The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry's law constants for oxygen and nitrogen are 3.30 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.
Percentage of oxygen (O2) in air = 20%
Percentage of nitrogen (N2) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg
= 7600 mm Hg
Therefore, Partial pressure of oxygen, pO2 = \(\frac{20}{100}\) x 7600 mm Hg
= 1520 mm Hg
Partial pressure of nitrogen, pN2 = \(\frac{79}{100}\) x 7600 mm Hg
= 6004 mmHg
Now, according to Henry's law:
p = KH.x
For oxygen:
pO2 = KH.xO2
⇒ xO2 = \(\frac{p_{O_2}}{K_H}\)
= \(\frac{1520mmHg}{3.30 \times 10^7 mm Hg}\) (Given KH = 3.30 x 107 mm Hg)
= 4.61 x 10-5
pN2 = KH.xN2
⇒ xN2 = \(\frac{p_{N_2}}{K_H}\)
= \(\frac{6004 \ mmHg}{6.51 \times 10^7 mm Hg}\)
= 9.22 x 10-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10-5 and 9.22 × 10-5 respectively.