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Chapter 13: Problem 48

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.

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