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 the Henry's law constants for oxygen and nbitrogen are `3.30xx10^(7)` and `6.51xx10^(7)` respectively, calculate the composition of these gases in water.

Explanation

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 `(10xx760)` mm Hg = 7600 mm Hg
Therefore,
Partial pressure of oxygen,
`P_(O_(2))=(20)/(100)xx7600` mm Hg
= 1520 mm Hg
Partial pressure of nitrogen,
`P_(N_(2))=(79)/(100)xx7600` mm Hg
= 6004 mm Hg 79
`= 79xx760` mm
Now, according to Henry.s law :
`p=K_(H)x` for oxygen :
`p_(O_(2))=KH.x_(O_(2))`
`x_(O_(2))=(P_(O_(2)))/(K_(H))`
`= ("1520 mm Hg")/(3.30xx10^(7)mm Hg)` (Given `K_(H)=330xx10^(7)` mm Hg)
`= 461xx10^(-5)`
For nitrogen :
`P_(N_(2))=KH.x_(N_(2))`
`x_(N_(2))=(P_(N_(2)))/(K_(H))`
`= ("6004 mm Hg")/(6.51xx10^(7)mm Hg)`
`= 9.22xx10^(-5)`
Hence, the mole fractions of oxygen and nitrogen in water are `4.61xx10^(-5)` and `9.22xx10^(-5)`.