, so its molar absorptivity is greatest there. Since A = εbc, the slope of absorbance against concentration, εb, is greatest at 620 nm, so measurements there are the most sensitive to concentration.
2. BAbsorbance is proportional to concentration only at 620 nm
   A student who thinks Beer's law holds only at the wavelength of maximum absorbance picks this. At any fixed wavelength the dye absorbs, absorbance is proportional to concentration; 620 nm is chosen because it gives the greatest sensitivity.
3. CThe most light passes through the solution at 620 nm
   A student who thinks absorbance measures the light passing through the sample reads the peak as the wavelength of greatest transmission. The peak shows the wavelength at which the solution absorbs the most light, so the least light passes through there.
4. DPhotons of 620 nm have more energy than shorter-wavelength photons
   A student who thinks longer-wavelength photons carry more energy picks this. E = hc/λ, so 620 nm photons carry less energy than photons of shorter wavelength; the peak at 620 nm reflects the dye's energy differences, not a high photon energy.

CED 3.13.A.2 · Read this in Fix

1. A0.90
   A student who thinks dilution does not change concentration, because the amount of dye is unchanged, picks this. The same amount of dye is spread through 30.0 mL instead of 20.0 mL, so the concentration and the absorbance fall.
2. B0.45
   A student who takes the dilution factor as the volume of water over the volume of solution, 10.0/20.0, picks this. The concentration changes by the original volume over the total volume, 20.0/30.0, so A = 0.60.
3. C0.60 Correct
   At constant path length and wavelength, absorbance is proportional to concentration. Mixing 20.0 mL of solution with 10.0 mL of water gives 30.0 mL, so the concentration and the absorbance fall to 20.0/30.0 of their values: 0.90 × 2/3 = 0.60.
4. D1.35
   A student who thinks absorbance measures the light passing through the sample expects the more dilute solution, which lets more light through, to have a higher absorbance: 0.90 × 30.0/20.0. Absorbance is proportional to concentration, so it decreases.

Working With path length and wavelength unchanged, A is proportional to c. Total volume = 30.0 mL, so c falls by 20.0/30.0: A = 0.90 × 20.0/30.0 = 0.60. Distractors: no change in concentration, 0.90; factor 10.0/20.0, 0.45; absorbance treated as rising when the solution is more dilute, 0.90 × 30.0/20.0 = 1.35.

CED 3.13.A.2 · Read this in Fix

1. AThe same straight line, since the same solutions and cuvette are used
   A student who thinks absorbance depends only on concentration and path length, and not on the wavelength, picks this. ε is a property of the dye at a particular wavelength; at 540 nm it is smaller than at 620 nm, so each standard gives a lower absorbance.
2. BA straight line with a smaller slope, since the dye absorbs less light at 540 nm Correct
   At 540 nm the path length is still constant, so absorbance is still proportional to concentration: the graph is a straight line through the origin. Because 620 nm is the wavelength of maximum absorbance, ε is smaller at 540 nm, so the slope, εb, is smaller.
3. CA curved line, since Beer's law holds only at the wavelength of maximum absorbance
   A student who thinks Beer's law holds only at the wavelength of maximum absorbance picks this. At any fixed wavelength that the dye absorbs, absorbance is proportional to concentration, so the graph is still a straight line.
4. DA straight line with a greater slope, since 540 nm photons carry more energy
   A student who thinks light of shorter wavelength is absorbed more strongly by any substance because it carries more energy picks this. 620 nm is where the dye absorbs most strongly, so at 540 nm ε, and the slope, is smaller.

CED 3.13.A.2 · Read this in Fix

1. A3.00
   A student who thinks absorbance depends only on concentration and path length picks this, multiplying 0.60 by 5 for the longer path. The molar absorptivity at 540 nm is one-fourth of its value at 620 nm, so the absorbance is also multiplied by 1/4.
2. B0.15
   A student who thinks the path length does not affect absorbance picks this, multiplying 0.60 by 1/4 for the smaller molar absorptivity. The light also passes through 5 times as much solution, so the absorbance is multiplied by 5 as well.
3. C0.75 Correct
   The concentration is the same in both measurements: c = 0.60 ÷ (2.0 × 10⁴ M⁻¹ cm⁻¹ × 1.00 cm) = 3.0 × 10⁻⁵ M. At 540 nm, A = εbc = (5.0 × 10³ M⁻¹ cm⁻¹)(5.00 cm)(3.0 × 10⁻⁵ M) = 0.75.
4. D0.48
   A student who thinks absorbance measures the light that gets through picks this, raising the value 4 times for the weaker absorption at 540 nm and cutting it to 1/5 for the longer path. Absorbance increases with ε and with b: A = εbc.

Working At 620 nm: c = A/(εb) = 0.60 ÷ (2.0 × 10⁴ M⁻¹ cm⁻¹ × 1.00 cm) = 3.0 × 10⁻⁵ M. At 540 nm: A = εbc = (5.0 × 10³ M⁻¹ cm⁻¹)(5.00 cm)(3.0 × 10⁻⁵ M) = 0.75.

CED 3.13.A.1 · Read this in Fix

1. A4.0 × 10⁻⁶ mol Correct
   c = A/(εb) = 0.60 ÷ (1.5 × 10⁴ M⁻¹ cm⁻¹ × 2.00 cm) = 2.0 × 10⁻⁵ M. The flask held 0.2000 L of solution of this concentration, so n = (2.0 × 10⁻⁵ mol/L)(0.2000 L) = 4.0 × 10⁻⁶ mol.
2. B8.0 × 10⁻⁶ mol
   A student who leaves the path length out of the calculation picks this, using c = A/ε = 4.0 × 10⁻⁵ M. The light passes through 2.00 cm of solution, so c = A/(εb) = 2.0 × 10⁻⁵ M.
3. C2.0 × 10⁻⁵ mol
   A student who thinks absorbance measures the total amount of solute picks this, reading A/(εb) = 2.0 × 10⁻⁵ as moles of dye. A/(εb) is a concentration in mol/L; the amount in the flask is that concentration multiplied by 0.2000 L.
4. D2.7 × 10⁻⁶ mol
   A student who takes b as the 3.0 mL of solution in the cuvette picks this, using c = 0.60 ÷ (1.5 × 10⁴ × 3.0). b is the distance the light travels through the solution, 2.00 cm, whatever volume the cuvette holds.

Working c = A/(εb) = 0.60 ÷ (1.5 × 10⁴ M⁻¹ cm⁻¹ × 2.00 cm) = 2.0 × 10⁻⁵ M. The solution in the cuvette has the same concentration as the solution in the flask. n = cV = (2.0 × 10⁻⁵ mol/L)(0.2000 L) = 4.0 × 10⁻⁶ mol.

CED 3.13.A.1 · Read this in Fix