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?

solubility: MM

At 298 K, what is the solubility of oxygen in water exposed to air at 0.891 atm?

solubility: MM

If atmospheric pressure suddenly changes from 1.00 atm to 0.891 atm at 298 K, how much oxygen will be released from 3.70 L of water in an unsealed container?

amount: mol

Answer 1

Solution-
In the atmosphere the partial pressure of oxygen is 1 atm * 21% O2

= 0.21 atm

Now 1 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 1 atm

= 0.000273 mole / L

Again at 0.891 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 0.891 atm

= 0.000243 mole / L

If as the pressure changes, the change in solubility = 0.000273 M - 0.000243 M

= 0.00003 M

So the concentration = mole / L, mole = concentration * volume

= 0.00003 mole / L * 3.70 L

= 0.000111 mole