Question

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 nitrogen at 298 K are \(3.30 × 107 mm\) and \(6.51 × 107 mm\) respectively, calculate the composition of these gases in water

Answer 1

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, \(p^o_2=\frac{20}{100} \times 7600mmHg\)
= 1520 mm Hg
Partial pressure of nitrogen, \(p_{N_2}=\frac{79}{100} \times 7600mmHg\)
= 6004 mmHg 
Now, according to Henry's law: 
\(p = K_H.x \)
For oxygen:
\(p^o_2=K_H.x_{O_2}\)
\(⇒x_{O_2}=\frac{p_{O_2}}{K_H}\)
\(=\frac{1520mmHg}{3.30 \times 10^7mmHg}\)  (Given \(K_H=3.30 \times 10^7mmHg\))
\(=461 \times 10^{-5}\)
For nitrogen: 
\(p_{N_2}=K_H.x_{N_2}\)
\(⇒x_{N_2}=\frac{p_{N_2}}{K_H}\)
\(=\frac{6004mm\,Hg}{6.51 \times 107mm\,Hg}\)
\(=9.22 \times 10^{-5}\)
Hence, the mole fractions of oxygen and nitrogen in water are \(4.61 ×10^{-5}\) and \(9.22 × 10^{-5}\) respectively.