# Dissolving oxygen into water

I was wondering how one would actually calculate how much oxygen would dissolve into water given the necessary initial conditions, and what those initial conditions would need to be. I assume they would be pressure, and initial concentration, but I really don't know where I would go from there. Clearly air and water have different concentrations of gases and liquids, despite having been in contact for thousands of years. And once in water, is oxygen still considered gaseous? I assume it is, but why is it called gaseous-what quality of it deems it a gas despite being surrounded by liquid?

• As concentrations of O$_2$ and N$_2$ are low in water, one can use Henry's law to calculate the mole fraction of the solute (the gas) $x_g = p_g/k_H$ where $x_g$ is the mole raction of gas and $p_g$ its pressure above solution. This assumes ideal, dilute solutions. The constants $k_H$ are known and are $5.10^4$ bar for oxygen and $9.10^4$ bar for nitrogen gas in water and at 298 K. The smaller the $k_H$ the stronger the intermolecular forces between solute and solvent. The values for both these gases are very large indicating week intermolecular forces between these gases and oxygen. – porphyrin Jul 6 '16 at 20:52

Air is a mixture of gases, and the concentration of oxygen dissolved in solution (in the water) is proportional to the partial pressure of oxygen in the air. Raoult's law states: The vapour pressure of an ideal solution is directly dependent on the vapour pressure of each chemical component and the mole fraction of the component present in the solution.

• I'm not sure you answered my question-If oxygen constitutes a large population of air, or rather nitrogen for that matter, why is it that water doesn't consist of more nitrogen and oxygen? – user24082 Jun 16 '13 at 7:54

1) Calculate the Gibbs free energy for (in principle, by quantum mechanics and statistical methanics)

$\mathrm{O}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_2 (\mathrm{aq})$

(g, l, and aq stand for gas, liquid, and aqueous solution)

2) Using the the reaction isotherm equation (linked with ref 4) to obtain the equilibrium constant, $K$,

3) Input the initial pressure and concentration into the equilibrium constant related equation. E.g. for $$A + B \rightarrow C$$, the equilibrium constant relates to $$\frac{[C]}{[A][B]}=K \tag{1}$$. If we have initial concentration $c$ for both $A$ and $B$, 0 for $C$, Eq. (1) reads $$\frac{ x }{(c-x)(c-x)}=K$$

There may be experimental data available for the Gibbs free energy and equilibrium constant in steps 1 and 2.

In water the gas is in aqueous state, because gas molecules are interacting with molecules. At equilibrium the concentration of the gas in water and above water are different, with the ratio being a constant of Henry's Law.

Check out http://en.wikipedia.org/wiki/Henry's_law for a list of constants.

The most anomalous constant is for CO2, which despite being 1/500 as abundant in air as O2, in water it is 50/1. this is due to CO2 actually being able to chemically react with water. Granted, this is a special case but for other gas molecules the dominant forces are polar or van der Waals'.

O2 molecules have two free electron pairs for intermolecular interactions while N2 only have one pair. Electron density is a key factor in vdW forces, so this is the reason O2 and N2 have similar equilibrium concentrations in water despite N2 being 4/1 in atmospheric abundance/partial pressure

Well you wouldn't "calculate" it so much as measure it. You'd have some water, change the O2 partial pressure above it and measure how much dissolves. Then you'd have a chart where you can calculate the chemical potential of O2 in water vs. partial pressure, Henry's law coefficient, ect.

You could try to do a simulation. For example a Monte Carlo simulation. The problem here is that you'd need a very accurate and very cheap model of the energy. And then you'd have to do a proper job with the simulation itself.

Finally, there are some things that you can deduce analytically. Like, that the O2 will leave the solution with increasing temperature, or that a hydrophobic liquid (i.e. perfluorinated alkanes) might dissolve more O2.