Using Henry's Law At 0 °C and an [latex]O_2[/latex] pressure of 1.00 atm, the aqueous solubility of [latex]O_2(g)[/latex] is 48.9 mL [latex]O_2[/latex] per liter. What is the molarity of [latex]O_2[/latex] in a saturated water solution when the [latex]O_2[/latex] is under its normal partial pressure in air, 0.2095 atm? Analyze Think of this as a two-part problem. (1) Determine the molarity of the saturated [latex]O_2[/latex] solution at 0 °C and 1 atm. (2) Use Henry's law in the manner just outlined.

Determine the molarity of [latex]O_2[/latex] at 0 °C when [latex]P_{O_2} = 1 \text{ atm}[/latex]. [latex]\text { molarity } = \frac{0.0489 L O _2 \times \frac{1 mol O _2}{22.4 L O _2( STP )}}{1 L \text { soln }} = 2.18 \times 10^{-3} \text{ M} O _2[/latex] Evaluate the Henry's law constant. [latex]k = \frac{C}{P_{\text {gas }}} = \frac{2.18 \times 10^{-3} \text{ M} O _2}{1.00 atm }[/latex] Apply Henry's law. [latex]C = k \times P_{\text {gas }} = \frac{2.18 \times 10^{-3} \text{ M} O _2}{1.00 atm } \times 0.2095 atm = 4.57 \times 10^{-4} \text{ M} O _2[/latex] Assess When working problems involving gaseous solutes in a solution in which the solute is at very low concentration, use Henry's law.

Question 13.5: Using Henry's Law At 0 °C and an O2 pressure of 1.00 atm, th...

General Chemistry: Principles and Modern Applications

Using Henry’s Law

At 0 °C and an $O_2$ pressure of 1.00 atm, the aqueous solubility of $O_2(g)$ is 48.9 mL $O_2$ per liter. What is the molarity of $O_2$ in a saturated water solution when the $O_2$ is under its normal partial pressure in air, 0.2095 atm?

Analyze
Think of this as a two-part problem. (1) Determine the molarity of the saturated $O_2$ solution at 0 °C and 1 atm. (2) Use Henry’s law in the manner just outlined.

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Determine the molarity of $O_2$ at 0 °C when $P_{O_2} = 1 \text{ atm}$ .

$\text { molarity } = \frac{0.0489 L O _2 \times \frac{1 mol O _2}{22.4 L O _2( STP )}}{1 L \text { soln }} = 2.18 \times 10^{-3} \text{ M} O _2$

Evaluate the Henry’s law constant.

$k = \frac{C}{P_{\text {gas }}} = \frac{2.18 \times 10^{-3} \text{ M} O _2}{1.00 atm }$

Apply Henry’s law.

$C = k \times P_{\text {gas }} = \frac{2.18 \times 10^{-3} \text{ M} O _2}{1.00 atm } \times 0.2095 atm = 4.57 \times 10^{-4} \text{ M} O _2$

Assess
When working problems involving gaseous solutes in a solution in which the solute is at very low concentration, use Henry’s law.