Problem 81 - Chemistry: The Central Science in SI Units, Global Edition

Question

Common lab spectrometers can detect absorbance down to 0.0002 with good reliability. Consider a dissolved harmful organic substance with a molar mass of \(120.5 \mathrm{~g} / \mathrm{mol}\), which can be detected in this spectrometer. It shows an extinction coefficient of \(\varepsilon=1.43 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at 320 \(\mathrm{nm}\), its absorption maximum (A Closer Look, p. 620). (a) Calculate the minimum concentration of the organic substance detectable by this spectrometer (path length \(1 \mathrm{~cm}\) ). (b) Convert the minimum observable molarity to ppb.

Step-by-Step Solution

Answer

The minimum detectable concentration of the organic substance by the spectrometer is \(1.40\times10^{-7}\mathrm{M}\), which is equivalent to approximately 16.87 ppb.

Step 1: Apply Beer's Law to calculate the minimum detectable concentration

Beer's Law is given by the formula A = εcl, where A is the absorbance, ε is the molar absorptivity (extinction coefficient), c is the concentration, and l is the path length. In this case, the absorbance is given as 0.0002, the extinction coefficient (ε) is \(1.43\times10^3 \mathrm{M}^{-1} \mathrm{~cm}^{-1}\), and the path length is 1 cm. Rearrange the formula to solve for the concentration (c): \[c = \frac{A}{εl}\] Now plug the values of A, ε, and l into the equation and solve for c.

Step 2: Calculate the minimum detectable concentration

Substitute the values (A = 0.0002, ε = \(1.43\times10^3 \mathrm{M}^{-1} \mathrm{~cm}^{-1}\), and l = 1 cm) into the equation. \[c = \frac{0.0002}{(1.43\times10^3 \mathrm{M}^{-1} \mathrm{~cm}^{-1})(1\mathrm{~cm})}\] Calculate the concentration: \[c = 1.40\times10^{-7}\mathrm{M}\] The minimum detectable concentration of the organic substance is \(1.40\times10^{-7}\mathrm{M}\).

Step 3: Convert the minimum observable molarity to ppb

To convert molarity to ppb (parts per billion), follow the steps below: 1. Convert the molarity to moles per liter: \(c=1.40\times10^{-7}\mathrm{M}\) is already in moles per liter. 2. Convert moles per liter to grams per liter using the molar mass: Molar mass = \(120.5 \mathrm{~g} / \mathrm{mol}\) \[1.40\times10^{-7}\mathrm{~moles}\times \frac{120.5\mathrm{~g}}{1\mathrm{~mole}}\] 3. Converting grams per liter (g/L) to micrograms per liter (µg/L) which is equivalent to parts per billion (ppb), by multiplying with \(10^6\). \[1.40\times10^{-7}\times 120.5 \times 10^6\] Calculate the value in ppb: \[ppb = 16.87\] The minimum detectable concentration of the organic substance in ppb is approximately 16.87 ppb.

Key Concepts

Molar Absorptivity

When dealing with solutions and their ability to absorb light, we encounter a key concept known as molar absorptivity, also referred to as the extinction coefficient. It is denoted by the Greek letter epsilon (\(\varepsilon\)). This parameter is essential as it indicates how strongly a chemical species absorbs light at a particular wavelength. A higher molar absorptivity means that the substance is more effective at absorbing photons, resulting in a higher absorbance for a given concentration and path length.
Molar absorptivity is typically expressed in units of \(\text{M}^{-1} \text{cm}^{-1}\), which beautifully illustrates its dependence on both the concentration of the solution and the path length through which light travels. For instance, in our given problem, the substance has a molar absorptivity of \(1.43 \times 10^3\) \(\text{M}^{-1} \text{cm}^{-1}\), meaning it heavily absorbs light at 320 nm. Understanding molar absorptivity is critical when applying Beer's Law in spectrophotometry discussions.

Spectrophotometry

Spectrophotometry is a fascinating analytical method used to measure how light interacts with a material to assess its concentration. It quantifies the intensity of light passing through a sample by determining the absorbance, employing Beer's Law as a foundation. Beer's Law is expressed as \(A = \varepsilon cl\), where \(A\) is absorbance, \(\varepsilon\) is molar absorptivity, \(c\) is the concentration, and \(l\) is the path length. This formula highlights a direct relationship between absorbance and concentration, given a fixed path length and molar absorptivity.
In practical applications, spectrophotometry can detect small concentrations of a substance if it strongly absorbs light at a specific wavelength, as evident in the case of the organic substance in our exercise. This method is widely favored in various scientific fields due to its precision and reliability in determining concentrations of solutes in a solution.

Concentration Conversion

Converting concentrations between different units can seem daunting but becomes straightforward with the right understanding. In our exercise, we converted from molarity, a measure of moles per liter, to parts per billion (ppb), which are micrograms per liter. This conversion is necessary when communicating findings at very low concentrations, like those in environmental analysis or food safety evaluations.
To convert molarity to ppb, follow these steps:

• First, calculate the mass in grams by multiplying the molarity by the molar mass of the substance.
• Secondly, convert this mass to micrograms by multiplying by \(10^6\), since there are 1,000,000 micrograms in a gram.

By applying this conversion, the exercise showed that a molarity of \(1.40 \times 10^{-7} \text{M}\) corresponds to approximately 16.87 ppb. Understanding and performing these conversions is crucial in adapting lab results to real-world applications and standards.