Question

The solubility of oxygen, O2, in water is 7.20 ✕ 10−4 mol/L at 0°C when the nitrogen pressure above water is 0.554 atm. Calculate the solubility of oxygen in water when the partial pressure of oxygen above water is 1.083 atm at 0°C? The Henry's Law constant for oxygen is 1.30 ✕ 10−3 mol/ L atm.

Answer #1

Henry's Law expression:

Solubility of a gas (S) = K Pgas ....where K = Henry's law constant, P=pressure of a gas above solution, S=solubility in mol/L

Given, K = 1.30*10^-3 mol/Latm, P_{O2} = 1.083 atm, S=
?

S = (1.3*10^-3)*1.083 = 1.41*10^-3 mol/L

Thus the solubility of O2 at 1.083 atm and 0^{o}C =
**1.41*10^-3 mol/L**

Please note that the other information may be given to calculate Henry's law constant. But I am not sure about nitrogen, I think it is typographic, instead of oxygen they gave nitrogen. Because if we calculate K using 7.20 ✕ 10−4 mol/L and 0.554 atm,

K = 1.3*10^-3 mol/Latm.

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