I think I got it figured out using electrochemical means, thanks to Aditya's suggestion to consult Ka values rather than Kb values. 

Consider these two half reactions:

$\ce{2e^- +H_2->2H^-}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~E^o=-2.25 ~V$

$\ce{H_2 + 2H_2O -> 2H_3O^+ + 2e^-}$ $~~~~~~~~~~~~~~~~~E^o=0.00~V$

Coupling these two half reactions results in:

$\ce{2H_2 +2H_2O ->2H_3 O^+ +2H^- }$ $~~~~~~~~~~~~~E^o=-2.25~V$

Application of the Nernst equation can help us find an equilibrium constant for this reaction. 

$ΔG^o = -nFE^o = -(2~e^-)(96,500~ C/mol)(-2.25~V) = +434,250$

Value makes sense; we'd expect the reaction of hydrogen gas as an acid with water to be highly disfavorable. 

$ΔG^o = -RTlnK=-(8.31~J/(mol*K))(298~K)lnK=+434,250$

$K = 6.97464*10^{-77}$

Now, this K correspond to this equilibrium: 

$\ce{2H_2 +2H_2O ->2H_3 O^+ +2H^- }$ 

So we must take the square root of the found equilibrium constant to generate a value for $\ce{K_a(H2)}$ $= 8.35*10^{-39}$. 

And finally this lines up well with Aditya's finding that the pKa of H2 is 35; the -log of the above Ka value I found is 38. Nice.