Henry's Law constants for aqueous solutions at 25 °C are 8.20 × 10^-7 m/mmHg for N₂ and 1.62 × 10^-6 m/mmHg for O₂.Determine the solubility of oxygen in water under an atmospheric pressure of 760 mmHg,assuming that air is 80% N₂ and 20% O₂. A)6.2 × 10^-3 m B)2.5 × 10^-4 m C)1.3 × 10^-4 m D)5.0 × 10^-4 m E)1.2 × 10^-3 m

Question

Quizplus · Accepted Answer

Using Henry's Law, we can calculate the concentration of oxygen in water under atmospheric pressure as:

PO2 = Kh × CO2

where PO2 is the partial pressure of oxygen (760 mmHg), Kh is the Henry's Law constant for oxygen (1.62 × 10-6 m/mmHg), and CO2 is the concentration of dissolved oxygen.

Since air is 80% nitrogen and 20% oxygen, we can assume that the partial pressure of oxygen in air is 0.2 × 760 = 152 mmHg.

Using the same equation for nitrogen,

PN2 = Kh × CN2

where PN2 is the partial pressure of nitrogen (608 mmHg), Kh is the Henry's Law constant for nitrogen (8.20 × 10-7 m/mmHg), and CN2 is the concentration of dissolved nitrogen.

Since we know that the total pressure of air is 760 mmHg and that nitrogen makes up 80% of air, we can calculate that PN2 = 0.8 × 760 = 608 mmHg.

Finally, since the total pressure is the sum of the partial pressures of oxygen and nitrogen, we have:

PO2 + PN2 = 760 mmHg

152 mmHg + PN2 = 760 mmHg

PN2 = 608 mmHg

Now we can solve for the concentration of oxygen:

CO2 = PO2 / Kh
CO2 = 152 mmHg / 1.62 × 10-6 m/mmHg
CO2 = 2.5 × 10-4 m

Therefore, the solubility of oxygen in water under atmospheric pressure is 2.5 × 10-4 m or 2.5 × 10-4 mol/L. Converted to scientific notation, this is 2.5 × 10-4 m = 2.5 × 10-7 km = 2.5 × 10-4 cm = 2.5 × 10-3 mm or 2.5 × 10-4 µm. The answer is closest to choice B (2.5 × 10-4 m or 2.5 × 10-4 mol/L).

Henry's Law Constants for Aqueous Solutions at 25 °C Are