Problem 46 - General Chemistry: Principles and Modern Applications

Question

At 1.00 atm, the solubility of \(\mathrm{O}_{2}\) in water is \(2.18 \times 10^{-3} \mathrm{M}\) at \(0^{\circ} \mathrm{C}\) and \(1.26 \times 10^{-3} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) What volume of \(\mathrm{O}_{2}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm},\) is expelled when \(515 \mathrm{mL}\) of water saturated with \(\mathrm{O}_{2}\) is heated from 0 to \(25^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

11.6 mL

Step 1: Understanding the solubility

Solubility at \(0^{\circ} \mathrm{C}\) is \(2.18 \times 10^{-3}\) M and at \(25^{\circ} \mathrm{C}\) it is \(1.26 \times 10^{-3} \mathrm{M}\). The molarity difference implies how much \(\mathrm{O}_{2}(\mathrm{g})\) is expelled when the temperature rises.

Step 2: Calculate the change in molarity

The change in solubility (change in molarity) is \(2.18 \times 10^{-3} \mathrm{M} - 1.26 \times 10^{-3} \mathrm{M} = 0.92 \times 10^{-3}\) M or \(9.2 \times 10^{-4} \mathrm{M}\). This is the amount of \(\mathrm{O}_{2}(\mathrm{g})\) that is expelled from each liter of water when it is heated from \(0^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).

Step 3: Find the moles of \(\mathrm{O}_{2}(\mathrm{g})\) expelled

Use the change in molarity and the volume of the water to find the moles of \(\mathrm{O}_{2}(\mathrm{g})\) expelled. Use the formula for molarity \(M = \frac{n}{V}\), where V is in liters. Solve for n (moles of gas): \(n = M \times V = 9.2 \times 10^{-4} \mathrm{M} \times 0.515 \mathrm{L} = 4.74 \times 10^{-4} \mathrm{mol}\)

Step 4: Calculate the volume of gas expelled

Under the conditions of \(25^{\circ}C\) and \(1.00 atm\), 1 mole of any gas occupies 24.5 L. Therefore, the volume of the \(\mathrm{O}_{2}\) gas expelled when the solution is heated is: \(V = 4.74 \times 10^{-4} \mathrm{mol} \times 24.5 \mathrm{L/mol} = 0.0116 \mathrm{L} = 11.6 \mathrm{mL}\)

Key Concepts

Molarity Calculations

Molarity is a key concept in chemistry that refers to the concentration of a solute in a solution. It is expressed in moles per liter (M). This enables chemists to quantitatively describe the concentration of molecules or atoms in a given volume of solution. The formula for calculating molarity is very simple:
\[ M = \frac{n}{V} \]
Here, \( M \) represents molarity, \( n \) signifies the number of moles of solute, and \( V \) denotes the volume of the solution in liters.

• At \(0^{\circ} \text{C}\), the solubility of \(\text{O}_2\) in water is \(2.18 \times 10^{-3} \text{ M}\).
• At \(25^{\circ} \text{C}\), it decreases to \(1.26 \times 10^{-3} \text{ M}\).

From this difference, we can determine the amount of oxygen that is expelled from the solution when the temperature is increased. The change in solubility is a direct indicator of how much gas leaves the solution.
Calculating the expelled gas involves subtracting the lower molarity value from the higher one, which gives us the change in molarity or concentration change of oxygen dissolved in water. For this exercise, the change is: \[2.18 \times 10^{-3} \text{ M} - 1.26 \times 10^{-3} \text{ M} = 9.2 \times 10^{-4} \text{ M}\]
Understanding molarity increases our ability to manage and predict reactions in solutions.

Expelled Gas Volume

Understanding expelled gas volume is important in calculating how much gas comes out of a solution when conditions change, like temperature. When water is heated from \(0^{\circ} \text{C}\) to \(25^{\circ} \text{C}\), the solubility of gases like oxygen decreases, meaning fewer gas molecules are held in the solution, thus expelling some gas.
The number of moles of a gas expelled can be calculated by multiplying the change in molarity by the volume of the water in liters. For example, if we have a 0.515 L container of water:

• Concentration change: \(9.2 \times 10^{-4} \text{ M} \)
• Volume = 0.515 L

Using the molarity formula: \[ n = M \times V = 9.2 \times 10^{-4} \text{ M} \times 0.515 \text{ L} = 4.74 \times 10^{-4} \text{ mol} \]
To find out how much space or volume this amount of \(\text{O}_2\) gas occupies we use the ideal gas law assumption. At \(25^{\circ} \text{C}\) and \(1 \text{ atm}\), 1 mole of gas occupies approximately 24.5 liters. Thus, the expelled gas volume is: \[ V = 4.74 \times 10^{-4} \text{ mol} \times 24.5 \text{ L/mol} = 0.0116 \text{ L} = 11.6 \text{ mL} \]
Understanding this conversion is crucial in various chemical processes where gas volume measurements are needed.

Temperature Effect on Solubility

The effect of temperature on solubility is a fundamental concept, especially in understanding how it influences gas solubility in liquids such as water. Generally, for gases dissolved in liquids, increasing the temperature reduces their solubility. This phenomenon occurs because higher temperatures provide more energy to gas molecules, allowing them to escape the liquid phase more readily.
In our exercise, when water is heated from \(0^{\circ} \text{C}\) to \(25^{\circ} \text{C}\), the molarity of \(\text{O}_2\) in water drops from \(2.18 \times 10^{-3}\) M to \(1.26 \times 10^{-3}\) M. This demonstrates that less oxygen remains dissolved as the temperature rises, making the gas leave the solution, or desorb, into the air.
An easy way to remember this relationship is to think about a can of soda left out in the sun: as the liquid warms, the soda seems to "go flat" because the carbon dioxide gas is less soluble in the warm drink and escapes.
This knowledge is valuable not just in chemistry labs but also has practical applications like understanding aquatic life dynamics in different climates or the operation of industrial processes involving gases and liquids.

• Increases in temperature = Decrease in gas solubility in liquid
• Gas molecules receive more kinetic energy and tend to escape.

By grasping how temperature affects solubility, one can predict and control various reactions and processes efficiently.