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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.

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.

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 equilibrum constant to generate a value for $\ce{K_a(H2)}$ $= 8.35*10^{-39}$.

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.

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.

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 equilibrum constant to generate a value for $\ce{K_a(H2)}$ $= 8.35*10^{-39}$.