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 \(1.0\) 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. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Short answer

The molarity of each solution is: a. \(2.5 \times 10^{-8} \mathrm{M}\) b. \(8.39 \times 10^{-9} \mathrm{M}\) c. \(1.33 \times 10^{-7} \mathrm{M}\) d. \(2.82 \times 10^{-10} \mathrm{M}\)

Step-by-step solution

Convert ppb to mass per volume

Since \(1\) ppb \(\mathrm{solute} = 1 \times 10^{-9} \frac{\mu\mathrm{g} \mathrm{solute}}{1\mathrm{mL}}\), we have \(5.0 \mathrm{ppb} \mathrm{Hg} = 5.0 \times 10^{-9} \frac{\mu \mathrm{g}}{1\mathrm{mL}}\).

Convert mass to moles

Using the molar mass of Hg as \(200.59 \frac{\mathrm{g}}{\mathrm{mol}}\), we convert mass of Hg to moles: \[ \frac{5.0 \times 10^{-9} \mu \mathrm{g}}{200.59 \frac{\mathrm{g}}{\mathrm{mol}}} = 2.5 \times 10^{-11} \mathrm{mol} \]

Find molarity

Since molarity is defined as moles of solute per liter of solution, \(\mathrm{Molarity} = \frac{2.5 \times 10^{-11} \mathrm{mol}}{0.001 \mathrm{L}} = 2.5 \times 10^{-8} \mathrm{M}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\)

Convert ppb to mass per volume

In this case, \(1.0 \mathrm{ppb} \mathrm{CHCl_3} = 1.0 \times 10^{-9} \frac{\mu\mathrm{g}}{1\mathrm{mL}}\).

Convert mass to moles

Using the molar mass of CHCl3 as \(119.37 \frac{\mathrm{g}}{\mathrm{mol}}\), we convert mass of CHCl3 to moles: \[ \frac{1.0 \times 10^{-9} \mu\mathrm{g}}{119.37 \frac{\mathrm{g}}{\mathrm{mol}}} = 8.39 \times 10^{-12} \mathrm{mol} \]

Find molarity

In this case, \(\mathrm{Molarity} = \frac{8.39 \times 10^{-12} \mathrm{mol}}{0.001 \mathrm{L}} = 8.39 \times 10^{-9} \mathrm{M}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\)

Convert ppm to mass per volume

Since \(1\) ppm \(\mathrm{solute} = 1\frac{\mu\mathrm{g}}{1 \mathrm{mL}}\), we have \(10.0 \mathrm{ppm} \text{As} = 10.0 \frac{\mu\mathrm{g}}{1\mathrm{mL}}\).

Convert mass to moles

Using the molar mass of As as \(74.92 \frac{\mathrm{g}}{\mathrm{mol}}\), we convert mass of As to moles: \[ \frac{10.0 \mu\mathrm{g}}{74.92 \frac{\mathrm{g}}{\mathrm{mol}}} = 1.33 \times 10^{-10} \mathrm{mol} \]

Find molarity

In this case, \(\mathrm{Molarity} = \frac{1.33 \times 10^{-10} \mathrm{mol}}{0.001 \mathrm{L}} = 1.33 \times 10^{-7} \mathrm{M}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Convert ppm to mass per volume

In this case, \(0.10 \mathrm{ppm}\) DDT \(= 0.10 \frac{\mu\mathrm{g}}{1\mathrm{mL}}\).

Convert mass to moles

Using the molar mass of DDT as \(354.49 \frac{\mathrm{g}}{\mathrm{mol}}\), we convert mass of DDT to moles: \[ \frac{0.10 \mu\mathrm{g}}{354.49 \frac{\mathrm{g}}{\mathrm{mol}}} = 2.82 \times 10^{-13} \mathrm{mol} \]

Find molarity

In this case, \(\mathrm{Molarity} = \frac{2.82 \times 10^{-13} \mathrm{mol}}{0.001 \mathrm{L}} = 2.82 \times 10^{-10} \mathrm{M}\) In conclusion, we have found the molarity of each solution: a. \(2.5 \times 10^{-8} \mathrm{M}\) b. \(8.39 \times 10^{-9} \mathrm{M}\) c. \(1.33 \times 10^{-7} \mathrm{M}\) d. \(2.82 \times 10^{-10} \mathrm{M}\)

Key concepts

Understanding Parts Per Million (ppm)

The concept of 'parts per million' (ppm) is a way of expressing very dilute concentrations of substances. Just as a percentage is an amount out of a hundred, ppm denotes the amount of substance in a million parts of the total. For example, 1 ppm means 1 part in 1,000,000 parts, often thought of as one drop of water diluted into about 50 liters (roughly a small bathtub).

In chemistry, this can be applied to the mass of a substance per mass of the solution, or mass per volume for aqueous solutions. Converting ppm to a molar concentration involves first converting the mass of solute to moles using the molar mass, then dividing by the volume of the solution in liters. This is useful when dealing with pollutant levels, mineral concentrations in water, or any situation where a substance is present in minute quantities.

Parts Per Billion (ppb) - The Ultra-dilute Measure

While parts per million (ppm) can measure everyday chemical concentrations, some substances are present in even smaller amounts. Parts per billion (ppb) is one such measure, quantifying the number of units of solute per billion units of total solution. One ppb corresponds to one microgram of solute per liter of solution or about one pinch of salt in an Olympic-sized swimming pool.

The process to find molarity from ppb is parallel to ppm but at a thousandfold finer scale; first, you convert ppb to micrograms per milliliter, then convert micrograms to moles using the substance's molar mass, and finally express this as moles per liter (molarity). Precision is key when working with such tiny quantities, as they are often involved in environmental standards and safety limits.

Molar Mass Calculation: The Foundation of Stoichiometry

The molar mass of a substance is the weight in grams of one mole (Avogadro's number, or approximately 6.022 x 10^23 particles) of that substance. Molar mass can be calculated by summing up the atomic masses (found on the periodic table) of each element present in the compound multiplied by the number of atoms of each element. For instance, water (H2O) has a molar mass of approximately 18 g/mol, which is calculated by adding twice the atomic mass of hydrogen (approximately 1 g/mol each) to the atomic mass of oxygen (approximately 16 g/mol).

Molar mass is vital for converting between mass and moles of a substance, which is a fundamental step in finding molarity, as shown in the step-by-step solution provided. Without this calculation, it would be impossible to communicate how much of a substance is present in a solution.

Concentration Conversion: From ppm/ppb to Molarity

Concentration conversion is essential for working across different units of concentration in chemistry, such as converting from ppm or ppb to molarity, and vice versa. To do this, you calculate how many moles of a substance are present per liter of solution. Here's a simplified process:

• Convert the concentration to mass per volume eg., ppm or ppb to micrograms per milliliter.
• Utilize the molar mass of the solute to convert the mass to moles.
• Since molarity is moles per liter, convert the volume from mL to L if necessary, and find the molarity.

When you understand these conversions, you are better equipped to grasp how solutions interact in everything from chemical reactions to pharmaceutical formulations.