Suggested Response - Q1

**/*** Determine the equilibrium solubilities of oxygen (mg/L) in water at 298 K and 1 atm a, using atmospheric air (21% v/v oxygen) and pure oxygen. The Henry’s law constant for the solubility of oxygen in water at 298 K is 770 atm/mole per L.

Response:

Henry’s law describes the relationship between the partial pressure p of a gas and its solubility x (mol fraction) in the liquid phase:

p space equals space H x

Re-arranging for x:

x space equals space p over H

Recall that x is the molar fraction of oxygen dissolved in water.

But what value of partial pressure p shall we take for air and pure oxygen respectively? We know that air has a volume concentration of 21% oxygen. For pure oxygen, the partial pressure is the system pressure, so in this case 1 atm a. The concept of molar volumes allows us to state that the volume fraction of a gas in a mixture is equal to the molar fraction. If we combine this with Dalton’s law, the partial pressure of oxygen pO2 in air would be given by:

p subscript O subscript 2 space equals space P subscript T x subscript O subscript 2

x subscript O subscript 2 space equals space 0.21

Note that x subscript O subscript 2 end subscript is the molar fraction of oxygen in air. Don’t get confused with the x above, which is the mol fraction of oxygen in solution as stated in Henry’s Law.

Hence the partial pressure of oxygen in air is:

p subscript O subscript 2 space equals space 1 space cross times space 0.21 space equals space 0.21 space a t m space a

Similarly, Henry’s law for air is:

x space equals space fraction numerator 0.21 over denominator 770 end fraction space equals space 2.73 space cross times space 10 to the power of negative 4 end exponent space m o l e space p e r space L

Similarly, for pure oxygen:

x space equals space 1 over 770 space equals space 1.3 space cross times space 10 to the power of negative 3 end exponent space m o l e space p e r space L

Note that the units of the Henry’s law constant are atm/mole per L. So, we need to account for a gram mole, which for oxygen is given by 2 x 16 = 32 g/mole. Hence the equilibrium solubility in mg/L for pure oxygen is:

x space equals space 32 space cross times space 1.3 space cross times space 10 to the power of negative 3 end exponent space equals space 0.0416 space g divided by L

x space equals space 41.6 space m g divided by L

So, with 32 g/mol as above, we have: 

x space equals space 2.73 space x space 10 to the power of negative 4 end exponent space x space 32 space x space 1 comma 000 space equals space 8.73 space m g divided by L

Out of interest, and as we have predicted from calculations above, this latter figure is the level of oxygen we might expect in waterways at normal atmospheric pressure and temperature (less if warmer, more if colder) and it supports aquatic plant and fish life. Pollution or stagnation may cause these levels to be lower and therefore aquatic life may not be supported.

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