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.