I am wondering about the reaction $\ce{H+ + O^2- -> OH-}$. Is it possible?

I think this is not a possible reaction, because here a strong acid plus another particle reacts to a strong base. If it is possible, what would this mean for $\ce{O^2-}$ in the acid-base system? Would it be a very, very strong base?

  • $\begingroup$ Don't use mathjax in title. $\endgroup$
    – Mithoron
    Oct 2, 2015 at 19:03
  • $\begingroup$ Well, let's look from this POV. You take a strong acid $\ce{HCl}$, subtract a proton, and get $\ce{Cl-}$, which is not an acid at all. You take a weak acid $\ce{CH3COOH}$, subtract a proton, and get $\ce{CH3COO-}$, which is in fact a weak base. You take a very weak acid $\ce{H2O}$, subtract a proton, and get $\ce{OH-}$, which is a strong base. See the trend? $\endgroup$ Oct 2, 2015 at 19:12
  • $\begingroup$ So, yes it's very strong - one of strongest common bases. $\endgroup$
    – Mithoron
    Oct 2, 2015 at 19:29

1 Answer 1


You can try this in the lab: Take some $\ce{CaO}$, add $\ce{HCl}$ and see what happens. Actually, no don’t. Don’t try this unless you know exactly what you are doing.

A single proton does not exist, especially not in aquaeous solution. Using $\ce{H3O+}$ as a proton-donating species is a much better way to formulate reactions. A professor once said, the smallest actually proven proton-equivalent be $\ce{H9O4+}$, so four water molecules that shift the additional proton between them, but since it was first-year, he didn’t go into details. Also in organic solutions or others, protons will always be bound to something, however weakly.

Now that I got that out of the way, let’s take a look at the $\ce{H3O+}$ particle I just postulated. It has three hydrogens connected to an oxygen and in a very, very, very broad way of arguing, every hydrogen can be considered a potentially acidic proton. Therefore, we can think of hydronium as a triprotic acid with the following conjugate bases:

$$\ce{H3O+ ->C[- H+] H2O ->C[- H+] HO- ->C[- H+] O^2-}$$

Aha! Oxide ions can be considered as fully deprotonated water. Of course, the entire reaction series is reversible, so add ‘protons’ to oxide ions (whatever the source) and find that they get protonated to hydroxide and finally to water. You knew of the second step (hydroxide to water). And here’s the big surprise: The first one is identical, only the equilibrium is much, much further on the hydroxide’s side.

Adding excess acid to oxides would just create water. However, the acid-base neutralisation reaction is exothermic. Very exothermic. Which is precisely why I wrote do not try this unless you know exactly what you are doing at the top. If you have ever neutralised semi-concentrated hydrochloric acid and sodium hydroxide, you know how much heat can be created. Think more than twice that.

In fact, water has what is termed a nivelling effect: Water itself is too strong an acid to allow oxide bases to exist in its vicinity; so $\ce{O^2-}$ cannot be dissolved as is, it will immediately get protonated to hydroxide. Find some $\ce{CaO}$, dissolve it in water, and you would immediately get $\ce{Ca(OH)2}$. Oxides are only able to exist in aquaeous solutions if they are either complexed entirely by cations ($\ce{Al}$ can form oxido-hydroxido multi-core complexes at $\ce{pH 7}$ which include internal oxide ions) or if they are in a bound state, so not technically oxides, such as in the vanadyl cation $\ce{(VO)^2+}$.

Finally, you stated something in your question saying that a strong acid would react with something to a strong base which you think is impossible. That is not the case; if the product of the reaction of something with a strong acid is a strong base, then ‘something’ just needs to be an even stronger base. Which is precisely the case of oxide ions. Just to recall for a second that strong acids and even stronger bases react very, very exothermicly. I feel I have not stressed it enough: Do not try this unless you know exactly what you are doing!


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