At one atmosphere the solubility of pure oxygen in water is 46 mg/L if the water

Question

At one atmosphere the solubility of pure oxygen in water is 46 mg/L if the water contains no dissolved solids. (The solubility at this temperature and 1 atm can be used as the Henry's constant, K). (a) what would be the quantity of dissolved oxygen in pure water if the gas above the water is oxygen (A)? (b) What would be the solubility if the oxygen was replaced by nitrogen (B)? (c) And if it was then replaced by air? (C) Assume that in all three cases the total pressure of the gases above the water is one atmosphere and the temperature is 20 degreeC.

Explanation / Answer

6. According to Henry's Law the solubility(S) of a gas in water is directly proportional to the partial pressure of the gas above water.

S = KxPO2  

where K = Henry law constant = 46 mg/L.atm

(a) In part-A the gas above water is pure oxygen. Hence PO2 = 1 atm

=> S = (46 mg/L.atm) x 1 atm = 46 mg/L (answer)

(b) In part-B when the gas above water is pure nitrogen, partial pressure due to pure oxygen will be 0. Hence

PO2 = 0 atm

Hence S = (46 mg/L.atm) x 0 atm = 0 mg/L (answer)

(c) In part-C  when the gas above water is air, the partial pressure due to oxygen will be 0.2 atm. Hence

PO2 = 0.2 atm

Hence S = (46 mg/L.atm) x 0.2 atm = 9.2 mg/L (answer)

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