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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Extinction Coefficient
Imagine you're examining the absorbance of a solution to identify impurities. The extinction coefficient is a crucial value catering to such experiments.
The extinction coefficient, often symbolized by \( \varepsilon \), represents how strongly a substance absorbs light at a particular wavelength. The higher the \( \varepsilon \), the more absorbance occurs, even with a small concentration.
In the context of Beer's Law, where \( A = \varepsilon \cdot l \cdot C \), \( \varepsilon \) quantifies how much light the solution absorbs.
  • It is measured in units of \( \text{M}^{-1} \cdot \text{cm}^{-1} \).
  • Determines the efficiency of using spectroscopy to measure specific substances.
Hence, understanding the extinction coefficient of a substance can greatly enhance the precision in detection methods, so higher coefficients allow for detection of very diluted samples.
Spectrophotometry
Spectrophotometry is a technique used to measure the intensity of light absorbed by a solution.
This method is vital in chemistry because it allows us to determine the concentration of a solute within a solution.
One fascinating aspect of spectrophotometry is its reliance on Beer's Law, \( A = \varepsilon \cdot l \cdot C \), ensuring a direct relationship between absorbance and concentration.
  • Light passes through the solution in a cuvette, usually 1 cm in length.
  • The spectrophotometer measures how much light the solution absorbs.
  • It covers various wavelengths, including ultraviolet (UV) and infrared (IR).
Applications of spectrophotometry range from analyzing biochemical substances to ensuring water safety, making it an indispensable tool in analytical laboratories.
Threshold Concentration
When we talk about threshold concentrations, we refer to the minimal concentration level of a substance that could pose a potential risk or detectability limit. This concept is crucial in contexts such as environmental safety and health sciences.
In our exercise, the focus is on a 50 parts per billion (ppb) threshold for an impurity in water.
  • Below this threshold, the substance is considered harmless.
  • Detecting concentrations above the threshold is significant to maintain public health standards.
With spectrophotometry's ability to detect concentrations much lower than this critical threshold, it suggests such a method is suitable for identifying when a substance surpasses safe levels. By comparing our spectrophotometer's detection capabilities with known thresholds, we can decide the adequacy of such methods for specific safety needs.

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Most popular questions from this chapter

A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, (b) greenhouse gas, \((\mathrm{c})\) photochemical smog, \((\mathbf{d})\) ozone depletion. Give a brief explanation of each term and identify one or two of the chemicals associated with each.

The water supply for a midwestern city contains the following impurities: coarse sand, finely divided particulates, nitrate ion, trihalomethanes, dissolved phosphorus in the form of phosphates, potentially harmful bacterial strains, dissolved organic substances. Which of the following processes or agents, if any, is effective in removing each of these impurities: coarse sand filtration, activated carbon filtration, aeration, ozonization, precipitation with aluminum hydroxide?

(a) Why is the fluorine present in chlorofluorocarbons not a major contributor to depletion of the ozone layer? (b) What are the chemical forms in which chlorine exists in the stratosphere following cleavage of the carbon-chlorine bond?

One mystery in environmental science is the imbalance in the "carbon dioxide budget." Considering only human activities, scientists have estimated that 1.6 billion metric tons of \(\mathrm{CO}_{2}\) is added to the atmosphere every year because of deforestation (plants use \(\mathrm{CO}_{2},\) and fewer plants will leave more \(\mathrm{CO}_{2}\) in the atmosphere). Another 5.5 billion tons per year is put into the atmosphere because of burning fossil fuels. It is further estimated (again, considering only human activities) that the atmosphere actually takes up about 3.3 billion tons of this \(\mathrm{CO}_{2}\) per year, while the oceans take up 2 billion tons per year, leaving about 1.8 billion tons of \(\mathrm{CO}_{2}\) per year unaccounted for. This "missing" \(\mathrm{CO}_{2}\) is assumed to be taken up by the "land." What do you think might be happening? [Sections \(18.1-18.3]\)

The hydroxyl radical, \(\mathrm{OH}\), is formed at low altitudes via the reaction of excited oxygen atoms with water: $$ \mathrm{O}^{*}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{OH}(g) $$ (a) Write the Lewis structure for the hydroxyl radical. (Hint: It has one unpaired electron.) Once produced, the hydroxyl radical is very reactive. Explain why each of the following series of reactions affects the pollution in the troposphere: (b) \(\mathrm{OH}+\mathrm{NO}_{2} \longrightarrow \mathrm{HNO}_{3}\) (c) \(\mathrm{OH}+\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{OOH}\) \(\mathrm{OOH}+\mathrm{NO} \longrightarrow \mathrm{OH}+\mathrm{NO}_{2}\) (d) \(\mathrm{OH}+\mathrm{CH}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3}\) \(\mathrm{CH}_{3}+\mathrm{O}_{2} \longrightarrow \mathrm{OOCH}_{3}\) \(\mathrm{OOCH}_{3}+\mathrm{NO} \longrightarrow \mathrm{OCH}_{3}+\mathrm{NO}_{2}\) (e) The concentration of hydroxyl radicals in the troposphere is approximately \(2 \times 10^{6}\) radicals per \(\mathrm{cm}^{3}\). This estimate is based on a method called long path absorption spectroscopy (LPAS), similar in principle to the Beer's law measurement discussed in the Closer Look essay on p. 564 , except that the path length in the LPAS measurement is \(20 \mathrm{~km}\). Why must the path length be so large?
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