Question

Calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm (a typical value at high altitude). Atmospheric Gas Mole Fraction kH mol/(L*atm) N2 7.81 x 10-1 6.70 x 10-4 O2 2.10 x 10-1 1.30 x 10-3 Ar 9.34 x 10-3 1.40 x 10-3 CO2 3.33 x 10-4 3.50 x 10-2 CH4 2.00 x 10-6 1.40 x 10-3 H2 5.00 x 10-7 7.80 x 10-4

Answer 1

Answer : The solubility of nitrogen in water at an atmospheric pressure will be, $2.5125\times 10^{-4}mole/L$

Explanation :

First we have to calculate the partial pressure of nitrogen.

Formula used :

$p_{N_2}=X_{N_2}\times P_{atm}$

where,

$p_{N_2}$ = partial pressure of nitrogen = ?

$X_{N_2}$ = mole fraction of nitrogen = $7.81\times 10^{-1}$

$p_{atm}$ = atmospheric pressure = 0.480 atm

Now put all the given values in the above formula, we get :

$p_{N_2}=7.81\times 10^{-1}\times 0.480 atm$

$p_{N_2}=0.375atm$

Now we have to calculate the solubility of nitrogen in water.

Formula used :

$s_{N_2}=p_{N_2}\times K_H$

where,

$p_{N_2}$ = partial pressure of nitrogen = 0.375 atm

$s_{N_2}$ = solubility of nitrogen in water = ?

$K_H$ = Henry's constant = $6.70\times 10^{-4}mole/L.atm$

Now put all the given values in the above formula, we get :

$s_{N_2}=0.375atm\times 6.70\times 10^{-4}mole/L.atm$

$s_{N_2}=2.5125\times 10^{-4}mole/L$

Therefore, the solubility of nitrogen in water at an atmospheric pressure will be, $2.5125\times 10^{-4}mole/L$