I was comparing the acidity of methane $\ce{CH4}$ and water $\ce{H2O}$, and a lot of things weren't making sense. Generally, the process of turning $\ce{HX}$ into $\ce{H+}$ and $\ce{X^-}$ can be broken into 2 hypothetical steps:

  1. Homolytic cleavage of the $\ce{H-X}$ bond, followed by
  2. Electron donation from $\ce{H}$ to $\ce{X}$.

The energy required for step 1 is correlated with bond length, while the energy released by step 2 is correlated with electronegativity of $\ce{X}$.

The way I see it, electronegativity comes as a direct consequence of atomic size. Adding an electron to smaller element means the electron is added to an orbital that's relatively close to the nucleus, which is more stable. Thus, smaller elements are generally more electronegative, particularly across a row in the periodic table.

Now we consider $\ce{CH4}$ and $\ce{H2O}$. Oxygen, being smaller than carbon, will be more electronegative; thus, $\ce{OH-}$ is more stable than $\ce{CH3-}$ and step 2 will be more favorable for $\ce{H2O}$. But oxygen being smaller than carbon also means the $\ce{O-H}$ bond length is smaller than the $\ce{C-H}$ bond length, and thus harder to homolytically break. Overall, I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark. Yet the $\mathrm{p}K_\mathrm{a}$ for $\ce{H2O}$ is 15.7, while the $\mathrm{p}K_\mathrm{a}$ for $\ce{CH4}$ is approximately 55!

How can I make sense of this? What are other factors that impact electronegativity and homolysis that I am not considering? Or are there other aspects to this problem that I am not considering entirely?

  • 3
    $\begingroup$ Your assumption that the process can be broken into two steps might be theoretically helpful, but it isn't what happens in the real world for hydrogen-based acids. And your assumptions about the drivers of bond length and electronegativity are also wrong. You should start with actual observations not crude theory. $\endgroup$
    – matt_black
    Aug 28, 2020 at 13:55

3 Answers 3


Generally, the process of turning $\ce{HX}$ into $\ce{H+}$ and $\ce{X^-}$ can be broken into 2 hypothetical steps [...]

Indeed, this process is slightly simplified because acidities are typically measured in solution, whereas your thermodynamic cycle only deals with gas-phase energetics.

However, in the present case, it's actually enough to look at gas-phase energetics (as will be shown below). The reason for this is because the difference between water and methane is so stark, that even ignoring solvation effects, we can arrive at the same result.

I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark.

Well, considering that water and methane have quite different acidities, it's almost certainly going to be the case that these two factors don't cancel out. The analysis below is probably not going to be very thorough. But again, because the difference is so huge, even some inaccuracies are going to be OK.

Let's look at factor 1, the bond dissociation enthalpy. Blanksby, S. J.; Ellison, G. B. Acc. Chem. Res. 2003, 36 (4), 255–263 quotes values of $\pu{104.99 kcal/mol}$ for methane, and $\pu{118.82 kcal/mol}$ for water. Indeed, the bond in methane is easier to break, which is in favour of methane being a stronger acid.

For factor 2, the electron gain, the NIST WebBook has (computationally obtained) values of $\pu{0.12983 eV}$ for the methyl radical and $\pu{1.80321 eV}$ for the hydroxyl radical. As far as I can tell, these are electron affinities (i.e. $\Delta U$, not $\Delta H$). So, one must note that a more positive value actually means that adding an electron is more favourable, as the Wikipedia page on electron affinities points out (as well as several textbooks). With this in mind, it's clear that this step favours the hydroxyl radical, suggesting that water should be a stronger acid.

The remaining question, of course, is which of these two factors is more important. We need to do some unit conversion, but it is not too difficult, especially if you use this handy online calculator. It turns out that

$$\begin{align} \overbrace{\pu{(118.82 - 104.99) kcal/mol}}^{\large \text{Factor 1}} &= \pu{13.83 kcal/mol} \\ &\approxeq \pu{0.600 eV} \\ &< \pu{1.67338 eV} = \underbrace{\pu{(1.80321 - 0.12983) eV}}_{\large \text{Factor 2}} \end{align}$$

Factor 2 is almost three times more important than factor 1. These values are quite rough, but certainly enough to draw the right conclusion.


Here is a bigger chunk of the periodic table with acid dissociation constants of the hydrides:

enter image description here

Source: Libretexts

The authors claim that the two factors are electronegativity (how much of a positive partial charge the hydrogen has in the covalent bond) and bond strength (expressed as estimated homolytic cleavage energy).

There is no strong anomaly for the three elements (N, O, F) that are implicated in strong hydrogen bonding.

The energy required for step 1 is correlated with bond length

The correlation is not that strong if you compare methane, ammonia, water and hydrogen fluoride. The correlation with electronegativity is better.

enter image description here

Source: Libretexts


As far as the acidity of water goes, it is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\ce{H2O}$, while none are observed between $\ce{CH^-_3}$ and $\ce{CH4}$

This factor might not be significant enough to provide explanation for all of the difference that you are seeing, so there is still another culprit out there.

  • $\begingroup$ I mean, if we're talking about aqueous pKa's, then yes, OH- is stabilised by hydrogen bonding; but the difference in pKa is still very large even in aprotic solvents, so there's still an additional factor. I also don't get how CH4 has anything to do with stability of CH3-. Your CH3- is going to be either in the gas phase (where it's surrounded by effectively nothing) or dissolved in something (where it's surrounded by solvent molecules), and that solvent isn't gonna be methane. $\endgroup$ Mar 30, 2020 at 14:49
  • $\begingroup$ I am not referring to hydrogen bonding with the solvent, because in this case water is not necessarily the solvent, just the acid. So, assuming you have more than one molecule of water, this would still apply, even if it is in an aprotic solvent. Also, the fact that CH4 has nothing to do with stability of CH3- actually is part of what my argument is based on. $\endgroup$ Mar 30, 2020 at 14:53
  • 2
    $\begingroup$ Uhm - let me get this right. You're saying that if we were to dissolve two molecules of water in DMSO and one dissociated, then the OH- would stick to the remaining undissociated H2O? What evidence do you have for this? $\endgroup$ Mar 30, 2020 at 14:55
  • $\begingroup$ It probably wouldn't work like that if you had too much DMSO because the H2O and the OH- would never come into contact, but yeah, that's the basic principle. Also, like I said, it's called "homoassociation." look it up. $\endgroup$ Mar 30, 2020 at 14:59
  • $\begingroup$ The examples provided on Wikipedia, such as HF, involve concentrated solutions, where you have lots of undissociated molecules – I definitely agree with you that these can influence observed acidities. But I'm really not sure they are relevant to the pKa values we're talking about. These are usually measured at rather dilute concentrations – for a discussion of water in DMSO see e.g. pubs.acs.org/doi/pdf/10.1021/ed086p864. No mention of association is made. But, I will let other people judge. $\endgroup$ Mar 30, 2020 at 15:08

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