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Chapter 13: Problem 37

The Henry's law constant for hydrogen gas \(\left(\mathrm{H}_{2}\right)\) in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{M} / \mathrm{kPa}\) and the constant for argon \((\mathrm{Ar})\) at \(25^{\circ} \mathrm{C}\) is \(1.4 \times 10^{-5} \mathrm{M} / \mathrm{kPa}\). If the two gases are each present at 253 kPa pressure, calculate the solubility of each gas.

Short Answer

Expert verified
The solubility of hydrogen gas is \(1.9481 \times 10^{-3} \, \mathrm{M}\) and for argon is \(3.542 \times 10^{-3} \, \mathrm{M}\).

Step by step solution

01

Understanding Henry's Law

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. It is expressed by the equation \(S = k_H \, P\), where \(S\) is the solubility, \(k_H\) is the Henry's law constant, and \(P\) is the pressure of the gas.
02

Calculating Solubility for Hydrogen Gas

The solubility \(S_{H_2}\) of hydrogen gas can be calculated using its Henry's law constant \(k_{H_2} = 7.7 \times 10^{-6} \, \mathrm{M/kPa}\) and the pressure \(P = 253 \, \mathrm{kPa}\). Use the formula \(S_{H_2} = k_{H_2} \times P\): \(S_{H_2} = 7.7 \times 10^{-6} \, \mathrm{M/kPa} \times 253 \, \mathrm{kPa}\).
03

Solving for Hydrogen Gas Solubility

Carrying out the multiplication for hydrogen gas gives: \(S_{H_2} = 7.7 \times 10^{-6} \times 253 = 1.9481 \times 10^{-3} \, \mathrm{M}\). Thus, the solubility of hydrogen gas in water at \(25^{\circ} \mathrm{C}\) and \(253 \, \mathrm{kPa}\) pressure is \(1.9481 \times 10^{-3} \, \mathrm{M}\).
04

Calculating Solubility for Argon

The solubility \(S_{Ar}\) of argon can be calculated similarly. Use the Henry's law constant for argon, \(k_{Ar} = 1.4 \times 10^{-5} \, \mathrm{M/kPa}\), and the same pressure \(P = 253 \, \mathrm{kPa}\). Apply the formula \(S_{Ar} = k_{Ar} \times P\): \(S_{Ar} = 1.4 \times 10^{-5} \, \mathrm{M/kPa} \times 253 \, \mathrm{kPa}\).
05

Solving for Argon Solubility

Performing the multiplication for argon gives us: \(S_{Ar} = 1.4 \times 10^{-5} \times 253 = 3.542 \times 10^{-3} \, \mathrm{M}\). Therefore, the solubility of argon in water at \(25^{\circ} \mathrm{C}\) and \(253 \, \mathrm{kPa}\) pressure is \(3.542 \times 10^{-3} \, \mathrm{M}\).

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Key Concepts

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

Solubility Calculation
Calculating the solubility of a gas using Henry's Law involves a simple process whereby you need to have the Henry's Law constant and the pressure of the gas.
This calculation is crucial in determining how much gas will dissolve in a liquid under a given pressure.
Here's the formula for Henry's Law, which is:
  • \[ S = k_H \times P \]
Where:
  • \( S \) is the solubility of the gas in the liquid (usually in molarity, M).
  • \( k_H \) is the Henry's Law constant specific to each gas and expressed in \( \, \mathrm{M / kPa} \).
  • \( P \) is the partial pressure of the gas above the liquid in kilopascals (kPa).
To perform a solubility calculation:
  • Substitute the known values of \( k_H \) and \( P \) into the formula.
  • Multiply these values to find \( S \).
This approach allows you to determine the molarity of the dissolved gas, which can be fundamental for experiments or industrial processes.
Through this method, we calculated the solubility of hydrogen and argon gases at a pressure of 253 kPa.
Gas Solubility
Gas solubility refers to the extent to which a gas dissolves in a solvent, like water. The principle that governs this is Henry's Law, which suggests that the solubility of a gas is directly proportional to its partial pressure above the liquid.
In simpler terms, if you increase the gas pressure, more gas will dissolve in the liquid.

Why is Gas Solubility Important?

Knowing the solubility of gases is crucial in many fields:
  • In environmental science, it helps understand how gases like carbon dioxide behave in water bodies.
  • In medicine, it supports processes like oxygen delivery in blood.
  • In industries, it allows the creation of carbonated beverages and efficient chemical reactions.

Factors Affecting Gas Solubility

Gas solubility is not just about pressure; it is also influenced by other factors:
  • Temperature: Generally, gases tend to dissolve more in liquids at lower temperatures.
  • Nature of the Gas and Solvent: Different gases have unique solubility levels in various liquids.
Being aware of gas solubility enables better control over these variables in laboratory and industrial applications.
Temperature Effect on Solubility
The temperature significantly affects the solubility of gases in liquids.
According to Henry's Law, while pressure remains a dominant factor, temperature plays an intriguing role too.

Understanding Temperature Impact

  • At higher temperatures, gas molecules gain kinetic energy, making them more likely to escape from the liquid.
  • This means, as temperature rises, gas solubility typically decreases, contrary to solubility trends for solids.

Practical Implications

  • In climate science, warmer waters tend to release dissolved gases back into the atmosphere, impacting ocean chemistry.
  • In industries, controlling the temperature is crucial in processes that require specific gas concentrations in solutions.
Thus, temperature factors in notably when calculating and understanding solubility. This consideration is essential for consistent results in both natural and artificial chemical processes.

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Most popular questions from this chapter

Which two statements about gas mixtures are true? (a) Gases always mix with other gases because the gas particles are too far apart to feel significant intermolecular attractions or repulsions. (b) Just like water and oil don't mix in the liquid phase, two gases can be immiscible and not mix in the gas phase. (c) If you cool a gaseous mixture, you will liquefy all the gases at the same temperature. (d) Gases mix in all proportions in part because the entropy of the system increases upon doing so.

(a) A sample of hydrogen gas is generated in a closed container by reacting \(1.750 \mathrm{~g}\) of zinc metal with \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) hydrochloric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 150 \(\mathrm{mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\). Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) boils at \(80.1^{\circ} \mathrm{C}\) and has a density of \(0.876 \mathrm{~g} / \mathrm{mL}\). (a) When \(0.100 \mathrm{~mol}\) of a nondissociating solute is dissolved in \(500 \mathrm{~mL}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\), the solution boils at \(79.52^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) (b) When \(10.0 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(500 \mathrm{~mL}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\), the solution boils at \(79.23^{\circ} \mathrm{C}\). What is the molar mass of the unknown?

The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(101.3 \mathrm{kPa}\). (a) What volume of \(\mathrm{CH}_{4}(g)\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C} ?\) (b) The solubilities (in water) of the hydrocarbons are as follows: methane \(<\) ethane \(<\) ethylene. Is this because ethylene is the most polar molecule? (c) What intermolecular interactions can these hydrocarbons have with water? (d) Draw the Lewis dot structures for the three hydrocarbons. Which of these hydrocarbons possess \(\pi\) bonds? Based on their solubilities, would you say \(\pi\) bonds are more or less polarizable than \(\sigma\) bonds? (e) Explain why \(\mathrm{NO}\) is more soluble in water than either \(\mathrm{N}_{2}\) or \(\mathrm{O}_{2}\). (f) \(\mathrm{H}_{2} \mathrm{~S}\) is more water-soluble than almost all the other gases in table. What intermolecular forces is \(\mathrm{H}_{2} \mathrm{~S}\) likely to have with water? \((\mathbf{g}) \mathrm{SO}_{2}\) is by far the most water-soluble gas in table. What intermolecular forces is \(\mathrm{SO}_{2}\) likely to have with water? $$ \begin{array}{lc} \hline \text { Gas } & \text { Solubility (mM) } \\ \hline \mathrm{CH}_{4} \text { (methane) } & 1.3 \\ \mathrm{C}_{2} \mathrm{H}_{6} \text { (ethane) } & 1.8 \\ \mathrm{C}_{2} \mathrm{H}_{4} \text { (ethylene) } & 4.7 \\ \mathrm{~N}_{2} & 0.6 \\ \mathrm{O}_{2} & 1.2 \\ \mathrm{NO} & 1.9 \\ \mathrm{H}_{2} \mathrm{~S} & 99 \\ \mathrm{SO}_{2} & 1476 \\ \hline \end{array} $$

Indicate whether each statement is true or false: (a) The higher the temperature, the more soluble most gases are in water. (b) The higher the temperature, the more soluble most ionic solids are in water. (c) As you cool a saturated solution from high temperature to low temperature, solids start to crystallize out of solution if you achieve a supersaturated solution. (d) If you take a saturated solution and raise its temperature, you can (usually) add more solute and make the solution even more concentrated.
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