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Chapter 18: Problem 83

An impurity in water has an extinction coefficient of \(3.45 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(280 \mathrm{nm}\), its absorption maximum Closer Look, p. 564). Below 50 ppb, the impurity is not a problem for human health. Given that most spectrometers cannot detect absorbances less than 0.0001 with good reliability, is measuring the absorbance of water at \(280 \mathrm{nm}\) a good way to detect concentrations of the impurity above the 50 -ppb threshold?

Short Answer

Expert verified
Using Beer's Law, we determined that the minimum detectable absorbance of the spectrometer corresponds to a concentration of 2.898 ppb at 280 nm. Since this concentration is lower than the harmful threshold (50 ppb), measuring the absorbance of the impure substance at 280 nm is an effective method for detecting concentrations above the threshold.

Step by step solution

01

Understand Beer's Law

Beer's Law states that the absorbance (A) of a solution is directly proportional to its concentration (C) and the path length (l), the latter being the distance the light travels through the solution. Mathematically, it is represented as \(A = \varepsilon \cdot l \cdot C\), where \(\varepsilon\) is the molar absorptivity or extinction coefficient. It's given in the problem that \(\varepsilon = 3.45 \times 10^{3} \mathrm{M}^{-1} \mathrm{cm}^{-1}\) and the minimum detectable absorbance is \(A=0.0001\). In most typical lab measurements, a cuvette with \(l = 1\,\mathrm{cm}\) is used.
02

Determine the concentration that results in the minimum detectable absorbance

First rearrange Beer's law to solve for the concentration: \(C = \frac{A}{\varepsilon \cdot l}\). By substituting the given values: \(C = \frac{0.0001}{3.45 \times 10^{3} \, \mathrm{M}^{-1} \, \mathrm{cm}^{-1} \cdot 1\, \mathrm{cm}} = 2.898 \times 10^{-8} \, \mathrm{M}\). This concentration can be converted to ppb (parts per billion) using the factor 1M = \(1 \times 10^{9}\) ppb. Thus, \(C = 2.898 \times 10^{-8} \, \mathrm{M} \cdot \frac{1 \times 10^{9} \, \mathrm{ppb}}{1 \, \mathrm{M}} =2.898\, \mathrm{ppb}\).
03

Compare to the harmful threshold concentration

The harmful threshold concentration is given as 50 ppb. The calculated concentration that will result in the minimum detectable absorbance (2.898 ppb) is less than the harmful threshold (50 ppb).
04

Conclude

Since the measured absorbance at 280 nm corresponds to a concentration below the harmful threshold, this implies that the spectrometer is capable of detecting concentrations of the impurity much lower than 50 ppb. Consequently, measuring the absorbance of water at 280 nm is indeed a good way to detect concentrations of the impurity above the 50 ppb threshold.

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