# Why doesn't hydroxide associate with water analogously to proton forming hydronium?

I'm a novice at chemistry. I've heard that an acid needs water to show its acidic properties, that's why nitric acid needs to be associated with water to show its acidic properties:

$$\ce{HNO3 + H2O <=> NO3- + H3O+}$$

$$\ce{H+}$$ does not exist freely and is always associated with another molecule like $$\ce{H2O}$$.

But then you would expect that the hydroxide anion would need to associate with water to show its basic properties. However, reactions never show that. In the case of sodium hydroxide, it just shows the $$\ce{Na+}$$ cation and the $$\ce{OH-}$$ anion dissociate in solution. But by that logic, nitric acid in water should be $$\ce{NO3- + H+}$$, which shouldn't be the case.

Why does $$\ce{H+}$$ need to associate with $$\ce{H2O}$$ to become $$\ce{H3O+}$$, but $$\ce{OH-}$$ can remain as it is and not as

$$\ce{NaOH + H2O <=> Na+ + H3O2-}?$$

• It does, and so what? Like every single ion is hydrated, but it's only shown when there's a reason for that. Oct 7, 2022 at 23:35
• Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in titles. Oct 8, 2022 at 0:25
• See also en.wikipedia.org/wiki/Hydronium // Note that strong acids need not water to manifest their acidic properties. They manifest it by reacting with water as a base, because they are stronger acids than H3O+(aq). Oct 8, 2022 at 1:26
• $\ce{H}_3\ce{O}_2^-$ is documented here. Note the two parts aren't joined by a covalent bond.
– J.G.
Oct 9, 2022 at 10:09

One could argue that, even though all the following representations are wrong in aqueous solution, $$\ce{H+}$$ is much more wrong than $$\ce{H3O+}$$, while $$\ce{OH-}$$ is only somewhat more wrong than $$\ce{H3O2-}$$.
Almost uniquely in Chemistry, the idealized chemical species $$\ce{H+}$$ contains no electrons - it's a bare proton. Because a bare proton has a +1 charge concentrated in a volume around $$10^{15}$$ times smaller than virtually all other ions, it has a comparatively gigantic charge density, and therefore behaves very differently. Due to its extreme polarizing power, true isolated $$\ce{H+}$$ is supremely reactive and can force the formation of a covalent bond with virtually anything, even unreactive or non-basic species such as $$\ce{CH4}$$, $$\ce{N2}$$, $$\ce{H2}$$, $$\ce{He}$$, etc. Therefore, if $$\ce{H+}$$ is ever generated outside of a dilute gas phase, it will instantaneously combine with something, and whatever the resulting ion is, it will almost always have lost the vast majority of the reactivity inherent to isolated $$\ce{H+}$$. Therefore, it kind of makes sense to avoid writing $$\ce{H+}$$ (with no solvation implied) except in very special circumstances.
In the real world, it's most convenient to write $$\ce{H+\color{red}{(aq)}}$$ and $$\ce{OH^-\color{red}{(aq)}}$$ in chemical reactions to make stoichiometry simple and to get the point of the chemistry across, recognizing that solvation is a factor and whose details can be brought into the discussion if required.