I think I got it figured out using electrochemical means, thanks to Aditya’s suggestion to consult $K_\text{a}$ values rather than $K_\text{b}$ values.
Consider these two half reactions:
$\ce{2e- +H2->2H-}\tag{$E^\circ=-2.25~\mathrm{V}$}$
$\ce{H2 + 2H2O -> 2H3O+ + 2e-}\tag{$E^\circ=0.00~\mathrm{V}$}$
Coupling these two half reactions results in:
$\ce{2H2 +2H2O ->2H3O+ +2H- }\tag{$E^\circ=-2.25~\mathrm{V}$}$
Application of the Nernst equation can help us find an equilibrium constant for this reaction.
$\Delta G^\circ = -nFE^\circ = -(2)(96\,500~\mathrm{C/mol})(-2.25~\mathrm{V}) = +434\,250~\mathrm{J/mol}$
Value makes sense; we’d expect the reaction of hydrogen gas as an acid with water to be highly disfavorable.
$\Delta G^\circ = -RT\ln K=-\left(8.31~\mathrm{J/(mol\cdot K)}\right)(298~\mathrm{K})\ln K=+434\,250~\mathrm{J/mol}$
$K = 6.97464\times10^{-77}$
Now, this $K$ correspond to this equilibrium:
$\ce{2H2 +2H2O ->2H3O+ +2H- }$
So we must take the square root of the found equilibrium constant to generate a value for $K_\text{a}(\ce{H2})= 8.35\times10^{-39}$.
And finally this lines up well with Aditya’s finding that the $\mathrm{p}K_\text{a}$ of $\ce{H2}$ is 35; the −log of the above $K_\text{a}$ value I found is 38. Nice.
