Question

The maximum allowable concentration of lead in drinking water is \(9.0\) ppb. (a) Calculate the molarity of lead in a 9.0-ppb solution. What assumption did you have to make in your calculation? (b) How many grams of lead are in a swimming pool containing \(9.0\) ppb lead in \(60 \mathrm{~m}^{3}\) of water?

Short answer

The molarity of lead in a 9.0-ppb solution is approximately \(4.35 \times 10^{-11}\) mol/L, assuming the density of water is 1 g/mL. There are approximately \(5.4 \times 10^{-4}\) grams of lead in a swimming pool containing 9.0 ppb lead in 60 cubic meters of water.

Step-by-step solution

Convert ppb to moles per liter

Given the concentration of lead in drinking water as 9.0 ppb, we first need to convert this to molarity (moles per liter). One assumption we have to make here is that the density of water is approximately 1 gram per milliliter (g/mL) or 1000 grams per liter (g/L). 1 ppb = 1/1,000,000,000 9.0 ppb = \(\frac{9.0}{1,000,000,000}\) g/L Now, we need to convert grams to moles using the molar mass of lead (Pb), which is 207.2 g/mol.

Calculate molarity of lead

To calculate the molarity of lead, divide the grams of lead per liter by the molar mass of lead. Molarity (M) = \(\frac{9.0 \times 10^{-9} \text{ g/L}}{207.2 \text{ g/mol}}\) M = \(\frac{9.0}{207.2 \times 10^9}\) mol/L ≈ \(4.35 \times 10^{-11}\) mol/L So, the molarity of lead in a 9.0-ppb solution is approximately \(4.35 \times 10^{-11}\) mol/L.

Calculate volume of water in the swimming pool

It's given that the volume of the swimming pool is 60 cubic meters. We need to convert this into liters. We know that 1 cubic meter equals 1000 liters. Volume = 60 m³ × 1000 L/m³ = 60,000 L

Calculate the moles of lead in the swimming pool

To find the moles of lead in the swimming pool, multiply the molarity of lead by the volume of water in the swimming pool: Moles of lead = Molarity × Volume Moles of lead = \(4.35 \times 10^{-11}\) mol/L × 60,000 L ≈ \(2.61 \times 10^{-6}\) mol

Calculate the grams of lead in the swimming pool

Now, we will convert the moles of lead back to grams using the molar mass: Grams of lead = Moles of lead × Molar mass Grams of lead = \(2.61 \times 10^{-6}\) mol × 207.2 g/mol ≈ 5.4 × 10^{-4} g Therefore, there are approximately 5.4 × 10^{-4} grams of lead in a swimming pool containing 9.0 ppb lead in 60 cubic meters of water.

Key concepts

PPB to Molarity Conversion

Understanding the conversion of parts per billion (ppb) to molarity is essential in the analysis of substances like pollutants in environmental chemistry. Molarity, denoted as M, represents the number of moles of a solute per liter of solution. To convert ppb, which is a measure of concentration by mass, to molarity, two key steps are involved.

Firstly, recognize that 1 ppb corresponds to 1 part of the substance in 1 billion parts of the total solution. To contextualize, if you have 1 gram of solute in 1 billion grams of solution, that's 1 ppb. Since 1 billion grams of solution is equivalent to 1 million liters (assuming the density of water is roughly 1 g/mL), we start by converting the ppb value to grams per liter (g/L).
For example, a concentration of 9.0 ppb is equal to \(\frac{9.0 g}{1,000,000,000 L}\) or \(9.0 \times 10^{-9} g/L\).

The second step involves converting the mass in grams to moles, which requires knowledge of the molar mass of the substance. By dividing the grams per liter by the molar mass (in g/mol), you obtain the concentration in molarity (mol/L). This is a critical step in molarity calculation and highlights the importance of accurate molar mass utilization.

Molar Mass Utilization

The molar mass of a substance is used to convert between the amount of substance in moles and its mass in grams, thus playing a vital role in chemical calculations. It is defined as the mass of one mole of a substance and is usually expressed in grams per mole (g/mol).

To use molar mass effectively, you need to understand the atomic or molecular weight of the substance in question and be aware that the molar mass of an element is numerically equivalent to its atomic weight from the periodic table. For instance, the molar mass of lead (Pb) is 207.2 g/mol, so one mole of lead weighs 207.2 grams.
With this information, you can perform the second critical step in converting ppb to molarity: dividing the mass of the substance (in grams) by its molar mass will give you the number of moles in that mass. For example, using the molar mass of lead,\( \frac{9.0 \times 10^{-9} g/L}{207.2 g/mol} \) gives a molarity of approximately \(4.35 \times 10^{-11} mol/L\).

Accurate determination of molar mass is essential because it ensures the precise conversion from mass to moles, important for all quantitative analyses in chemistry, especially when dealing with minuscule concentrations such as ppb levels of contaminants.

Lead Contamination Quantification

Quantifying lead contamination becomes particularly significant in environmental monitoring, where regulations specify the maximum allowable levels of such toxic substances in drinking water or other resources. Lead, with its potential to cause severe health problems, has its contamination levels closely regulated.

To quantify lead contamination in a given volume of water, it's not enough to state the concentration in ppb; it must be expressed in a mass or molar quantity for practical applications. As seen in the exercise, the process involves a series of conversions – from ppb to grams per liter, then to molarity, and finally to total mass in the volume of water in question.
For example, to ascertain how many grams of lead are present in a swimming pool, one would calculate the molarity of lead in the water as demonstrated previously, then multiply that molarity by the volume of the pool in liters. The resulting moles of lead are then converted back into grams using the molar mass of lead. Such calculations aid in evaluating whether the contamination is within legal safety limits and determining the necessary action to remediate the pollution if those limits are surpassed.

Therefore, understanding ppb to molarity conversion, molar mass utilization, and contaminant quantification not only fulfills academic objectives but also equips students with the knowledge to deal with real-world environmental concerns.