calculate [N2] and [O2] in water under a barometric pressure of 735 torr?

2014-12-12 3:47 pm

Nitrogen and Oxygen comprise 78% and 20% respectively, of normal air. Given the Henrys law constants k=6.8×10^-4 mol/l × atm for (N2) and k=1.3×10^-3 mol/L × atm for (O2). Calculate [N2] and [O2] in water under barometric pressure of 735 Torr

Please can anyone make head or tails of this problem?

回答 (1)

hcbiochem

2014-12-12 3:53 pm

The concentration of a gas in solution is directly related to the pressure of the gas above the solution. The proportionality constant is the Henrys Law constant for that gas. So, answering the problem is really quite easy.

Since air is 78% N2, the partial pressure of N2 is 735 X 0.78 = 573.3 torr. However, your Henry's Law constant is given in terms of atmospheres, so the partial pressure of N2 in atm is 573.3/760 = 0.754

[N2] = 6.8X10^-4 (0.754) = 5.1X10^-4 mol/L

For O2, pressure O2 = 0.2 X 735 / 760 = 0.193 atm
[O2] = 1.3X10^-3 (0.193) = 2.5X10^-4 mol/L

Hope that helps...

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