Question

Some soft drinks contain up to 85 ppm oxygen. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain \(85 \mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C} ?\) (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(\left.1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa} .\right)\)

Short answer

(a) The concentration of \(85 \mathrm{ppm} \mathrm{O}_{2}\) in water in mol/L is \(2.66 \times 10^{-3} \mathrm{mol/L}\). (b) The required partial pressure of \(\mathrm{O}_{2}\) above water at \(10^{\circ} \mathrm{C}\) is \(4.49 \times 10^{-5} \mathrm{Pa}\).

Step-by-step solution

Part (a): Convert 85 ppm to mol/L

To find the concentration in mol/L, we need to convert the given concentration from ppm to mol/L. The given concentration is 85 ppm of oxygen, which means there are 85 grams of O2 in 1 million grams of water: Concentration in g/L = \(\frac{85 \mathrm{g}}{10^6 \mathrm{g}} * 10^3 \mathrm{g} = 0.085 \mathrm{g/L}\) Now, we will convert the concentration in g/L to mol/L. The molar mass of oxygen, O2, is \(2 * 16.0 = 32.0\) grams/mol. Concentration in mol/L = \(\frac{0.085 \mathrm{g/L}}{32.0 \mathrm{~g/mol}} = 2.66 \times 10^{-3} \mathrm{mol/L}\)

Part (b): Find the partial pressure of O2 using Henry's Law constant

To find the partial pressure of O2, we will use Henry's law formula, which states that: \(P_\mathrm{O2} = K_\mathrm{H} * C\) Where: - \(P_\mathrm{O2}\): Partial pressure of O2 above water - \(K_\mathrm{H}\): Henry's law constant - C: Concentration of O2 in the water In this exercise, we are given the Henry's law constant for O2 at 10°C as \(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\) To find the partial pressure of \(\mathrm{O}_{2}\) above water, we will plug the Henry's law constant value of O2 and the concentration of O2 in the water: \(P_\mathrm{O2} = \left(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\right) * \left(2.66 \times 10^{-3} \mathrm{mol/L}\right)\) To match the units on both sides, we need to convert the concentration of O2 from mol/L to mol/\(\mathrm{m}^{3}\): \(2.66 \times 10^{-3} \mathrm{mol/L}\) = \(2.66 \times 10^{-3} \mathrm{mol/L} * \frac{10^3 \mathrm{L}}{1 \mathrm{m}^3} = 2.66 \mathrm{mol/m}^3\) Now, we can calculate the partial pressure of \(\mathrm{O}_{2}\): \(P_\mathrm{O2} = \left(1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\right) * \left(2.66 \mathrm{mol/m}^3\right) = 4.49 \times 10^{-5} \mathrm{Pa}\) The partial pressure of \(\mathrm{O}_{2}\) above water to obtain \(85 \mathrm{ppm}\) of \(\mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C}\) is \(4.49 \times 10^{-5} \mathrm{Pa}\).

Key concepts

Understanding Concentration Conversion

When dealing with concentrations like parts per million (ppm), it's essential to convert these measurements into more practical units for chemical calculations, such as molarity (mol/L). PPM denotes the amount of solute (in grams) per one million grams of solvent. In our example, 85 ppm of \( ext{O}_2\) means that there are 85 grams of oxygen in a million grams of water.

• First, it is helpful to convert ppm to grams per liter (g/L). This is achieved by recognizing that there are 1,000 grams in a liter of water.


• Then, convert the outcome in grams per liter to moles per liter using the molar mass of the solute.

In short, understanding these conversion steps helps in shifting between units, which is vital for precise scientific calculations.

Calculating Molar Mass

The molar mass of a compound is the mass of one mole of that substance and is expressed in grams per mole (g/mol). To find the molar mass of a molecule, add up the atomic masses of its constituent elements. For diatomic oxygen (\(\text{O}_2\)), each oxygen atom has an atomic mass of approximately 16.0 g/mol.

• Thus, for \(\text{O}_2\), the molar mass is calculated as \(2 \, \times \, 16.0 \, \text{g/mol} = 32.0 \, \text{g/mol}\).

Knowing how to calculate and apply molar mass allows one to accurately convert concentrations from g/L to mol/L. This step is crucial for solving problems related to chemical concentrations and reactions.

Understanding and Using Partial Pressure

Partial pressure refers to the pressure that a gas would exert if it were the only gas in a certain volume. To find the partial pressure of a gas above a liquid, Henry's Law can be used. Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

• The formula is \(P_{\text{gas}} = K_H \times C\), where \(P_{\text{gas}}\) is the partial pressure, \(K_H\) is the Henry's Law constant, and \(C\) is the concentration of the gas in the liquid.


• In the context of the exercise, to determine the partial pressure of oxygen above water, convert the concentration from mol/L to mol/m³ to match the units of Henry's Law constant, and simply multiply.

This concept is vital in predicting how gases behave and dissolve in liquids under different conditions.

Exploring Gas Solubility in Liquids

Gas solubility in liquids is a critical chemistry concept determined by the interactions between molecules in the gas and liquid phases. Solubility can be affected by several factors such as temperature, pressure, and the nature of both the solvent and solute.

• An increase in the partial pressure of a gas above a liquid generally increases the gas's solubility in the liquid, according to Henry's Law.


• Temperature also affects solubility; gases are typically less soluble in liquids as the temperature rises.

Understanding the dynamics of gas solubility aids in comprehending and predicting processes such as carbonation in beverages and respiratory functions in organisms. It is an important concept for practical applications in daily life and industrial processes.