# Are there actual H^+ or H_3O^+ ions in aqueous solution?

Aug 13, 2017

It would not be my rationalization.......

#### Explanation:

Water is a so-called protic solvent, and it can exchange protons with itself, and with other species in solution. We represent this reaction by the so-called autoprotolysis reaction:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

Under standard conditions of temperature and pressure, we measure the extent of this equilibrium:

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$

And taking ${\log}_{10}$ of each side......

${\log}_{10} {K}_{w} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$, and on rearrangement.....

$- {\log}_{10} {K}_{w} = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left({10}^{-} 14\right)$,

and thus.....

$14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{p H} - {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{p O H}$, i.e. $14 = p H + p O H$

So back to your question (finally!), we CONCEIVE of the acidium species in water as ${H}^{+}$ or ${H}_{3} {O}^{+}$ and represent it like that. The actual acidium species may be a cluster of water molecules with an extra proton, i.e. ${H}_{3} {O}^{+}$, or ${H}_{5} {O}_{2}^{+}$, or ${H}_{7} {O}_{3}^{+}$, with the protium ion exchanged between clusters. We happily use ${H}^{+}$ or ${H}_{3} {O}^{+}$ as labels of convenience.

Are you happy with this treatment?