Background: Henry's law states that:

At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

And mathematically:

$$S=k\times p_{\text{gas}}$$ where $S$ is solubility, $k$ is Henry's constant and $p$ is the partial pressure of the gas.

Question: I want to calculate the oxygen solubility down the sea. Is partial pressure the atmospheric one? If it is so, we have same solubility on all sea deepness, which is contradictory. What am I missing?


I know this law can be thought as depending on this equilibrium:

$$\ce{A {(g)} <=> A {(s)}}$$


1 Answer 1


What does it even mean to calculate or measure solubility of something down the sea?

Why, there is a precise and well-defined meaning, and it is hinged on equilibrium between the gas and the solution, which supposedly should be established at the said conditions. Our imaginary gas bubble, like everything else, must withstand the immense pressure of all those layers of water up above.

So no, the partial pressure is not just the atmospheric one.

Deep abyss of the seas is undersaturated in all gases, including oxygen, and that by a wide margin.

  • $\begingroup$ Do you mean deep on the sea there is no equilibrium with atmospheric gases? (I don't know why) so this law is just valid for a small surface layer of the sea? $\endgroup$
    – user43021
    Commented Jun 6, 2018 at 18:46
  • $\begingroup$ To be in equilibrium, two things have to be in touch to begin with. The gas of the atmosphere is not in touch with the deep waters, hence no luck. $\endgroup$ Commented Jun 6, 2018 at 18:54
  • $\begingroup$ I don't think so, we can suppose that there is a dynamic equilibrium between the oxygen down the water and on the top...; Suppose the oil water system, there you have no touch between liquid and water vapor, and a equilibrium, isn't it? $\endgroup$
    – user43021
    Commented Jun 6, 2018 at 19:02
  • $\begingroup$ OK, well, skip the "in touch" part. (Indeed, I have mentioned the oil/water system in a closed vessel on an earlier occasion.) The counterparts in equilibrium have to be in the same conditions, which they are not, thanks to the gravity potential. In a laboratory setup yon can neglect it; in 100m down the sea you can't. $\endgroup$ Commented Jun 6, 2018 at 19:11
  • $\begingroup$ Just in case, do you know any expression to estimate solubility in the example I ask? Taking int account gravity maybe $\endgroup$
    – user43021
    Commented Jun 6, 2018 at 19:59

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