Question

Pure water in equilibrium with air at 0C has an oxygen concentration equal to 14.6 mg/L. Compute KH,O2. (Henry's law)

Answer 1

This is Hnery Law

this states the law:

the concnetration of a species "i" in any solvent (mol per liter) of a gas is proportional to its partial pressure in the vapor/Gas phase

Then

M = Hc*Pi

M = concnetration of i in solvent

Hc = henry constant

Pi = partial pressure of i

then

PO2 = Pair*0.21

assume 1 atm so

PO2 = 0.21 atm

then

change henry constant to mol per liter

C = 14.6 mg/L

1 mol of O2 = 32 g

mol = mass/MW = (14.6*10^-3)/32 = 0.00045625 mol of O2 in 1 L

then

M = 0.00045625

and

M = Hc*Pi

Hc = M/Pi

Hc = (0.00045625)/(0.21) = 0.00217 M/atm