Question

At 298 K, the Henry\'s law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.

At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm? _____M

At 298 K, what is the solubility of oxygen in water exposed to air at 0.893 atm? ______M

If atmospheric pressure suddenly changes from 1.00 atm to 0.893 atm at 298 K, how much oxygen will be released from 5.00 L of water in an unsealed container? _______mol

Answer 1

Partial preesure of O2 = P = (21/100) x 1 atm = 0.21 atm

from henry Law Kh = C/P

when atmosphere Pressure = 1atm , PO2 = 0.21 atm    , we have Kh = 0.0013M/atm we find C

0.0013 M/atm = C/0.21 atm

C = solubility = 0.000273 M

when atmosphere = 0.893 atm ,   PO2 = 0.21 x 0.893   = 0.18753 atm

now Kh = 0.0013 M/atm = C/0.18753 atm

C = 0.0002438 M

Now when atmospheric pressure drops from 1 atm to 0.893 atm the change in solubility

             = 0.000273 - 0.0002438 = 0.0000292 M = 0.0000292 moles /L

volume given = 5L

Moles of O2 released = M x V = 0.0000292 mol/L x 5L = 0.000146 = 1.46 x 10^ - 4 mol