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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?

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

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FreezingFire
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    It hardly makes sense to compare the pKa of methanol in water with that of water in water. I've heard this question before, and to me it is like splitting hair. – Karl Aug 25 '16 at 19:54
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    So the super uber ultimate answer is: it depends – Mithoron Aug 26 '16 at 00:21
  • @Karl But the question I have linked to, has compared the acidities of alcohols in aqueous solutions with water itself! And the same answer has indeed used pKa values to compare their acidities. My concern is they might be using the wrong pKa for water. – FreezingFire Aug 26 '16 at 06:23
  • @Mithoron I already realised that when I read that question I have mentioned. But right now, I just want to compare acidities of their aqueous solutions! – FreezingFire Aug 26 '16 at 06:25
  • I'm sorry, but this discussion is nonsense, imho. You can formally run the calculation that gives the pKa of something dissolved in water also for pure water, but the result has no sensible meaning at all. – Karl Aug 26 '16 at 09:23
  • @Karl So how would you compare the acidities of water and methanol? And then do you consider the concept of pKa not applicable for water or what? I do not understand your reasoning...I am guessing that you want to calculate the pKa values of water and methanol using equilibrium constants, in some other solvent like DMSO. But that doesn't explain why measuring pKa of water directly, with the pKa of methanol in water. – FreezingFire Aug 26 '16 at 10:25
  • The concept of pKa is NOT applicable to the solvent, it is about the solute! If your actual question is "can a methoxy anion exist in water", then say so. – Karl Aug 26 '16 at 11:34
  • $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ – Jan Aug 26 '16 at 23:10
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    If you want a clear-cut answer, get rid of all those complicated intermolecular interactions and study the compounds' acidity in the gas phase. The hydroxide anion releases more energy when protonated, compared to a methoxide anion. Thus, excluding all but the most fundamental factors, methanol is more acidic than water. – Nicolau Saker Neto Aug 26 '16 at 23:46
  • @NicolauSakerNeto Why not instead compare the pure substances in the liquid phase, since that's the more common application? I.e., why not compare the compare the $pK_a$ for pure liquid MeOH with that for pure liquid water? I.e., compare the $pK_a$ for $\ce{CH3OH_{(l)} <=> CH3O^-{(l)} + H^+{(l)}}$ to that for $\ce{H2O_{(l)} <=> OH^-{(l)} + H^+{(l)}}$. If the former is measurable, the complications from the interactions aren't an issue—they're subsumed into the measurement. It's easy enough to find the $pK_a$ for MeOH in water. But I've not been able to find a $pK_a$ for pure MeOH. – theorist May 01 '22 at 21:53
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    @theorist There is some confusion regarding whether the equilibria you mention are represented by the acid dissociation or autoionization constants, which has been discussed several times across multiple Chem.SE posts, though it seems they are, in fact, one and the same. – Nicolau Saker Neto May 01 '22 at 22:51
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    (cont.) You could certainly compare them (the autoionization constant of MeOH at 25 °C is $10^{-16.6}$ ref), though there is room for ambiguity - what if MeOH is actually a stronger acid than water, but a much weaker base? That would also lead to a smaller autoionization constant. It would also be interesting to compare the $pK_a$ of water in methanol and vice-versa. – Nicolau Saker Neto May 01 '22 at 22:51

2 Answers2

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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.

andselisk
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SCH
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  • Thank you for your answer! Just one more thing, can I then assume the methoxide ion to be a stronger base than hydroxide ion, (in DMSO)? I was intending to use this answer for that purpose also, but I realise that maybe that would be too simplistic! – FreezingFire Aug 27 '16 at 06:05
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    No, it's the other way around. Methanol being more dissociated means that the methoxide is a weaker base than the hydroxide (in DMSO). In water the trend is reversed because of better hydrogen bonding with the smaller hydroxide ion. – SCH Aug 27 '16 at 10:38
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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.

Karsten
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  • It seems the OP was curious about the acidity of methanol in water, which you addressed. But if I myself were to compare the acidity of methanol and water, I think I'd want to compare the $pK_a$ for pure methanol with that for pure water. It's easy enough to find the $pK_a$ for methanol in water—for a dilute soution, it's 15.5. But I've not been able to find a $pK_a$ for pure methanol, i.e., for $\ce{CH3OH_{(l)} <=> CH3O^-{(l)} + H^+{(l)}}$ – theorist May 01 '22 at 21:42
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    @theorist Not methanol but ethanol, and just the autoprotolysis constant: DOI: 10.1016/S1658-3655(12)60002-8 – Karsten May 02 '22 at 01:01