The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 $\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\( in \)\mathrm{H}_{2} \mathrm{O}$

Step by step solution

01

Convert ppb to mass

5.0 ppb Hg is 5.0 μg Hg in 1.0 kg of the solution.

02

Calculate moles of solute

For Hg, the molar mass is 200.59 g/mol. We can find the moles of Hg using the mass: moles of Hg = \(\frac{5.0\,\mathrm{\mu g}}{200.59\,\mathrm{g/mol}} \times \frac{1\,\mathrm{g}}{10^6\,\mathrm{\mu g}}\) = \(2.49 \times10^{-11}\,\mathrm{mol}\)

03

Calculate the volume of the solution

We are given a mass of 1.0 kg. We can assume the solution is mainly water, so the density is 1 g/mL, therefore Volume = \( \frac{1.0\,\mathrm{kg}}{1\,\mathrm{g/mL}} \times \frac{10^3\,\mathrm{g}}{1\,\mathrm{kg}} \) = \(1000\,\mathrm{mL}\)

04

Calculate the molarity

Divide the moles of Hg by the volume of the solution (in liters): Molarity = \(\frac{2.49 \times 10^{-11}\,\mathrm{mol}}{1.0\,\mathrm{L}}\) = \(2.49 \times 10^{-11}\,\mathrm{M}\) #Solution b. - 1.0 ppb CHCl₃ in H₂O# Repeat the steps from solution a:

05

Convert ppb to mass

1.0 ppb CHCl₃ is 1.0 μg CHCl₃ in 1.0 kg of the solution.

06

Calculate moles of solute

For CHCl₃, the molar mass is 119.38 g/mol. Calculate the moles of CHCl₃: moles of CHCl₃ = \(\frac{1.0\,\mathrm{\mu g}}{119.38\,\mathrm{g/mol}} \times \frac{1\,\mathrm{g}}{10^6\,\mathrm{\mu g}}\) = \(8.38 \times10^{-12}\,\mathrm{mol}\)

07

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

08

Calculate the molarity

Molarity = \(\frac{8.38 \times 10^{-12}\,\mathrm{mol}}{1.0\,\mathrm{L}}\) = \(8.38 \times 10^{-12}\,\mathrm{M}\) #Solution c. - 10.0 ppm As in H₂O# Repeat the steps from solution a:

09

Convert ppm to mass

10.0 ppm As is 10.0 mg As in 1.0 kg of the solution.

10

Calculate moles of solute

For As, the molar mass is 74.92 g/mol. Calculate the moles of As: moles of As = \(\frac{10.0\,\mathrm{mg}}{74.92\,\mathrm{g/mol}} \times \frac{1\,\mathrm{g}}{10^3\,\mathrm{mg}}\) = \(1.34 \times10^{-4}\,\mathrm{mol}\)

11

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

12

Calculate the molarity

Molarity = \(\frac{1.34 \times 10^{-4}\,\mathrm{mol}}{1.0\,\mathrm{L}}\) = \(1.34 \times 10^{-4}\,\mathrm{M}\) #Solution d. - 0.10 ppm DDT (C₁₄H₉Cl₅) in H₂O# Repeat the steps from solution a:

13

Convert ppm to mass

0.10 ppm DDT is 0.10 mg DDT in 1.0 kg of the solution.

14

Calculate moles of solute

For DDT, the molar mass is 354.48 g/mol. Calculate the moles of DDT: moles of DDT = \(\frac{0.10\,\mathrm{mg}}{354.48\,\mathrm{g/mol}} \times \frac{1\,\mathrm{g}}{10^3\,\mathrm{mg}}\) = \(2.82 \times10^{-7}\,\mathrm{mol}\)

15

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

16

Calculate the molarity

Molarity = \(\frac{2.82 \times 10^{-7}\,\mathrm{mol}}{1.0\,\mathrm{L}}\) = \(2.82 \times 10^{-7}\,\mathrm{M}\)

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