Calculate the solubility of oxygen in water at the top of Mt. Everest where the atmospheric pressure is 0.320 atm.

Question

Calculate the solubility of oxygen in water at the top of Mt. Everest where the atmospheric pressure is 0.320 atm. The mole fraction of O₂ in air is 0.209. Assume the k_H of O₂ is 1.30 ×× 10^–3 mol/(L • atm)

The Henry's law constant for oxygen at 20°C is 1.3 ×× 10^–3 mol/(L • atm)

Answer 1

Henry’s Law:

C = kP

C = concentration = ?

k = Henry’s constant = 1.0 × 10^–3 mol/(L • atm)

P = partial pressure of O₂ = 0.0631 atm (see below)

The partial pressure of O₂ can be determined using Raoult’s law that the partial pressure of a gas is the mole fraction of that gas times the total pressure. Thus.

P = 0.209 x 0.302 = 0.0631 atm

C = 1.30 × 10^–3 mol/(L • atm) x 0.0631 atm

C = 8.21x10^-5 mol /L