Answer:
[tex]1.75\cdot 10^{-4} M[/tex]
Explanation:
Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:
[tex]S = k_H p^o[/tex]
Where [tex]k_H[/tex] 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:
[tex]S_1 = 2.67\cdot 10^{-4} M[/tex]
Also, at sea level, we have an atmospheric pressure of:
[tex]p = 1.00 atm[/tex]
Given mole fraction:
[tex]\chi_{O_2} = 0.209[/tex]
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:
[tex]p^o = \chi_{O_2} p[/tex]
Then the equation becomes:
[tex]S_1 = k_H \chi_{O_2} p[/tex]
Solve for [tex]k_H[/tex]:
[tex]k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm[/tex]
Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:
[tex]p = 0.657 atm[/tex]
Apply Henry's law using the constant we found:
[tex]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[/tex]
