Problem 12 - Chemistry and Chemical Reactivity

Question

Silver ion has an average concentration of 28 ppb (parts per billion) in U.S. water supplies. (a) What is the molality of the silver ion? (b) If you wanted \(1.0 \times 10^{2} \mathrm{g}\) of silver and could recover it chemically from water supplies, what volume of water in liters, would you have to treat? (Assume the density of water is \(1.0 \mathrm{g} / \mathrm{cm}^{3} .\) )

Step-by-Step Solution

Answer

Molality is ≈ 2.6 × 10^{-7} mol/kg, and the required volume is ≈ 3.56 million liters.

Step 1: Understanding Parts Per Billion

Parts per billion (ppb) is a unit of concentration that measures the amount of solute in one billion parts of the solution. Here, it means 28 grams of silver ions in one billion grams of water.

Step 2: Converting ppb to Molality

To find the molality, we need the number of moles of silver ions per kilogram of water. The formula to convert from ppb to molality is: \( \text{Molality} = \frac{\text{ppb} \times 10^{-9}}{\text{Molar mass of solute in g/mol}} \times 1000 \). The molar mass of silver (Ag) is approximately 107.87 g/mol.

Step 3: Calculate Number of Moles in 1 kg Water

Using the formula from the previous step, we calculate the number of moles: \( \text{Number of moles} = \frac{28 \times 10^{-9} \times 1000}{107.87} \approx 2.6 \times 10^{-7} \text{ mol/kg} \).

Step 4: Volume Calculation for Required Mass of Silver

We need to find how many grams of silver 1 liter of water contains. Since 1 liter of water is 1 kg (due to density), the concentration is \(2.6 \times 10^{-7}\) moles per kg of water. Multiply by the molar mass of Ag to find grams per L: \( 2.6 \times 10^{-7} \times 107.87 \approx 2.80862 \times 10^{-5} \) g/L.

Step 5: Calculating Required Volume of Water

Given you want 100 g of silver, find the volume of water: \( \frac{100}{2.80862 \times 10^{-5}} \approx 3.56 \times 10^{6} \) L.

Key Concepts

Parts Per Billion (ppb)

Understanding the concept of parts per billion (ppb) is crucial when dealing with very dilute solutions, like the concentration of silver ions in water. Essentially, ppb is a way to express tiny amounts of a substance in a large volume of solution. When we say there are 28 ppb of silver ions in water, it means there are 28 grams of silver ions in one billion grams of water.

This unit of measurement is most commonly used in environmental science and chemistry when the concentrations are so low that other measures like parts per million (ppm) would not be practical. For quick conversion, remember this: 1 ppb is equivalent to 1 microgram (µg) of solute per liter of solution, assuming the density is similar to water.

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For silver (Ag), the molar mass is approximately 107.87 g/mol.

Think of molar mass as the bridge between the microscopic scale (atoms and ions) and the macroscopic scale (grams). It's calculated by summing the atomic masses of all atoms in a molecule. For an element like silver, its molar mass is simply the atomic weight found on the periodic table.

Understanding molar mass is essential for tasks like converting between grams and moles, especially when determining concentrations such as molality.

Density

Density is a measure of how much mass an object contains within a given volume. It's typically expressed in units like grams per cubic centimeter (g/cm³) or kilograms per liter (kg/L). In this exercise, the density of water is given as 1.0 g/cm³, which conveniently equals 1 kg/L.

This concept is helpful when you want to determine how much a liquid weighs, given its volume. For example, knowing that 1 liter of water has a mass of 1 kg simplifies calculations in aqueous solutions.

When dealing with solutions, density allows you to convert between mass and volume, crucial for solving problems involving concentrations and dilutions.

Mole Calculation

Mole calculations involve converting between the number of particles (atoms, molecules, ions) and the amount of substance in moles, a fundamental unit in chemistry. We use the formula involving molar mass to relate grams to moles: \( ext{moles} = \frac{ ext{mass in grams}}{ ext{molar mass in g/mol}} \).

In our exercise, to find the molality of silver ions, we must calculate the number of moles in one kilogram of water. A clear understanding of this conversion allows us to express concentrations in a meaningful way, such as molality, which is moles of solute per kilogram of solvent. This method helps solve problems involving substances in solutions, making it easier to handle chemical reactions, concentrations, and other properties in aqueous solutions.

Remember, mole calculations give you a better grasp on how much of a substance is present beyond just its mass.