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Chapter 17

Q. 17.2

Absorbance, Transmittance, and Beer’s Law

Find the absorbance and transmittance of a 0.002 40 M solution of a substance with a molar absorptivity of 313 M^{−1}  cm^{−1} in a cell with a 2.00-cm pathlength.

Step-by-Step

Verified Solution

Equation 17-6 gives us the absorbance.

A = εbc = (313  M^{−1}  cm^{−1})(2.00  cm)(0.002  40  M) = 1.50

Transmittance is obtained from Equation 17-5 by raising 10 to the power equal to the expression on each side of the equation:

A = log(\frac{P_0}{P} )= – log T                (17-5)

\log T = −A

T = 10^{\log T} = 10^{−A} = 10^{−1.50} = 0.031  6

Just 3.16% of the incident light emerges from this solution.

Test Yourself     The transmittance of a 0.010 M solution of a compound in a 0.100-cm-pathlength cell is T = 8.23%. Find the absorbance (A) and the molar absorptivity (ε).
(Answer: 1.08, 1.08 × 10^{3}  M^{−1}  cm^{−1})

If x = y, 10^{x} = 10^{y}.