It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions. Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion. If I am correct at this point, how would the dynamics be different in these cases?
0
$\begingroup$
$\endgroup$
-
1$\begingroup$ It certainly wouldn't be $\ce{H_3O_2^{3-}}$. If anything, then $\ce{H_3O_2^{-}}$. $\endgroup$ – wythagoras Feb 26 '16 at 18:17
4
$\begingroup$
$\endgroup$
It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions.
This is not quite right. Because of the hydrogen bonding a wide variety of $\ce{H3O+(H2O)$_n$}$ ions are in dynamic equilibrium, where $n$ can take values as large as few tens. It has been shown that $\ce{H3O+}$ can donate three hydrogen bonds, so that it's first solvation shell of 3 water molecules, and the hydrogen bonds between $\ce{H3O+}$ and $\ce{H2O}$ are quite strong. This means that the $\ce{H3O+}$ cation can actually be considered as $\ce{H9O4+}$ in solution. For more details, see here.
Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion.
Reasoning by analogy is a very dangerous road. As mentioned by sheogorath there is no possibility for a coordinate covalent bond as in the case of $\ce{H3O+}$. But the hydrogen bonding is possible and the hydroxide ion is indeed strongly hydrated as well as $\ce{H3O+}$. The details of the hydration are not well known yet, but recent studies indicate $\ce{OH- (H2O)4}$ as the prominent complex in an aqueous solution. For details, see here.
1
$\begingroup$
$\endgroup$
$\ce{H3O}$+ is formed by $\ce{H2O + H+ = H3O+}$ where $\ce{O}$ share $2$ electrons to $\ce{H+}$ with no electron to form coordinate covalent bond .There is no possibility for oxygen to form such bond with $\ce{H2O}$ and $\ce{OH-}$
