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Chapter 13: Problem 37

The Henry's law constant for helium gas in water at \(30^{\circ} \mathrm{C}\) is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and the constant for \(\mathrm{N}_{2}\) at \(30^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) . If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas.

Short Answer

Expert verified
The solubility of helium gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(5.55 \times 10^{-4} \mathrm{M}\), and the solubility of nitrogen gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(9.0 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Identify given data

We are given the Henry's law constants for helium gas and nitrogen gas in water at 30掳C. The constant for helium is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and for nitrogen gas is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\). The partial pressure of both gases is given as 1.5 atm.
02

Apply Henry's law

According to Henry's law, the solubility of a gas (C) in a liquid is directly proportional to the partial pressure of the gas (P) above the liquid. Mathematically, it can be represented as: \[ C = k_H \times P \] where, \(C\) is the solubility of the gas, \(k_H\) is the Henry's law constant and \(P\) is the partial pressure of the gas.
03

Calculate solubility of helium gas

Using the Henry's law constant and the partial pressure given for helium gas, let's calculate the solubility. We will use the following formula: \[ C_\mathrm{He} = k_\mathrm{He} \times P_\mathrm{He} \] Substitute the given values into the equation: \[ C_\mathrm{He} = (3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm}) \] After performing the calculations, we get: \[ C_\mathrm{He} = 5.55 \times 10^{-4} \mathrm{M} \]
04

Calculate solubility of nitrogen gas

Similarly, we can calculate the solubility of nitrogen gas using the Henry's law constant and the partial pressure given for nitrogen gas. We will use the following formula: \[ C_\mathrm{N_2} = k_\mathrm{N_2} \times P_\mathrm{N_2} \] Substitute the given values into the equation: \[ C_\mathrm{N_2} = (6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}) \times (1.5 \mathrm{atm}) \] After performing the calculations, we get: \[ C_\mathrm{N_2} = 9.0 \times 10^{-4} \mathrm{M} \]
05

Write the final answer

The solubility of helium gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(5.55 \times 10^{-4} \mathrm{M}\), and the solubility of nitrogen gas in water at \(30^{\circ} \mathrm{C}\) and 1.5 atm pressure is \(9.0 \times 10^{-4} \mathrm{M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility
Solubility refers to the ability of a substance to dissolve in a solvent. In the context of gases dissolving in liquids, solubility is the concentration of the gas in the liquid when the gas phase is in equilibrium with the liquid phase.
When a gas dissolves in a liquid, it disperses throughout the liquid, forming a solution. This process is influenced by factors like temperature and pressure.
In relation to Henry's Law, solubility determines how much of a gas will dissolve in a liquid at a certain pressure. Understanding solubility helps us predict and control how gases behave in different situations, such as carbonated beverages or natural bodies of water. Knowing the solubility of gases like helium and nitrogen is essential in fields like medicine and diving, where pressure conditions vary.
Partial Pressure
Partial pressure is the pressure exerted by a single component within a mixture of gases. Each gas in a mixture behaves independently and contributes to the total pressure proportionally to its amount.
According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of all the individual gases.
This concept is important for understanding how each gas in a mixture dissolves differently based on its own partial pressure.In the exercise, both helium and nitrogen gases have a given partial pressure of 1.5 atm in the scenario.
  • The higher the partial pressure of a gas, the more it will dissolve in the liquid according to Henry's Law.
  • Partial pressure helps in calculating the solubility through the equation: \( C = k_H \times P \).
Applications of partial pressure are vital in diverse areas including respiratory physiology and industrial gas separation processes.
Gas Solubility Calculation
Calculating the solubility of a gas in a liquid involves using Henry's Law, which connects the solubility of the gas with its partial pressure and Henry's Law constant.
Henry's Law states that the solubility \(C\) of a gas in a liquid is directly proportional to the partial pressure \(P\) of the gas above the liquid. In mathematical terms, this relationship is described by:\[ C = k_H \times P \]Where:
  • \(C\) is the solubility of the gas in the liquid.
  • \(k_H\) is the Henry's Law constant specific to the gas and solvent at a particular temperature.
  • \(P\) is the partial pressure of the gas.
In the original problem, this principle was used to find:
  • The solubility of helium: \( C_{\text{He}} = (3.7 \times 10^{-4} \, \text{M/atm}) \times 1.5 \, \text{atm} = 5.55 \times 10^{-4} \, \text{M} \)
  • The solubility of nitrogen: \( C_{\text{N}_2} = (6.0 \times 10^{-4} \, \text{M/atm}) \times 1.5 \, \text{atm} = 9.0 \times 10^{-4} \, \text{M} \)
These calculations help in designing systems where precise gas dissolution is critical, such as beverage carbonation and medical oxygen delivery.

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Most popular questions from this chapter

The solubility of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is 70 \(\mathrm{g}\) per 100 \(\mathrm{mL}\) of water. (a) Is a 1.22 \(\mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

At \(63.5^{\circ} \mathrm{C},\) the vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) is 175 torr, and that of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 400 torr. A solution is made by mixing equal masses of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) . (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal-solution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C} ?\) (c) What is the mole fraction of ethanol in the vapor above the solution?

The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in \(1.0 \times 10^{3}\) gal of seawater.

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C} ?\)
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