The dissociation of $\ce{HCl}$ in water is written as:

$$\ce{HCl (aq) -> H+ (aq) + Cl- (aq)} \tag{1}$$

Sometimes we also include $\ce{H2O}$ as a reactant to stress the fact that the protons are not isolated in water.

$$\ce{HCl (aq) + H2O (l)-> H3O+ (aq) + Cl- (aq)} \tag{2}$$

The last equation is quite useful to establish what is the conjugate acid and what is the conjugate base in Brønsted-Lowry's theory, as far as I understand.

Let's do the same thing for a strong base, say $\ce{NaOH}$. We write the dissociation in water:

$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)} \tag{3}$$

What if I wanted to write equation $(3)$ in a form analogous to equation $(2)$? I tried including water as a reactant in the same way but then it yields a chemical equation where $\ce{H2O}$ is on both sides, so we can elide it.

In Brønsted-Lowry theory we generally write $\ce{H2O}$ in the dissociation for acids and for some basis, say $\ce{NH3}$. Why don't we do it for Arrhenius bases like $\ce{NaOH}$ or other hydroxides?

  • $\begingroup$ Use a different strong base, like $\ce{NaOCH3}$ $\endgroup$
    – Karsten
    Commented Sep 7, 2020 at 23:48
  • $\begingroup$ Or write hydroxide hydrate as the product. $\endgroup$
    – Karsten
    Commented Sep 7, 2020 at 23:52
  • $\begingroup$ $NaOH+H_2O\rightarrow Na^+ +H_3O_2^-$ $\endgroup$ Commented Sep 8, 2020 at 11:55
  • 3
    $\begingroup$ Yes, that's what I meant. I don't know why we do that. At least, hydroxide by itself is fairly common in ionic solids, so it is more of a thing. $\ce{H3O+}$ is not that accurate either, there is also $\ce{H5O2+}$ (Zundel cation) and $\ce{H9O4+}$ (Eigen cation). See e.g. en.wikipedia.org/wiki/Hydrogen_ion $\endgroup$
    – Karsten
    Commented Sep 8, 2020 at 12:22
  • 1
    $\begingroup$ I would add that $\ce{H+}$ is a simple proton. It is not an atom. It deserves a special treatment. All atoms are about the same size, between $\pu{5 · 10^{-11}}$ m to $\pu{5 · 10^{-10}}$ m. A proton is at least thousand times smaller. It must react by addition on a water molecule yielding a "usual" ion, namely $\ce{H3O^+}$. The hydroxyl ion $\ce{OH-}$ does not have this property. $\endgroup$
    – Maurice
    Commented Nov 19, 2021 at 17:43

1 Answer 1


Not all strong acids are like HCl

There are unusual strong acids that have ionic compounds with hydronium as the cation. Hydronium perchlorate is an example. It's dissolution in water would be written like this:

$$\ce{H3OClO4(s) -> H3O+(aq) + ClO4-(aq)}$$

This looks very similar to the dissolution of $\ce{NaOH}$.

Not all strong bases are like NaOH

Bases like $\ce{NaOCH3}$ lack hydroxide as an anion. The dissolution (and then reacting as a base) would be written like this:

$$\ce{NaOCH3(s) -> Na+(aq) + OCH3-(aq)}\tag{1}$$ $$\ce{OCH3-(aq) + H2O (l) -> CH3OH(aq) + OH-(aq)}\tag{2}$$

Leaving out the second part would not explain why the pH changes upon dissolution, and would not reflect the major products.

What actually happens to NaOH

If you label the oxygen of $\ce{NaOH}$, it would make sense to write the following:

$$\ce{Na^17OH(s) + H2O(l) -> Na+(aq) + ^17OH-(aq)}\tag{1}$$ $$\ce{^17OH-(aq) + H2O(aq) <=> H2^17O(aq) + OH-(aq)}\tag{2}$$

The equilibrium of the second reaction would be on the right hand side because the concentration of water is higher than that of hydroxide. So in that sense, the hydroxide in NaOH does react with water, yielding water and hydroxide.

Why are they different?

In Brønsted-Lowry theory we generally write H2O in the dissociation for acids and for some basis, say NH3. Why don't we do it for Arrhenius bases like NaOH or other hydroxides?

Acids and bases can change the concentration of hydroxide and hydronium ions in aqueous solution either by reacting with water, or by providing these ions directly. Metal hydroxides do the latter, while $\ce{HCl}$ and $\ce{NH3}$ and other acids and bases to the former.


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