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 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 10^{7} mm and 6.51 × 10^{7} mm respectively, calculate the composition of these gases in water.

#### Solution

Percentage of oxygen (O_{2}) in air = 20 %

Percentage of nitrogen (N_{2}) 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, `p_(o_(2)) = 20/100xx7600` mm Hg

= 1520 mm Hg

Partial pressure of oxygen, `p_(N_(2)) = 79/100xx7600` mm Hg

= 6004 mmHg

Now, according to Henry’s law:

*p *= *K*_{H}.*x*

For oxygen:

`p_(o_(2)) =K_H.x_(o_2)`

`=> x_(o_2) = p_(o_(2))/K_H`

`= (1520 "mm Hg")/3.30xx10^(7) mm Hg` (Given `K_H = 3.30 xx 10^(7)` mm Hg)

= 4.61 x 10^{-5}

For nitrogen:

`p_(N_(2)) = K_H.x_(N_(2))`

`=>x_(N_(2)) = p_(N_(2))/K_H`

` = (6004 "mm Hg")/(6.51xx10^(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.