What is the Ka of OH- and Kb of H3O+?

Question

What is the $K_\mathrm{a}$ of $\ce{OH^-}$ and $K_\mathrm{b}$ of $\ce{H_3O^+}$? Have these constants been determined?

Answer 1

In water both equilibrium constants will be extremely small; the concentrations of $\ce{O^2-}$ or $\ce{H4O^2+}$ ions in water are essentially zero.

According to Cotton and Wilkinson's Advanced Inorganic Chemistry, for the hydrolysis of solid oxide ions you have

$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)} \qquad \qquad K_b > 10^{22}$$

so you can never have any significant concentration of oxide ions in aqueous solution. As DavePhD points out, $K_\mathrm{a}$ for the hydroxide ion is less than $K_\mathrm{w}/K_\mathrm{b} = 10^{-36}$.

Now, $\ce{H3O+}$ may have some capacity to act as a base in media like $\ce{HF:SbF5}$ (see this paper on the role of $\ce{H4O^2+}$ in isotopic exchange reactions between hydronium ions). $\ce{H4O^2+}$ can exist in sulfolane solution, too (see this paper).

If $\ce{H4O^2+}$ exists in water at all, it's probably going to actually be two protons bridged and hydrated by a lot of waters, rather than an actual $\ce{H4O^2+}$ ion. It will have a much shorter lifetime than even a hydronium ion. The molecule can exist in theory, though, and its electronic structure has been studied for at least half a century (see Rosenfeld's 1964 paper including SCF calculations on $\ce{H4O^2+}$).

Answer 2

For $\ce{O^2-}$, the $\mathrm{p}K_\mathrm{b}$ is approximately $-22$, so the $\mathrm{p}K_\mathrm{a}$ of hydroxide is about $36$.

sources: http://www.newworldencyclopedia.org/entry/Oxide and https://www.chem.tamu.edu/rgroup/hughbanks/courses/462/lecturenotes/class6-2.pdf

For hydronium, the reference Fred Senese cited

Protonated hydronium dication...

suggests that at acidities of Hammet acidity function $H_o = -25$ to $-28$, protonated hydronium can be present.