What is the property of a water molecule that allows it to split into $\ce{OH-}$ and $\ce{H+}$ ions?

And subsequently, why - when these ions are attracted towards each other in a neutralisation reaction, for example - can these ions then become a single covalently bonded molecule?


3 Answers 3


The splitting described doesn't actually happen in bulk water. There is just the autoionization reaction:

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

The net transfer of a proton from one water to another creates two ions. This is somewhat disfavored. Hence, the autoionization constant is quite small:

$$\mathrm{K}_{w} = \ce{[H3O+][OH-]} = 10^{-14}$$

Neutralization is a somewhat related process, whereby one species is able to create an excess of hydronium, and another is able to create an excess of hydroxide. This creates an imbalance in the autoionization equilibrium, and it rapidly shifts left to consume excesses of both ions.

  • 1
    $\begingroup$ The chemical reaction is absolutely correct as shown, but it belies the situation. The bare ions don't just float around in the water. Rather each is solvated. In other words The $\ce{H^+}$ attracts oxygen atoms of adjacent water molecules to form multiple "shells" of water molecules oriented around the $\ce{H^+}$ ion. Likewise the $\ce{OH^-}$ ion attracts H atoms of water molecules to form oriented layers. $\endgroup$
    – MaxW
    Sep 20, 2017 at 16:46
  • $\begingroup$ As @MaxW says, solutions are infinitely more complicated than what we can capture in a single line of text. The goal of this answer is to help the OP understand the nature of the "split." $\endgroup$
    – Zhe
    Sep 20, 2017 at 17:23
  • $\begingroup$ Well the point of discussing solvation is that it stabilizes the split. Small amounts of water in different solvents have vastly different dissociation constants. $\endgroup$
    – MaxW
    Sep 21, 2017 at 3:51

Water does not really split into ions. It transfers protons, reversibly, from one molecule to another:

$\ce{2 H2O = H3O+ + OH-}$

The ions represent, of course, a less stable state than molecular water, at least under ordinary conditions, so the equilibrium constant is low. Yet this reaction strongly affects the chemical prpoeryies and even some physical properties of aqueous solutions.

Such a transfer reaction is possible because of the polarity of the hydrogen-oxygen bonds in water. But if that is so, then might other solvents with similarly polar bonds to hydrogen transfer protons in a similar way?

Of course they do. Ammonia does it, forming ammonium and amide ions; although it does so only weakly because you have to chill it to make it liquid and the nitrogen-hydrogen bonds are only weakly polar. Sulfuric acid does it (and undergoes some other solvent reactions such as oligomerization) to such an extent that "100% sulfuric acid" contains a significant amount of things other than $\ce{H2SO4}$. Even some organic liquids like ethanol and glacial acetic acid undergo intrinsic proton transfer.

But wait, there's more. Solvents that contain no hydrogen but do contain halogens with polar bonding can transfer halide ions, like bromine trifluoride ($\ce{BrF3}$) forming $\ce{BrF2+}$ and $\ce{BrF4-}$ ions.

Dinitrogen tetroxide, $\ce{N2O4}$, is an exception, as it really does split molecules to form $\ce{NO+}$ and $\ce{NO3-}$. The relatively weak nitrogen-nitrogen bond allows that, but even here an oxygen atom must also be transferred between the fragments. Direct cleavage would produce species that are too strongly acidic ($\ce{NO2+}$) and too strongly basic ($\ce{NO2-}$) for the solvent.

Among common solvents with polar bonds, the one that most stands out is not water but sulfur dioxide. It fails to autoionize to any significant extent, because it would have to form radicals to make singly charged species and the alternative of forming doubly charged ions ($\ce{SO^{2+}, SO3^{2-}}$) is too highly disfavored for just about any solvent (even water).


The "bare" proton (H+) does exist. But not in equilibrium aqueous solution (near STP). Chemists talk about H+ and Physicists do too, but generally they are talking about very different things. The Physicists actually are (usually) talking about H+, and the Chemists are almost always talking about the various species which have been protonated. In the case of water, the two most common cations are H3O+ and H5O2+ which can be thought of as n(H2O)H+ with n a small integer. We chemists call all of these cations "the (aqueous) proton" although we know 'taint so. Its a short-hand. What is going on is that while liquid water has its molecules moving and dancing around, at any instant of time, most of the molecules are in one or more hydrogen bonds. With the H atom often bonded to one molecule's O atom and h-bonded to a second molecule's O atom. (a single O atom can be associated with more than one single H-bond at any given instant, due to its much larger size). Let me simplify this: any given HOH molecule h-bonded to an O of another molecule will have the h-bond looking like this O-H....O, but also like this O-H...O and this O-H..O. That is, it isn't correct to think of the three atoms O,H and O as "exactly" covalently bonded and "exactly" h-bonded but on a spectrum between O-H....O (neutral) and O....H+O ionized. (where I've ignored the other bonds in the two molecules). If 1 h-bond in 10,000,000 results in the second case (the ion) then the pH would be 7, and guess what? it is. So, we can say that the proton (although we don't mean a bare proton) engages in the H2O→OH- + H+ reaction rarely, but when you've got 6E23 molecules in 20grams of water, there are a LOT of "rare" events. The proton is able to "escape" its molecule because that molecule's negative charge is being partially neutralized by other h-bonds --I mean you would expect OH- to be h-bonding in water, wouldn't you?? So, while the h-bonds "explains" why water dissociates, you shouldn't look at these as stable bonds, it's more of a one night hook-up, or perhaps even an orgy of constantly changing hookups.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.