Problem 48 - General Chemistry: Principles and Modern Applications

Question

The aqueous solubility of \(\mathrm{CO}_{2}\) at \(20^{\circ} \mathrm{C}\) and 1.00 atm is equivalent to \(87.8 \mathrm{mL} \mathrm{CO}_{2}(\mathrm{g}),\) measured at STP, per \(100 \mathrm{mL}\) of water. What is the molarity of \(\mathrm{CO}_{2}\) in water that is at \(20^{\circ} \mathrm{C}\) and saturated with air at 1.00 atm? The volume percent of \(\mathrm{CO}_{2}\) in air is \(0.0360 \% .\) Assume that the volume of the water does not change when it becomes saturated with air.

Step-by-Step Solution

Answer

The molarity of CO2 in water that is at \(20^{\circ} \mathrm{C}\) and saturated with air at 1.00 atm is approximately 14.1 µM.

Step 1: Determine the Moles of CO2

First, use the known solubility to calculate the moles of CO2. Given that the solubility is \(87.8 \mathrm{mL} \mathrm{CO}_{2}\) per \(100 \mathrm{mL}\) of water, this tells us that 87.8 mL of CO2 are dissolved in 100 mL of water under the conditions of the problem. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 L. So, you can convert the volume of CO2 into moles using this relationship: \[\frac{87.8 \mathrm{mL}\ \mathrm{CO}_2}{1} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} \times \frac{1 \mathrm{mol}\ \mathrm{CO}_2}{22.4 \mathrm{L}} \approx 0.00392\ \mathrm{mol}\ \mathrm{CO}_2.\] This means there are approximately 0.00392 mol of CO2 in 100 mL of water.

Step 2: Determine the Molarity of CO2

Next, calculate the molarity of CO2 using the moles calculated in the previous step and the volume of water. The molarity (M) is defined as the number of moles of solute divided by the volume of solution (in liters): \[M = \frac{n}{V}\] Substituting in the moles of CO2 (n) and the volume of water (V), you find: \[M = \frac{0.00392\ \mathrm{mol}\ \mathrm{CO}_2}{0.100\ \mathrm{L}} \approx 0.0392\ \mathrm{M}\] So, the molarity of CO2 in the water is approximately 0.0392 M.

Step 3: Use Volume Percent to Calculate Final Molarity

The final step is to use the volume percent of CO2 in air to calculate the final molarity of CO2 in the water when it is saturated with air. The volume percent of CO2 in air is 0.0360%, which means that CO2 makes up 0.0360% of the volume of the air. Therefore, the final molarity is: \[0.0392\ \mathrm{M} \times 0.0360\% \approx 0.00001411\ \mathrm{M}\] or 14.1 µM.

Key Concepts

Molarity Calculation

Molarity is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This calculation is key to understanding how much of a substance, like carbon dioxide (\(\mathrm{CO}_2\), is dissolved in a given amount of water.
In the given exercise, we first convert the volume of dissolved \(\mathrm{CO}_2\) from milliliters to liters to find its molarity.

• Start with the volume of \(\mathrm{CO}_2\) dissolved: 87.8 mL in 100 mL of water.
• Convert 87.8 mL to liters: 87.8 mL \(\times \frac{1 \, \text{L}}{1000 \, \text{mL}} \approx 0.0878 \, \text{L}\).
• Use the relationship at STP (Standard Temperature and Pressure): 1 mole of gas = 22.4 L. Convert mL \(\mathrm{CO}_2\) to moles: \(0.0878 \, \text{L} \times \frac{1 \, \text{mol}}{22.4 \, \text{L}} \approx 0.00392 \, \text{mol}\).

Finally, the molarity is found by dividing mol by the volume of solution, which is 100 mL in liters (0.100 L): \(M = \frac{0.00392 \, \text{mol}}{0.100 \, \text{L}} = 0.0392 \, \text{M}\). This represents the concentration of \(\mathrm{CO}_2\) in water at the given conditions.

STP Conditions

STP stands for Standard Temperature and Pressure, which are baseline conditions for comparing gas behaviors. STP is defined as a temperature of 0°C (273 K) and a pressure of 1 atm.
Under these conditions, 1 mole of an ideal gas occupies 22.4 liters.
This volume at STP is crucial for gas calculations because it provides a consistent reference point.

• Allows us to directly convert the volume of gas to moles.
• Essential for calculations involving gas laws, as it standardizes experimental outcomes.

For the exercise, we used the STP volume of 22.4 L/mol to convert the given 87.8 mL of \(\mathrm{CO}_2\) into moles. Without specifying STP conditions, finding accurate gas concentrations would be much more complex.

Volume Percent

Volume percent expresses the concentration of a component in a mixture as a percentage of the total volume. It represents how much of the total volume is made up by the component in question.
In the context of the original problem, the volume percent of \(\mathrm{CO}_2\) in air is 0.0360%. This means that out of every 100 units of air, 0.036 units are \(\mathrm{CO}_2\).

• To calculate how this affects the final molarity in water, multiply the initial molarity by the volume percent.
• For example, multiply the \(0.0392 \, \text{M}\) molarity of \(\mathrm{CO}_2\) by 0.0360%: \(0.0392 \, \text{M} \times 0.0360\% \approx 0.00001411 \, \text{M}\).

This calculation adjusts the concentration to reflect the proportion of \(\mathrm{CO}_2\) in air that actually dissolves into the water.