Problem 30 - Chemistry The Molecular Science

Question

The partial pressure of \(\mathrm{O}_{2}\) in your lungs is about \(100 \mathrm{mmHg} .\) Calculate the concentration of \(\mathrm{O}_{2}\) (in grams per liter) that can dissolve in water at \(37^{\circ} \mathrm{C}\) when the \(\mathrm{O}_{2}\) partial pressure is \(100 . \mathrm{mmHg} .\) The Henry's law constant for \(\mathrm{O}_{2}\) at \(37{ }^{\circ} \mathrm{C}\) is \(1.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{mmHg}^{-1}\).

Step-by-Step Solution

Answer

The concentration of \( \mathrm{O}_2 \) in water is \( 4.8 \times 10^{-3} \mathrm{~g/L} \).

Step 1: Understand Henry's Law

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 given by \( C = k_H \times P \), where \( C \) is the concentration, \( k_H \) is the Henry's law constant, and \( P \) is the partial pressure.

Step 2: Substitute Values into Henry's Law Equation

Substitute the given values into the Henry's Law formula: \( k_H = 1.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{mmHg}^{-1} \) and \( P = 100 \mathrm{~mmHg} \). Thus, the concentration \( C \) is \( C = (1.5 \times 10^{-6}) \times 100 \).

Step 3: Calculate the Concentration in Moles per Liter

Perform the multiplication: \( C = 1.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \). This gives the concentration of \( \mathrm{O}_2 \) in moles per liter.

Step 4: Convert Moles to Grams

To convert moles of \( \mathrm{O}_2 \) to grams, use the molar mass of \( \mathrm{O}_2 \), which is approximately \( 32 \mathrm{~g/mol} \). The concentration in grams per liter is \( 1.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \times 32 \mathrm{~g/mol} \).

Step 5: Calculate the Final Concentration in Grams per Liter

Perform the conversion: \( C = 4.8 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1} \). This is the concentration of \( \mathrm{O}_2 \) that can dissolve in water at the given conditions.

Key Concepts

Partial Pressure

Partial pressure is a crucial concept in understanding how gases behave in mixtures. It's like the pressure one individual gas would exert if it were alone in the volume the gas mixture occupies.
When we talk about the partial pressure of oxygen in the lungs, we mean the pressure oxygen exerts within the mixture of gases in the pulmonary space.
This is essential because the partial pressure of a gas affects its solubility in liquids, such as how oxygen dissolves in blood.

• Partial pressure of \(\mathrm{O}_{2}\) in lungs: \(100 \,\mathrm{mmHg}\)
• Affects how much gas dissolves during respiration
• Key factor in deriving gas laws like Henry's Law

Remember, each gas in a mixture contributes independently to the total pressure, depending on its concentration and molecular nature.

Gas Solubility

Gas solubility is all about how much of a gas can be dissolved in a liquid at a specific pressure and temperature. It's a measure of a gas's ability to enter and stay in the liquid phase.
According to Henry's Law, as the partial pressure of a gas above a liquid increases, more gas dissolves in the liquid.
The relationship is linear and given by:\[ C = k_H \times P \]where \(C\) is the concentration of the gas, \(k_H\) is the Henry's law constant, and \(P\) is the partial pressure.

• Solubility depends on the nature of the gas and the solvent
• Higher pressure leads to more gas dissolution
• Henry's Law constant varies with each gas

Think of this as the initial step in understanding how gases dissolve, which is very relevant for gases like oxygen when considering breathing and metabolic processes.

Molar Concentration

Molar concentration is a way of expressing how much of a substance is present in a certain volume. For gases dissolved in liquids, it often refers to moles of the gas per liter of solution.
In the context of this exercise, we used Henry's Law to calculate the molar concentration of oxygen in water given its partial pressure.
The calculation involved substituting known values into Henry's Law:\[ C = (1.5 \times 10^{-6}) \times 100 = 1.5 \times 10^{-4} \,\mathrm{mol/L} \]This gives us the oxygen concentration in moles per liter before converting to grams.

• Makes it simpler to understand gas quantities in a solution
• Important for conversion between different units (moles to grams)
• Crucial in various chemical calculations and reactions

Understanding this concept helps in the broader picture of chemical solutions and gas-liquid interactions.

Temperature and Gas Solubility

Temperature is a key player in gas solubility. Generally, as temperature increases, the solubility of gases in liquids decreases.
This is because higher temperatures give gas molecules more energy to escape from the liquid phase back into the gas phase, reducing solubility.
In this exercise, we have assumed a constant temperature of 37°C, similar to human body temperature, which helps to model physiological conditions.

• Higher temperature typically reduces gas solubility
• Directly affects how concentrations are calculated
• Important in considering biological and environmental systems

Bear in mind, the temperature dependency is particularly important when considering dynamic systems such as the human body or aquatic environments where temperature variations can significantly alter gas solubility.