Chemistry: Equilibrium Math Question 2

Question

The equilibrium constant for dissolving oxygen gas in aqueous solution is 1.33E-4. Compute the equilibrium concentration of dissolved oxygen, in mg L^-1 when the partial pressure of O₂ above the solution is 0.89 bar.

Answer 1

Normally one can compute the concentration of dissolved gas in water by using Henry's Law:

C = kP where C is concentration, k is Henry's constant and P is the partial pressure of the gas above the liquid. Since the question gives the equilibrium constant and not Henry's constant, I'll approach it in a different way. But if the 1.33E-4 is supposed to be Henry's constant, then it would simply be [O2] = 2.33E-4 x 0.89 bar = 2.07E-4 and then convert to mg/L depending on units used for Henry's constant. Since it supposedly gives the Keq (and not Henry's constant), I did the following:

If we set up the following equilibrium of oxygen in air in equilibrium with oxygen in solution, we can write...

O_2(air) <-->O_2(aq) K= 1.33x10^-4

1.33x10^-4 = O_2(aq)/O_2(air) = x/0.8783 atm (converted 0.89 bar to atmospheres)

O_2(aq) = 1.17x10^-4 atm

Applying Henry's Law and using a constant of 1.3x10^-3 mol/L-atm, I calculate the following:

C = hP = 1.3x10^-3 mol/L-atm x 1.17x10^-4 atm = 1.5x10^-7 mol/L

1.5x10^-7 mol/L x 18 g/mol x 1000 mg/g = 0.0027 mg/L (this seems rather low, so not sure this is how they want the problem to be solved since they did not supply Henry's constant).