Question

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air is 2.67 ✕ 10-4 M at sea level and 25°C, what is the solubility of O2 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 O2 in air is 0.209 at both 12,000 ft and at sea level.

Answer 1

The solubility of oxygen at an elevation of 12,000 ft, with an atmospheric pressure of 0.657 atm, can be calculated using Henry's law. The solubility is determined by the partial pressure of the gas above the liquid, and can be calculated using the mole fraction of O2 and the Henry's law constant. Using these calculations, the solubility of oxygen at 12,000 ft is determined to be 3.66 x 10^-5 M.

Step-by-step explanation:

In order to determine the solubility of oxygen at an elevation of 12,000 ft, we can use Henry's law. Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula to calculate the solubility (S) is S = k x P, where k is the Henry's law constant and P is the partial pressure of the gas.

Given that the mole fraction of O2 in air is 0.209 at both sea level and 12,000 ft, we can use the mole fraction to calculate the partial pressure of oxygen at 12,000 ft. The partial pressure of oxygen at sea level is 0.209 x 1 atm = 0.209 atm. Using the equation P1V1/T1 = P2V2/T2, we can calculate the partial pressure of oxygen at 12,000 ft, which is 0.209 atm x (0.657 atm/1 atm) = 0.1372 atm.

Now, we can use Henry's law to calculate the solubility of oxygen at 12,000 ft. The solubility at sea level (S1) is 2.67 x 10^-4 M, and the partial pressure at sea level (P1) is 1 atm. Using the formula S1/P1 = S2/P2, we can calculate the solubility at 12,000 ft (S2) as (2.67 x 10^-4 M) x (0.1372 atm/1 atm) = 3.66 x 10^-5 M. Therefore, the solubility of oxygen at an elevation of 12,000 ft is 3.66 x 10^-5 M.

Answer 2

`1.75\cdot 10^(-4) M`

Step-by-step explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

`S = k_H p^o`

Where
`k_H` is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

`S_1 = 2.67\cdot 10^(-4) M`

Also, at sea level, we have an atmospheric pressure of:

`p = 1.00 atm`

Given mole fraction:

`\chi_(O_2) = 0.209`

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

`p^o = \chi_(O_2) p`

Then the equation becomes:

`S_1 = k_H \chi_(O_2) p`

Solve for
`k_H`:

`k_H = (S_1)/(\chi_(O_2) p) = (2.67\cdot 10^(-4) M)/(0.209\cdot 1.00 atm) = 0.001278 M/atm`

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

`p = 0.657 atm`

Apply Henry's law using the constant we found:

`S_2 = k_H \chi_(O_2) p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^(-4) M`