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

(a) Molality is \(2.595 \times 10^{-7} \text{ mol/kg}\); (b) Treat 3.57 million liters of water.

Step 1: Convert Concentration to Mass

The concentration of silver ion (Ag⁺) is 28 ppb, meaning there are 28 grams of silver in one billion grams of water. First, convert this concentration into grams per liter (g/L) using the definition of ppb, noting that 1 billion grams is equivalent to 1 million liters, given the density of water is 1 g/cm³ or 1000 g/L. Therefore, 28 ppb equals \(28 \times 10^{-6} \) g/L.

Step 2: Calculate Molality

Molality (m) is calculated using the formula \( m = \frac{n}{m_{solvent}} \), where \( n \) is the number of moles of solute and \( m_{solvent} \) is the mass of the solvent in kilograms. First, find the moles of silver ion using its molar mass \(107.87 \text{ g/mol} \):\[ n = \frac{28 \times 10^{-6} \text{ g/L}}{107.87 \text{ g/mol}} = 2.595 \times 10^{-7} \text{ mol/L}\]Considering 1 liter of water weighs approximately 1 kilogram, the molality is \(m = 2.595 \times 10^{-7} \text{ mol/kg} \).

Step 3: Calculate Moles of Silver Needed

To obtain 100 g of silver, use the formula \( n = \frac{m}{M} \), where \( m \) is the mass of silver needed and \( M \) is the molar mass of silver. This results in:\[n = \frac{100 \text{ g}}{107.87 \text{ g/mol}} \approx 0.927 \text{ mol}\]Thus, we need approximately 0.927 moles of silver.

Step 4: Find Volume of Water Required

Using the previously calculated moles of silver ion in water (\( 2.595 \times 10^{-7} \text{ mol/L} \)), find the volume of water needed to recover 0.927 moles:\[\text{Volume} = \frac{0.927 \text{ mol}}{2.595 \times 10^{-7} \text{ mol/L}} \approx 3.57 \times 10^{6} \text{ L}\]Therefore, you would need approximately 3.57 million liters of water.

Key Concepts

Concentration Conversion

When dealing with small concentrations, often recorded in parts per billion (ppb), understanding how to convert this to more manageable units is crucial. In the context of the exercise, the concentration of the silver ion is given as 28 ppb, which stands for 28 parts of solute (silver) per one billion parts of solvent (water). To work efficiently with this number, a useful strategy is to convert it into grams per liter (g/L), which is a more common concentration unit for laboratory calculations.
To convert from ppb to g/L, we first recognize that 1 billion grams of water translates to 1 million liters if we assume the density of water as 1 g/cm³, which equals 1000 g/L. This conversion then simplifies our task to transform 28 ppb into 28 ppm (parts per million) based on the earlier definition.** So, 28 ppb becomes:**

• \( \frac{28 \text{ g in 1 billion g of water}}{1 \text{ million liters}} = 28 \times 10^{-6} \text{ g/L} \)

This conversion is foundational to compute molality, as the subsequent steps rely on knowing the silver content per liter.

Moles Calculation

Once we have the concentration of silver in g/L, the next step is to translate grams into moles, since scientific chemistry calculations often use moles as the unit for measuring the amount of substance. Moles are powerful because they link the macroscopic and atomic worlds, letting us work with everyday quantities while thinking in terms of atoms and molecules.
To compute the number of moles from the given grams per liter, we use the formula:

• \( n = \frac{\text{given mass (g)}}{\text{molar mass (g/mol)}} \)

The molar mass of silver is 107.87 g/mol. With the concentration in grams per liter already known, we find:
\[ n = \frac{28 \times 10^{-6} \text{ g/L}}{107.87 \text{ g/mol}} = 2.595 \times 10^{-7} \text{ mol/L} \]
This conversion to moles is essential for determining the molality in the next step and for any further calculations involving chemical reactions or recovery processes.

Volume Calculation

Volume calculation becomes necessary when determining how much solvent is needed to recover a specific amount of solute. In the exercise, we aim to obtain 100 grams of silver. To determine the corresponding volume of water required to extract this quantity, we start by knowing the number of moles we want to extract.
The exercise tells us that obtaining 100 grams of silver requires:
\[ n = \frac{100 \text{ g}}{107.87 \text{ g/mol}} \approx 0.927 \text{ mol} \]
Next, using the molarity (moles per liter) we've already determined, the required volume of water can be calculated. This involves rearranging the molarity formula to solve for volume:

• \( \text{Volume} = \frac{\text{moles needed}}{\text{moles per liter}} \)

Plugging in the values:
\[ \text{Volume} = \frac{0.927 \text{ mol}}{2.595 \times 10^{-7} \text{ mol/L}} \approx 3.57 \times 10^{6} \text{ L} \]
This volume calculation is critical for practical applications, such as if you were to design a system to extract silver from a water supply, needing to know the physical space required for the process.