# Why does water dissociate to H3O+ and OH- rather than H+ and OH-?

Why does water dissociate to $\ce{H_{3}O^{+} + OH^{-}}$ instead of $\ce{H^{+} + OH^{-}}$?

This question came to surface when I was learning about acids and bases, and learned this definition:

$\mathrm{pH=}-\log_{10}[\ce H^{+}]$

Which after looking at wikipedia, looks like a simplified definition, but it got me thinking - how can water have a pH of around seven if it dissociates into $\ce{H_{3}O^{+}}$ instead of $\ce{H^{+}}$? Why does it do this instead of the expected dissociation?

• Something still needs to be mentioned, though I can't write an answer with the deserving care. Though of course it's important to underline the reactivity of bare $H^+$ and its tendency to be strongly solvated (by anything), the thing is that even $H_3O^{+}$ is incorrect in water. Exactly how one should represent the hydrated proton in the autodissociation of water is a bit of a nitpick; it doesn't really matter for most purposes, and honestly $H^+_{(aq)}$ is best because it does not erroneously suggest the nature of the solvation is entirely described by the equation. It's just a placeholder. – Nicolau Saker Neto Nov 2 '13 at 20:56
• @NicolauSakerNeto You are so right here, and it is very much of nitpicking. I think it is at least better to consider the first order approximation $\ce{H3+O}$ over the zeroth order $\ce{H+}$, since that is virtually impossible in condensed phase. And the again $\ce{H2O}$ is also only an approximation, the hydrogen bond network is much too strong to be neglected, but I have not yet seen $\ce{H2O_{(aq.)}}$. – Martin - マーチン Jul 23 '14 at 8:10

You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly self but a matter of "couple-ionization": you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen bonding. From there the electron from Hydrogen forms a lone pair with the oxygen of the same molecule forming $\ce{OH^-}$. Hydron ($\ce{H^+}$) is in fact a proton, a very reactive specie so is not so simple found it uncoupled and moreover the hydrogen is suppose to be bonded with an hydrogen bonding with the other water molecule so is more accurate use hydronium ($\ce{H_3O^+}$) instead of hydron...

However you don't have to worry about it for your pH calculation:

$\mathrm{pH=}-\log_{10}[\ce {H^{+}}]= -\log_{10}[\ce {H_3O^{+}}]$

Because, in fact, you have the same ratio between basic species ($\ce{OH^-}$) and acid species( $\ce{H_3O^+}$ or $\ce{H^+}$ as you prefer) in both cases!

Single water molecule dissociates into $\ce{H^+ + OH^-}$

As noted in @G M's answer, $\ce{H^+}$ is very reactive species and some people do not like it and write it as less provoking $\ce{H_3O^+}$, for discussion see first paragraph of http://en.wikipedia.org/wiki/Hydronium

In short - $\ce{H^+}$ organizes water molecules around itself and it's behavior is complicated and interesting.

Another point important point stated by @G M is

$\mathrm{pH=}-\log_{10}[\ce H^{+}]= -\log_{10}[\ce{H3O+}]$

which is correct, because the concentration of $\ce{H^+}$ is so low, that using $\ce{H_3O^+}$ does not change the concentration of $\ce{H_2O}$.