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Chapter 18: Problem 81

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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 \(1.40\times10^{-7}\mathrm{M}\).
03

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.

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