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Chapter 13: Problem 97

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. (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

Expert verified
The molarity of lead in a 9.0 ppb lead solution is \(4.35 \times 10^{-12} M\). There are \(5.40 \times 10^{-3} g\) of lead in a swimming pool containing 9.0 ppb lead in 60 m³ of water.

Step by step solution

01

Understanding ppb and Molarity

ppb (parts per billion) is a unit used to express the concentration of a substance in a solution. 1 ppb means 1 part of solute per billion (10⁹) parts of solution. Molarity (M) is the concentration of a substance in a given volume of solution, expressed as moles of solute per liter of solution. To find molarity, we need to divide the moles of solute by the volume of the solution in liters.
02

Convert the concentration of lead from ppb to molarity

Given, the concentration of lead in drinking water is 9.0 ppb. To convert ppb to molarity we can use the formula: Molarity (M) = (ppb × 10⁻⁹) / (Molar mass of solute × Volume of solution in L) We are given the concentration of lead in ppb and need the molarity of lead (Pb). The molar mass of lead (Pb) is 207.2 g/mol. Since the volume of the solution is not provided, we assume 1 L of the solution. Hence, the molarity of lead in a 9.0 ppb solution can be calculated as: Molarity (M) = (9.0 × 10⁻⁹) / (207.2 × 1)
03

Calculate the molarity of lead in a 9.0 ppb solution

Molarity (M) = (9.0 × 10⁻⁹) / 207.2 = 4.35 × 10⁻¹² M The molarity of lead in a 9.0 ppb lead solution is 4.35 × 10⁻¹² M.
04

Calculate the grams of lead in the swimming pool

We are given the volume of water in the swimming pool as 60 m³, which is equivalent to 60,000 L (since 1 m³ = 1,000 L). Now, to find the number of moles of lead in 60,000 L of water, we can use the molarity formula: Moles of lead = Molarity × Volume of solution in L Moles of lead = (4.35 × 10⁻¹² M) × (60,000 L) Next, to find the grams of lead in 60,000 L of water, we can use the formula: grams of lead = moles of lead × molar mass of lead grams of lead = (4.35 × 10⁻¹² M × 60,000 L) × 207.2 g/mol
05

Find the grams of lead in the swimming pool

grams of lead = (4.35 × 10⁻¹² M × 60,000 L) × 207.2 g/mol = 5.40 × 10⁻³ g There are 5.40 × 10⁻³ grams of lead in the swimming pool containing 9.0 ppb lead in 60 m³ of water.

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