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The oceans have dissolved gases as N2 or O2.

Henry's law establishes there is an equilibrium between the atmospheric concentration of a gas and its concentration in the ocean.

The atmosphere has H2O(g).

Is there H2O(g) dissolved in the oceans or Henry's law does not apply because the water is also H2O?

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    $\begingroup$ H2O dissolved in H2O is H20. There will be no separate gas phase, only liquid. When looking at the vapour pressure of water in mixtures containing a small amount of other components, you need to look up Raoult's Law. $\endgroup$
    – haresfur
    Feb 17 '20 at 23:40
  • $\begingroup$ I think this is a very fundamental question about matter that could be asked in Chemistry SE or Physics SE, but "Can Henry's law be applied to a substance dissolving in itself?" is not really specific to Earth Science. $\endgroup$
    – uhoh
    Feb 18 '20 at 9:04
  • $\begingroup$ @uhoh but the specialist here is the oceanographer. I ask if something happens at oceans. $\endgroup$
    – user18590
    Feb 18 '20 at 9:11
  • $\begingroup$ Okay let's see what other people think. The comment above about Raoult's law seems to offer some insight. I have a hunch that this is a general question about the thermodynamics of a gas mixture above a liquid mixture and is askable in Physics or Chemistry SE, but there's nothing specific about your question that makes it about an ocean per se. $\endgroup$
    – uhoh
    Feb 18 '20 at 9:24
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    $\begingroup$ If you look at a phase diagram of water, there is a domain where water shows both liquid and vapour properties (supercritical fluid): en.wikipedia.org/wiki/… Now the question is: could an ocean be at such P and T conditions? Maybe on a hot planet with a dense atmosphere, like Venus? $\endgroup$
    – Jean-Marie Prival
    Feb 18 '20 at 12:12
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The answer is no, yes and then, perhaps, no.

No: water vapour is not "dissolved" in the oceans, rather it becomes part of the oceans through a phase change from vapour to liquid (possibly via an intermediary stage as ice). The process of "dissolving" refers to a substance entering a "solution", which is defined as stable mixture of two or more substances in which the state of the solution is determined, generally, by the solvent (the liquid water in this case). For example, see the wikipedia article on "solution". When water vapour enters the ocean, it is just adding to the volume of liquid water. Water vapour bubbles suspended in the ocean is another issue. When a gas is dissolved in a liquid, it ceases to be a gas. So we do not have "O2 gas" dissolved in the ocean, we have O2, which is a gas in the atmosphere, present in the ocean as a solute.

Yes: on the other hand, there is continuous exchange of water between the atmosphere and the ocean: sometimes water passes from the atmosphere to the ocean, sometimes the other way around, depending on the weather conditions. This is nothing to do the the process of being "dissolved", but, if the question is motivated by an interest in the exchange of compounds between the atmosphere and the ocean, this cycling of water between the two, which is described more below, will be of interest.

Finally, perhaps, no: the water vapour in the atmosphere is not in equilibrium with liquid water in the ocean (see wikipeadia foe discussion of equilibrium between liquid and gas phases). Generally speaking, the partial pressure of water vapour above the ocean will be below equilibrium, so there will be an evaporative flux of water from the ocean into the atmosphere. The steady evaporation of water from the surface of the oceans is balanced by precipitation (rain, sleet and snow) out of the atmosphere which falls on the oceans or falls on land and then drains into the oceans (see The Fundamentals of the Water Cycle).

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  • $\begingroup$ Your second and third paragraphs are correct, there is continuous exchange, which is not in equilibrium, but that has nothing to do with what is "dissolved" in the normal sense of the word. A more exact geochemical description would use thermodynamics to define the phases present and the relationships between the composition of the phases. Thermodynamics will describe this in terms of the phase equilibria. If you want to look at disequilibrium, then you need to describe the kinetic reaction rates. $\endgroup$
    – haresfur
    Feb 19 '20 at 0:35
  • $\begingroup$ I agree, except that, in the case of water vapour above the ocean, we don't have an equilibrium between the gas and liquid phases. So it is a question of phase changes rather than phase equilibria, isn't it? I'll edit the answer to clarify that the third paragraph ("Yes: on the other hand ..." does not refer to anything being "dissolved"). $\endgroup$
    – M Juckes
    Feb 19 '20 at 7:54
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TL;DR: Henry's law describes an ideal linear relationship between the equilibrium amount of a low-concentration solute in a solution and the partial pressure of the solute in the gas phase above the solution. Raoult's Law describes the ideal linear relationship between the concentration of the dominant solvent in a solution and the partial pressure of the solvent in the gas phase above the mixture. Since water is the dominant solvent in the oceans, Henry's law does not apply to it. Any water vapour from the atmosphere that dissolves in the oceans simply adds to the solvent.

Henry's law describes the behaviour of solutes in a pure solvent. I will quote parts of the WikiPedia article you cited to bring out some relevant points.

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

One point neglected in the article is that it is assumed that the gas phase is an ideal mixture of different gases - that is that they follow the ideal gas law. This is a very good approximation for the earth's atmosphere. For ideal gases, the concentration in solution then is related to the concentration in air by the Henry's law constant. The value of the constant depends on how the equation is written and the concentration units used.

Henry's law has been shown to apply to a wide range of solutes in the limit of infinite dilution (x → 0), including non-volatile substances such as sucrose. In these cases, it is necessary to state the law in terms of chemical potentials. For a solute in an ideal dilute solution, the chemical potential depends only on the concentration. For non-ideal solutions, the activity coefficients of the components must be taken into account

Henry's law becomes exact at very low concentrations of the solute. At higher concentrations, there is deviation from ideality due to interactions between the dissolved species. This means Henry's law is only used for low-concentration solutes. At the other end, Raoult's law applies. So for H2O in the oceans the water is nearly pure so it behaves as an ideal liquid. In thermodynamic terms the ratio between the water partial pressure at equilibrium in air above the solution (the ocean) and the water partial pressure above pure water times the mole fraction of water in the ocean (the mole fraction is the number of moles of water divided by the total number of moles of the solution - so close to 1 because the ocean is mostly water).

If there was absolutely complete ideal mixing between two solutes, say ethanol and water, then Raoult's law would be followed for all mixtures between the two. This is the same as saying the Henry's law constant would be 1 if the concentrations are given in mole fractions. That is not usually the case and the slope of the linear Henry's law region is different. Here is a figure for ethanol water from a more rigorous answer to a question in chemistry stack exchange.

https://chemistry.stackexchange.com/questions/89872/what-are-the-key-differences-between-raoult-s-law-and-henry-s-law/90771

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I'm not a chemist, but I'm prepared to take a punt at the answer:

No, there is no water dissolved in the oceans. The water is the oceans.

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  • $\begingroup$ And yet I can heat some ocean water and extract as vapor the dissolved water that isn't there! ;-) $\endgroup$
    – uhoh
    Feb 18 '20 at 9:25
  • $\begingroup$ Sorry for unmarcking your answer. I thought the question was poor or obvious and I like your sentence "the water is the oceans". But haresfur developed it a bit and fully answering the things of the question. $\endgroup$
    – user18590
    Feb 19 '20 at 12:00
  • $\begingroup$ @Universal_learner quite right, the new accepted answer is much better! $\endgroup$
    – Semidiurnal Simon
    Feb 19 '20 at 12:25
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The answer to your question is yes and no.

You will not find dissolved H2O in water but water can have dissolved O2 and you can find dissolved hydrogen in water i do not know how common it is in nature but it is possible to dissolve hydrogen in water.

And it looks like hydrogenated water is a thing https://en.wikipedia.org/wiki/Hydrogen_water

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  • $\begingroup$ The part about dissolved gases in water isn't related to the question. Water is the solvent and the dissolved gases present at low concentrations are the solutes. All gases in the atmosphere can dissolve in the oceans to some degree. $\endgroup$
    – haresfur
    Feb 19 '20 at 0:27
  • $\begingroup$ @haresfur thank you for your input.the rules here is to upvote good questions and answers and to not vote/down vote bad ones not to attack the users. $\endgroup$
    – trond hansen
    Feb 19 '20 at 6:57
  • $\begingroup$ no attack intended $\endgroup$
    – haresfur
    Feb 21 '20 at 0:02

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