Question

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O 2 from the air is 2.67 × 10 –4 M at sea level and 25°C, what is the solubility of O 2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm?

Assume the temperature is 25°C, and that the mole fraction of O 2 in air is 0.209 at both 12,000 ft and at sea level. 1.75 × 10–4 M 2.67 × 10–4 M 3.66 × 10–5 M 4.06 × 10–4 M

Answer 1

Answer:

S = 1,75x10⁻⁴M

Explanation:

It is possible to answer this question using Henry's law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The formula is:

S = k×Pi

Where S is solubility, K is Henry's constant and Pi is partial pressure.

At sea level, Pi of O₂ is:

1atm×0,209 = 0,209 atm. Replacing:

2,67x10⁻⁴M = k×0,209atm

k = 1,28x10⁻³M/atm

At 12,000ft Pi of O₂ is:

0,657atm×0,209 = 0,137 atm. Replacing:

S = 1,28x10⁻³M/atm×0,137atm

S = 1,75x10⁻⁴M

I hope it helps!