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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}\( and the constant for argon (Ar) at \)25^{\circ} \mathrm{C}$ is \(1.4 \times 10^{-5} \mathrm{M} / \mathrm{kPa}\). If the two gases are each present at \(253 \mathrm{kPa}\) pressure, calculate the solubility of each gas.

Short Answer

Expert verified
The solubilities of hydrogen gas and argon at 25°C and 253 kPa are approximately \(1.95 \times 10^{-3} M\) and \(3.54 \times 10^{-3} M\), respectively.

Step by step solution

01

Write down the given information

We are given the following: - Henry's law constant for hydrogen gas (H₂) = \(7.7 \times 10^{-6} M/kPa\) - Henry's law constant for argon (Ar) = \(1.4 \times 10^{-5} M/kPa\) - Pressure of hydrogen and argon = 253 kPa
02

Apply Henry's law formula for hydrogen gas

Using the Henry's law formula, we can calculate the solubility of hydrogen in water: \(C_{H_{2}} = k_{H_{2}} \cdot P_{H_{2}}\) Plugging in the values: \(C_{H_{2}} = (7.7 \times 10^{-6} M/kPa) \cdot (253 kPa)\)
03

Calculate the solubility of hydrogen gas

By performing the calculation: \(C_{H_{2}} \approx 1.95 \times 10^{-3} M\) So, the solubility of hydrogen gas is approximately \(1.95 \times 10^{-3} M\).
04

Apply Henry's law formula for argon

Similarly, we can calculate the solubility of argon in water: \(C_{Ar} = k_{Ar} \cdot P_{Ar}\) Plugging in the values: \(C_{Ar} = (1.4 \times 10^{-5} M/kPa) \cdot (253 kPa)\)
05

Calculate the solubility of argon

By performing the calculation: \(C_{Ar} \approx 3.54 \times 10^{-3} M\) So, the solubility of argon is approximately \(3.54 \times 10^{-3} M\). Therefore, the solubilities of hydrogen gas and argon at 25°C and 253 kPa are approximately \(1.95 \times 10^{-3} M\) and \(3.54 \times 10^{-3} M\), respectively.

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Most popular questions from this chapter

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of $0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\( of a solution that is \)5.75 \%$ glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

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