Is methanol really more acidic than water?

Question

The question Why is methanol more acidic than water? deals with the reasoning of why methanol is more acidic than water. However, as mentioned in the comments of that question, the acidity constant of water is $14.0$, as confirmed by two sources$^{[1][2]}$, with one of them offering a very convincing explanation. The Wikipedia page$^{[3]}$ for methanol quotes its acidity constant to be $15.5$.

So it would seem, water is more acidic than methanol. But an answer to the question mentioned above gives a good reasoning for why methanol is more acidic than water (it doesn't deal with the pKa of water, though). Also, the answer mentions that water would be much worse an acid in DMSO. Another question implies that methanol and other alcohols are more acidic than water.

On the other hand, it is also true that we regard alkoxide ions as being more strongly basic than the hydroxide ion (thus we use them primarily for elimination reactions).

This leaves us in a fix. So what is the ultimate truth?

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_{[1]: What is pKa of water? - Chemistry ChemLibreTexts
[2]: Water (Data Page)
[3]: Methanol}

Answer 1

First, water’s $\mathrm{p}K_\mathrm{a}$ in water is $14$ as explained here. This means that $\ce{H2O}$ is slightly dissociated in liquid form, such that $[\ce{H+}] = [\ce{HO-}] = 10^{-7}$.

Second, I would say that caring too much about which compound is the best acid is a bit like arguing by definition. Acidity is a fuzzy concept used as a shorthand for deeper meaning. This means there is little point arguing about the shorthand when we have the actual properties it synthesizes.

In this case (sou rce):

$$ \begin{array} {lrrl} \hline \text{Solvent} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{MeOH}) & \mathrm{Interpretation} \\ \hline \ce{H2O} & 14.0 & 15.5 & \text{Water is more dissociated than methanol} \\ \text{DMSO} & 31.4 & 29.0 & \text{Methanol is more dissociated than water}\\ \hline \end{array} $$

So you can argue (as in the answer you linked) that methanol is more acidic than water in the abstract sense, because $\ce{DMSO}$ doesn’t stabilize the anions by hydrogen bonding and so is closer to the “ideal” case (whatever that may be).

You already have the $\mathrm{p}K_\mathrm{a}$ values, any practical question you might have about the concentrations of theses solutes in $\ce{H2O}$ and $\text{DMSO}$ is already answerable.

Answer 2

In water, methanol is more acidic than water

The apparent contradiction comes from using a different standard state for water (activity of pure water defined as 1) and methanol (activity of approximately 1 mol/L defined as 1). So it is inappropriate to directly compare equilibrium constants or $\mathrm{p}K_\mathrm{a}$ values if in one system, a species is the solvent and in another one it is not.

Instead, consider the concentrations at a pH of 14 (e.g. roughly 1 mol/L $\ce{NaOH}$). At this pH, the ratio of water to hydroxide is 55:1, or roughly 2% dissociated. We can calculate the ratio of methanol to methoxide using the Henderson-Hasselbalch relationship (or straight from the equilibrium constant expression). It comes out to about 30:1, or roughly 3% dissociated. So methanol is the stronger acid because at a given pH, the ratio of deprotonated species to protonated species is higher. You could do the same at any other pH with the same conclusion that methanol is the stronger acid with water as the solvent.

Using $\mathrm{p}K_\mathrm{a}$ values for comparison

If you try to use the Henderson-Hasselbalch equation for water, you have to plug in "1" for the activity of water because that is what was used in the definition of the $\mathrm{p}K_\mathrm{a}$. If you do that, you find that the ratio of activities is 1:1, corresponding to 1 mol/L hydroxide and almost pure water, i.e. 55 mol/L. This is all internally consistent, but to compare it to methanol without comparing apples to oranges, it is appropriate to compare the degree of dissociation as done above.

In SCH's answer, the claim is made that in aqueous solution, "water is more dissociated than methanol". If you look at the degree of dissociation (mol fraction of undissociated vs dissociated) as I did above, this statement is incorrect. What is correct is that, using the more logical value of $\mathrm{p}K_\mathrm{a} = 14$ for water, the $\mathrm{p}K_\mathrm{a}$ of water is lower than that of methanol.

Solvent-dependence

[OP] ...we regard alkoxide ions as being more strongly basic than the hydroxide ion...

This statement was made in the context of organic chemistry. In this context, you will have solvents other than water. So you could compare methanol's and water's acidity in e.g. acetonitrile, DMSO or methanol or in solvent mixtures. This makes things complicated (see @Mithoron in the comments: "It depends"), and you would have to limit your question to a specific situation. The blanket statement that alkoxide ions are (always) more strongly basic than hydroxide ions is incorrect.