The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at \(30^{\circ} \mathrm{C}\) is \(7.27 \times 10^{-3} \mathrm{M},\) what is the Henry's law constant for radon in water at this temperature? (b) A sample consisting of various gases contains \(3.5 \times 10^{-6}\) mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at \(30^{\circ} \mathrm{C} .\) Calculate the molar concentration of radon in the water.

Short Answer

Expert verified
The Henry's law constant for radon in water at 30°C is \(7.27 \times 10^{-3}\,\frac{\text{M}}{\text{atm}}\). The molar concentration of radon in water when the 32 atm gas mixture is shaken with water at 30°C is \(8.14 \times 10^{-7}\,\text{M}\).

Step by step solution

01

Part (a): Finding Henry's law constant

Given, the solubility of radon in water at 30°C and 1 atm is \(7.27 \times 10^{-3}\) M. We can now calculate the Henry's law constant using the formula for Henry's law: \(C = k_HP\) Rearrange the formula to solve for k_H: \(k_H = \frac{C}{P}\) The pressure P is given as 1 atm, so plug in the known values and calculate k_H: \(k_H = \frac{7.27 \times 10^{-3}\,\text{M}}{1\,\text{atm}} = 7.27 \times 10^{-3}\,\frac{\text{M}}{\text{atm}}\) Therefore, the Henry's law constant for radon in water at 30°C is \(7.27 \times 10^{-3}\,\frac{\text{M}}{\text{atm}}\).
02

Part (b): Calculating the molar concentration of radon in water

We are given the mole fraction of radon as \(3.5 \times 10^{-6}\) and the total pressure of the gas mixture as 32 atm. First, calculate the partial pressure of radon (P): \(P = \text{mole fraction of radon} \times \text{total pressure}\) \(P = (3.5 \times 10^{-6}) \times 32\,\text{atm} = 1.12 \times 10^{-4}\,\text{atm}\) With the partial pressure of radon and the Henry's law constant calculated in part (a), we can now calculate the molar concentration of radon in water (C): \(C = k_HP = (7.27 \times 10^{-3}\,\frac{\text{M}}{\text{atm}}) \times (1.12 \times 10^{-4}\,\text{atm}) = 8.14 \times 10^{-7}\,\text{M}\) Therefore, the molar concentration of radon in water when the 32 atm gas mixture is shaken with water at 30°C is \(8.14 \times 10^{-7}\,\text{M}\).

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Key Concepts

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

Solubility of Gases
Understanding the solubility of gases in liquids is crucial when considering environmental and health issues, such as the presence of radon in water. Solubility refers to the amount of a gas that can dissolve in a specific volume of liquid under certain conditions, primarily temperature and pressure. It's important to note that the solubility of gases increases with decreased temperature and increased pressure.

The solubility of radon, for instance, is measured in molar concentration (moles of gas per liter of liquid), which gives us a quantitative way of describing how much radon can dissolve in water. This solubility is not just a key concept in understanding environmental phenomena but also in various industrial processes where gases are dissolved in liquids for different applications.
Henry's Law
Henry's Law is a fundamental principle in chemistry that defines the relationship between the solubility of a gas and the pressure exerted by the gas above the liquid. According to Henry's Law, at a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of the gas in equilibrium with the liquid.

The equation for Henry's Law is given by: \[ C = k_H \times P \]
where \( C \) is the solubility (molar concentration of the gas), \( k_H \) is the Henry's law constant (unique for each gas-liquid combination at a given temperature), and \( P \) is the partial pressure of the gas. The constant integrates the nature of the gas, the solvent, and the temperature into one value, allowing us to calculate one of the other two if two are known.
Molar Concentration
Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It is typically expressed in moles per liter (M). When working with the solubility of gases, it's the amount of gas, in moles, that is dissolved in a liter of liquid. It's an essential concept in chemistry as it allows for the quantification of substances within a solution, enabling precise stoichiometric calculations in reactions.

In environmental studies, such as assessing the concentration of radon in water, understanding molar concentration helps in gauging the potential exposure levels to harmful substances, thus assisting in the analysis of public health risks.
Partial Pressure
The term 'partial pressure' relates to the pressure that a single gas in a mixture of gases would exert if it occupied the entire volume alone at the same temperature. Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.

This concept is central to the application of Henry's Law, as it directly influences the solubility of the gas in the liquid. As seen in our radon example, calculating the partial pressure of radon is a necessary step in determining its concentration when the gas is in a mixture, as it allows for the use of Henry's Law to calculate solubility in water. The higher the partial pressure of the gas, the greater its solubility in the solvent, assuming temperature remains constant.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

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