Question

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 \(\mathrm{H}_{2} \mathrm{O}\)

Short answer

The molarities of the given aqueous solutions are: a. \(2.49 \times 10^{-11}\,\mathrm{M}\) of Hg b. \(8.38 \times 10^{-12}\,\mathrm{M}\) of CHCl₃ c. \(1.34 \times 10^{-4}\,\mathrm{M}\) of As d. \(2.82 \times 10^{-7}\,\mathrm{M}\) of DDT

Step-by-step solution

Convert ppb to mass

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

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 = \(\frac{5.0\,\mathrm{\mu g}}{200.59\,\mathrm{g/mol}} \times \frac{1\,\mathrm{g}}{10^6\,\mathrm{\mu g}}\) = \(2.49 \times10^{-11}\,\mathrm{mol}\)

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 = \( \frac{1.0\,\mathrm{kg}}{1\,\mathrm{g/mL}} \times \frac{10^3\,\mathrm{g}}{1\,\mathrm{kg}} \) = \(1000\,\mathrm{mL}\)

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:

Convert ppb to mass

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

Calculate moles of solute

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

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

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:

Convert ppm to mass

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

Calculate moles of solute

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

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

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:

Convert ppm to mass

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

Calculate moles of solute

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

Calculate the volume of the solution

Volume = 1000 mL (same as solution a)

Calculate the molarity

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

Key concepts

Parts Per Million (ppm)

Parts per million, abbreviated as ppm, is a way to express very dilute concentrations of substances. Imagine you have a million tiny parts in a solution. In layman's terms, 1 ppm indicates that one of these tiny parts is the solute, while the remaining 999,999 parts are the solvent. It's like finding one specific grain of sand on a large beach. It’s a commonly used measurement in environmental chemistry, especially when dealing with contaminants or impurities in air, water, or soil.

To calculate ppm by mass, you use the formula:\[\text{ppm} = \frac{\mu \text{g solute}}{\text{g solution}} = \frac{\text{mg solute}}{\text{kg solution}}\]
The formula highlights how ppm relates the mass of the solute to the total mass of the solution. In very dilute aqueous solutions, 1.0 ppm is roughly equivalent to 1.0 microgram of solute per 1.0 milliliter of solution, which weighs nearly 1 gram.

This equivalency is particularly useful when testing the purity of water in environmental studies, where even a small amount of contaminant can be significant.

Parts Per Billion (ppb)

Parts per billion, or ppb, is another unit used for measuring even more dilute concentrations than ppm. To give you an idea of scale, 1 ppb implies one part of solute in a billion parts of the solution. It’s akin to finding a single leaf in a forest with a billion trees.

The conversion for ppb is similar to that for ppm, but it accounts for a greater level of dilution. Often, chemists use ppb to measure pollutants that might be harmful even at incredibly low levels. Such precision is essential for substances that have a dramatic effect even in minuscule amounts, like certain heavy metals or toxic organic compounds.

To put it in a formulaic perspective, 1 ppb is expressed similarly to ppm but the divisor is larger:\[\text{ppb} = \frac{\mu \text{g solute}}{\text{kg solution}}\]or\[\frac{\text{ng solute}}{\text{g solution}}\]
This allows for a precise and consistent way to quantify substances when they appear in extraordinarily low concentrations. Chemical safety and environmental guidelines often rely on ppb measurements for clarity and accuracy.

Aqueous Solutions

An aqueous solution is any solution where water acts as the solvent. Water, being the 'universal solvent', has unique properties that make it capable of dissolving a wide array of substances, ranging from salts to gases. This property is fundamental to numerous biological and chemical processes on Earth.

In the context of ppm and ppb, aqueous solutions often come up because we commonly analyze water for contaminants. Water's density, conveniently around 1 g/mL, simplifies calculations, as 1 liter of water approximately equals 1 kg. This relationship is typically leveraged to interpret concentration measurements like ppm and ppb with ease.

When calculating concentrations in aqueous solutions, it’s essential to account for the solute dissolved in water. Depending on the solute's properties, it might affect the physical or chemical characteristics of the solution. A thorough understanding of how aqueous solutions work is vital in disciplines like chemistry and environmental science, where water-based reactions are a common focus.