The Henry's law constant for the solubility of argon gas in water is \(1.0 \times 10^{-3} \mathrm{M} / \mathrm{atm}\) at \(30^{\circ} \mathrm{C}\). (a) Express the constant for the solubility of argon gas in \(M / \mathrm{mm} \mathrm{Hg}\). (b) If the partial pressure of argon gas at \(30^{\circ} \mathrm{C}\) is \(693 \mathrm{~mm} \mathrm{Hg}\), what is the concentration of dissolved argon in \(M\) at \(30^{\circ} \mathrm{C} ?\) (c) How many grams of argon gas can be dissolved in \(25 \mathrm{~L}\) of water at \(693 \mathrm{~mm} \mathrm{Hg}\) and \(30^{\circ} \mathrm{C} ?\) (Ignore the partial pressure of water.)

Short Answer

Expert verified
Answer: The concentration of dissolved argon gas is \(5.27 \times 10^{-1} M\) at \(30^{\circ} C\) and \(693 mmHg\), and \(5.27 \times 10^{2}\) grams of argon gas can be dissolved in \(25 L\) of water at those conditions.

Step by step solution

01

(a) Express the constant for solubility in \(M / mmHg\)

First, we need to convert the given constant from \(M/atm\) to \(M/mmHg\). We are given the constant: \(1.0\times10^{-3} M / atm\). Now, we can use the conversion factor, \(1 atm = 760 mmHg\) to convert this constant: \(new\_constant = \frac{1.0\times10^{-3} M / atm}{1 atm/760 mmHg} = \frac{(1.0\times10^{-3})(760)}{1} M / mmHg\) \(new\_constant = 7.6\times10^{-4} M / mmHg\) So, the constant for the solubility of argon gas in \(M / mmHg\) is \(7.6\times10^{-4} \frac{M}{mmHg}\).
02

(b) Determine the concentration of dissolved argon

We know Henry's law states that \(C = kP\), where \(C\) is the concentration of the dissolved gas, \(k\) is Henry's law constant, and \(P\) is the partial pressure of the gas. Using the Henry's law constant obtained in part (a) and the given partial pressure, \(693 mmHg\), we can find the concentration of dissolved argon gas: \(C_{argon} = (7.6\times10^{-4} \frac{M}{mmHg})(693 mmHg)\) \(C_{argon} = 5.27 \times 10^{-1} M\) Thus, the concentration of dissolved argon in \(M\) at \(30^{\circ} C\) and \(693 mmHg\) is \(5.27 \times 10^{-1} M\).
03

(c) Find the mass of dissolved argon in grams

First, we need to calculate the number of moles of argon gas dissolved in \(25 L\) of water at the given concentration: moles = concentration × volume moles_argon = $$(5.27 \times 10^{-1} M)(25 L) = 1.32 \times 10^{1} moles$$ Next, we can convert moles to grams using the molar mass of argon, which is \(39.95 \frac{g}{mol}\): mass_argon = moles_argon × molar_mass_argon mass_argon = $$(1.32 \times 10^{1} moles)(39.95 \frac{g}{mol}) = 5.27 \times 10^{2} g$$ Therefore, \(5.27 \times 10^{2}\) grams of argon gas can be dissolved in \(25 L\) of water at \(693 mmHg\) and \(30^{\circ} C\).

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Key Concepts

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

Solubility of Gases in Liquids
Understanding the solubility of gases in liquids is crucial for comprehending how substances like carbon dioxide in soda or oxygen in water bodies are dissolved. Solubility is defined as the maximum amount of gas that can dissolve in a liquid at a specific temperature and pressure.

Solubility is influenced by a gas's nature, and the temperature and pressure of the system. Generally, an increase in pressure increases gas solubility, while an increase in temperature typically decreases it. Henry's Law offers a quantitative relationship for the solubility of gases, stating that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.

Significance of Henry's Law

Henry's Law is widely used in chemistry and environmental sciences to predict how gases will behave when they come into contact with liquids. For instance, it helps in assessing the amount of a pollutant that will dissolve in bodies of water and also has practical applications in industries dealing with carbonated beverages or in the medical field where gas solubility is essential for respiratory therapy.
Partial Pressure
Partial pressure is a term that plays a crucial role in the solubility of gases. It refers to the pressure exerted by a single gas in a mixture of gases. The total pressure of a gas mixture is the sum of the partial pressures of all the gases present.

In the context of Henry's Law, the partial pressure of a gas above a liquid is what determines its solubility within that liquid. If the partial pressure increases, according to Henry's Law, the concentration of the dissolved gas also increases. For example, when you pump more air into a tire, the partial pressure of oxygen inside the tire increases, which in turn would increase the amount dissolved in the tire's liquid components, if there were any.

Understanding Partial Pressure in Daily Life

Partial pressure is not only a theoretical concept but is also observed in everyday life. Scuba divers must understand partial pressures to avoid decompression sickness, and it is crucial for understanding how our blood absorbs and transports gases like oxygen and carbon dioxide.
Concentration of Dissolved Gas
The concentration of dissolved gas in a solution is a measure of how much of a particular gas is contained in a given volume of liquid. It is typically expressed in molarity (M), which is moles of gas per liter of solution.

Henry's Law enables us to calculate the concentration of a dissolved gas if we know the partial pressure of the gas and the Henry's law constant for that gas in the given liquid at a specified temperature. The formula from Henry's Law, often written as \( C = kP \), where \( C \) is the concentration, \( k \) is Henry's law constant, and \( P \) is the partial pressure, illustrates this direct relationship.

Practical Application

In practical terms, this means that by manipulating the pressure of the gas, we can control its solubility in the liquid. This is common in industrial processes such as in the manufacturing of soft drinks, where carbon dioxide is dissolved under high pressure to create the fizz. In medical treatment, controlling the concentration of oxygen dissolved in blood is critical for ensuring patients receive the proper amount for respiration.

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