1
$\begingroup$

I can't find any data about the Kb value of the hydride anion except that it's huge. Does anyone know what it is, or how I can characterize it?

I know I can use electrochemical means but the only reaction I find with hydride ion is this one:

http://en.wikipedia.org/wiki/Hydride#Hydride_ion

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Recall that $\mathrm{p}K_\text{b}$ of hydride is actually $14-\mathrm{p}K_\text{a}$ of hydrogen gas. The value I found is $\mathrm{p}K_\text{a}=35$, so $\mathrm{p}K_\text{b}=14-35=-21$.

$\endgroup$
  • $\begingroup$ You mean Kw/Ka of hydrogen gas? $\endgroup$ – Dissenter Aug 13 '14 at 17:21
  • $\begingroup$ That 35 value also sounds like a pKa value rather than a Ka value. Can't imagine hydrogen gas being that acidic. $\endgroup$ – Dissenter Aug 13 '14 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.