Question

Silver ion has an average concentration of \(28 \mathrm{ppb}\) (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 \(\left.1.0 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

Short answer

Molality is \(2.597 \times 10^{-7} \text{ mol/kg}\); need \(3.57 \times 10^6 \text{ L}\) to recover 100 g.

Step-by-step solution

Understanding Parts per Billion (ppb)

Parts per billion (ppb) is a unit of concentration that indicates the number of parts of a component per billion parts of the whole. In this problem, 28 ppb indicates that there are 28 grams of silver ions per 1 billion (\(10^9\)) grams of water.

Calculate the Mass of Water

Since the concentration is 28 ppb, we have 28 grams of \(Ag^+\) in \(10^9\) grams of water. Given the density of water is \(1.0 \text{ g/cm}^3\), hence \(10^9\) grams of water is equivalent to \(10^9\) mL or \(10^6\) L.

Convert Mass to Moles

The molar mass of silver (Ag) is \(107.87 \text{ g/mol}\). To find the number of moles of silver ions, we convert 28 grams into moles: \[ n = \frac{28 \text{ g}}{107.87 \text{ g/mol}} \approx 0.2597 \text{ moles}. \]

Calculate Molality

Molality (\(m\)) is defined as moles of solute per kilogram of solvent. We have 0.2597 moles of \(Ag^+\) in \(10^6\) kg of water. Therefore, the molality is \[ m = \frac{0.2597 \text{ moles}}{10^6 \text{ kg}} \approx 2.597 \times 10^{-7} \text{ mol/kg}. \]

Determine Moles Needed for 100 g of Silver

To compute the amount of water required to recover 100g of silver, first convert 100 grams of silver to moles: \[ n = \frac{100 \text{ g}}{107.87 \text{ g/mol}} \approx 0.927 \text{ moles}. \]

Calculate Volume of Water for 100 g Silver

Given that 0.2597 moles are present in \(10^6\) L of water, we can set up a proportion to find the volume, \(V\), needed for 0.927 moles: \[ \frac{0.2597}{10^6} = \frac{0.927}{V}. \] Solving for \(V\) gives \[ V \approx \frac{0.927 \times 10^6}{0.2597} \approx 3.57 \times 10^6 \text{ L}. \]

Key concepts

Understanding parts per billion (ppb)

Parts per billion (ppb) is a way to express very small concentrations of solutes in solutions. This unit is especially useful when dealing with extremely dilute solutions, such as certain elements in water. The term "parts per billion" simply means that for every one billion parts of a solution, one part is the solute.

In this context, if you have a 28 ppb concentration of silver ions in water, it means there are 28 grams of silver ions within 10^9 grams of water. This unit helps chemists and environmental scientists measure the tiny amounts of substances found in large bodies of water.

Here's something to note:

• One billion is a large number: 1,000,000,000 (or 10^9). Thus, ppb is used for measuring small concentrations like pollutants or trace elements in a mixture.
• An alternative unit for low concentrations is "parts per million" (ppm), which is 1,000 times greater than ppb.

Using ppb can help effectively communicate and track the presence of elements within environments where precision is crucial.

Exploring Molality

Molality is a concept that describes the concentration of a solute relative to the mass of the solvent in a solution. Specifically, molality (\(m\)) is defined as the moles of solute per kilogram of solvent. The formula is:

\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]
Unlike molarity, molality is independent of temperature because it involves mass rather than volume, which can expand or contract with temperature changes.

• Use molality when you need to know how the concentration changes with temperature.
• In this exercise, the molality of silver ion in water was found to be approximately 2.597×10^-7 mol/kg.
• It involves using a large mass of water as the solvent since the concentration is stated as parts per billion.

Overall, understanding molality assists in determining the precise concentration of a substance within a solvent, allowing for accurate predictions and calculations in laboratory settings.

The Mechanics of Molar Conversion

Molar conversion is fundamental in chemistry, allowing the transition from mass of a substance to its amount in moles and vice versa. This conversion is key because chemical reactions often take place in terms of moles, according to the balanced chemical equations.

The key formula is:

• \[ n = \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \]
• Here, "n" is the number of moles. It's essential for determining how many particles (atoms or molecules) of a substance you have.

In this exercise, for example, the molar mass of silver is 107.87 g/mol, which was used to find the moles of silver needed from the given mass.
Each element has a unique molar mass, providing the basis to convert from grams to moles. This allows the accurate calculation of reactants and products in chemical reactions.
Developing an understanding of molar conversion assists in consistent and precise computations in various experimental and industrial chemical processes.

Density of Water: The Gold Standard

Density is a measure of how much mass is contained within a given volume. The density of water is crucial not only in chemistry but across various scientific fields. Water's density is typically approximated as 1.0 g/cm³ or 1000 kg/m³ at 4°C. This simplification allows for straightforward conversions between volume and mass.

Key points include:

• When the density is 1.0 g/cm³, then 1 liter (1000 mL) of water weighs 1000 grams (or 1 kilogram).
• This concept simplifies measurements in laboratory experiments, where water serves as the solvent.

In the context of this exercise, knowing the density of water allows us to convert massive weights to volumes accurately. For example, 10⁹ grams of water equate to 10⁶ liters due to this constant density.
Familiarity with the density of water ensures precise and reliable results in both practical and theoretical calculations, promoting consistency in experimental data.