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Chapter 10: Problem 50

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\) for Henry's law in the form \(C=k P\) where \(C\) is the gas concentration (mol/L).

Short Answer

Expert verified
The solubility of O鈧� in water at a partial pressure of 120 torr and a temperature of \(25^{\circ} \mathrm{C}\) is \(2.05 \times 10^{-4}\,\mathrm{mol} / \mathrm{L}\).

Step by step solution

01

List the given values

- Partial pressure of O鈧�, \(P = 120\) torr - Temperature, \(T = 25^{\circ} \mathrm{C}\) - Henry's Law constant for O鈧�, \(k = 1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\)
02

Convert the pressure to atm

We need to convert the partial pressure of O鈧� from torr to atm because the Henry's law constant is given in atm. The conversion factor is \(1 \mathrm{atm} = 760 \mathrm{torr}\). $$ P_{\text{atm}} = \frac{120 \mathrm{torr}}{760 \mathrm{torr} \cdot \mathrm{atm}^{-1}} = 0.1579 \mathrm{atm} $$
03

Apply Henry's Law

Now we can apply Henry's Law to calculate the concentration of O鈧� in water. The formula for Henry's Law is: $$ C = k \cdot P $$ Plugging in the given values: $$ C = (1.3 \times 10^{-3} \mathrm{mol/L}\cdot\mathrm{atm}) \cdot (0.1579 \mathrm{atm}) $$
04

Calculate the solubility

Multiply the Henry's Law constant and the partial pressure to find the solubility of O鈧� in water. $$ C = 1.3 \times 10^{-3} \mathrm{mol/L}\cdot\mathrm{atm} \cdot 0.1579 \mathrm{atm} = 2.05 \times 10^{-4} \mathrm{mol/L} $$ The solubility of O鈧� in water at a partial pressure of 120 torr and a temperature of \(25^{\circ} \mathrm{C}\) is \(2.05 \times 10^{-4}\,\mathrm{mol} / \mathrm{L}\).

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Key Concepts

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

gas solubility
The solubility of a gas in a liquid is a measure of how much of that gas can be dissolved in the liquid at a specific temperature and pressure. In the context of the exercise, we are interested in the solubility of oxygen (\(O_2\)) in water. Gas solubility can be influenced by several factors, including temperature, pressure, and the nature of the gas and the liquid.
  • As pressure increases, more gas molecules are "pushed" into the liquid, increasing solubility.
  • Conversely, if temperature increases, solubility typically decreases for gases because gas molecules gain kinetic energy and escape the liquid.
These competing factors are why it's crucial to know both temperature and pressure when calculating gas solubility using Henry's Law.
partial pressure
Partial pressure is the pressure a single component of a mixture of gases would exert if it occupied the entire volume alone. In the case of the exercise, we're looking at the partial pressure of oxygen (\(O_2\)), which is essential for applying Henry's Law.
Henry's Law states that the concentration of a gas dissolved in a liquid is directly proportional to its partial pressure above the liquid:
\[C = k \cdot P\]where \(C\) is the concentration of the gas, \(k\) is the Henry's law constant, and \(P\) is the partial pressure of the gas.
  • This concept helps us understand how gases like \(O_2\) behave in beverages, aquatic environments, and even in the blood.
  • Knowing the partial pressure, we can predict how much gas can dissolve at specific conditions.
conversion between units
Many scientific calculations require converting units, as different expressions often use different units. In the exercise, the given partial pressure is in torr, but the Henry's law constant is in atm. Thus, we first convert torr to atm for consistent calculations.
Using the conversion factor:
  • 1 atm = 760 torr
We convert 120 torr to atm by dividing by 760, resulting in:\[P_{\text{atm}} = \frac{120 \, \text{torr}}{760 \, \text{torr/atm}} = 0.1579 \, \text{atm}\]
  • This step ensures that our calculations are precise and compatible with the given constants.
  • Accurate conversions are vital in chemistry, ensuring that formulas and equations work correctly.
chemical calculations
For chemical calculations, understanding the relation between physical principles and mathematical formulas is vital. Once the correct values and units are established, proceeding with the computation is often straightforward.
In our exercise, applying Henry's Law involves:
  • Multiplying the Henry's law constant \(k = 1.3 \times 10^{-3} \, \text{mol/L}\cdot\text{atm}\) by the partial pressure \(P_{\text{atm}} = 0.1579 \, \text{atm}\).
  • Performing this multiplication, we find the solubility \(C\) of oxygen in water:\[C = 1.3 \times 10^{-3} \, \text{mol/L}\cdot\text{atm} \times 0.1579 \, \text{atm} = 2.05 \times 10^{-4} \, \text{mol/L}\]
This comprehensive approach to chemical calculations allows predictions and analyses of real-world phenomena involving gases and solutions. It's a key aspect of many scientific and engineering applications.

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Most popular questions from this chapter

In lab you need to prepare at least \(100 \mathrm{mL}\) of each of the following solutions. Explain how you would proceed using the given information. a. \(2.0 \mathrm{m} \mathrm{KCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) b. \(15 \%\) NaOH by mass in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) c. \(25 \%\) NaOH by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{g} / \mathrm{cm}^{3}\right)\) d. 0.10 mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\)

Rationalize the trend in water solubility for the following simple alcohols: $$\begin{array}{lc} \text { Alcohol } & \begin{array}{c} \text { Solubility } \\ \left(\mathrm{g} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { at } 20^{\circ} \mathrm{C}\right) \end{array} \\ \hline \text { Methanol, } \mathrm{CH}_{3} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Butanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{2} \mathrm{OH} & 8.14 \\ \text { Pentanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH} & 2.64 \\ \text { Hexanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH} & 0.59 \\ \text { Heptanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CH}_{2} \mathrm{OH} & 0.09 \\ \hline \end{array}$$

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C}\). c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) and hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) form an ideal solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of pentane and hexane are 511 and \(150 .\) torr, respectively. A solution is prepared by mixing \(25 \mathrm{mL} \text { pentane (density, } 0.63 \mathrm{g} / \mathrm{mL})\) with \(45 \mathrm{mL}\) hexane (density, 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?
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