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I am told that because the methyl group is electron donating in the conjugate base of acetic acid, this destabilizes the conjugate base by exacerbating the existing negative formal charge on the deprotonated oxygen, while in formic acid the electron donating methyl is absent in lieu of a hydrogen, which is neither withdrawing nor donating.

Is this reasoning correct, and are there any other reasons why formic acid might be stronger than acetic acid?

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  • $\begingroup$ Yes the reasoning is correct . The more the charge is increased on a molecule the more it gets destabilised. $\endgroup$
    – dsinghvi
    Commented Aug 28, 2014 at 16:33
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    $\begingroup$ I don't have an authoritative reference to the effect, but I strongly suspect that its small size is probably a major factor. That potentially results in formation of more efficient hydrogen-bond networks and superior solvation. $\endgroup$
    – Greg E.
    Commented Aug 28, 2014 at 17:08
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    $\begingroup$ Yes, the electron donation effect is pretty small. $\endgroup$
    – Dissenter
    Commented Aug 28, 2014 at 17:09
  • $\begingroup$ @GregE. You may want to look at my answer, which also mentions solvation and identifies a similar effect with oxalate ion. $\endgroup$
    – Oscar Lanzi
    Commented Mar 14, 2023 at 13:51

4 Answers 4

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We are discussing the following equilibrium

enter image description here

We can make the acid a stronger acid by pushing the equilibrium to the right. To push the equilibrium to the right we can

  1. destabilize the starting acid pictured on the left side of the equation, and \ or

  2. stabilize the carboxylate anion pictured on the right side of the equation.

Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen. Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon. This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge.

Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen. Consequently, acetic acid is a weaker acid than formic acid.

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  • $\begingroup$ I get the point you are trying to get across, but it seems weird that you chose exactly those resonance forms for the free acid and those different ones for the carboxylate. The resonance structure with the positive charge on $\ce{OH}$ should contribute more than the $\ce{C+ -O-}$ one. $\endgroup$
    – Jan
    Commented Jun 25, 2016 at 17:03
  • $\begingroup$ @Jan wrote "The resonance structure with the positive charge on $\ce{OH}$ should contribute more than the $\ce{C^{+}-O^{-}}$ one." Jan, why do you suspect that? Carbon is less electronegative than oxygen and I thought the position of the carbonyl carbon in $\ce{^13C}$ nmr suggests a fair amount of positive charge on carbon. $\endgroup$
    – ron
    Commented Jun 25, 2016 at 17:53
  • $\begingroup$ Because all-octet resonance structres as $\ce{^{- }O-C={O+}H}$ generally contribute more to the overall structure than those without. Also, the carboxyl carbon’s NMR shift is about $30$ to $40~\mathrm{ppm}$ lower than that of a ketone, showing that the carbon’s positive partial charge is lower thanks to the neighbouring hydroxyl group. $\endgroup$
    – Jan
    Commented Jun 25, 2016 at 18:04
  • $\begingroup$ What's the partial positive charge on the carbonyl carbon in acetone, around 40%? Reducing that value to account for replacing a methyl group with a hydroxyl would still (IMO) leave significant positive charge on the carbon. $\endgroup$
    – ron
    Commented Jun 25, 2016 at 19:09
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The strength of carboxylic acid can be discussed by positive inductive effect. Here, in ethanoic acid there is an electron releasing inductive effect from the alkyl group. The larger the positive inductive effect, the lesser will be the strength of an acid. Similarly in formic acid there is no alkyl group and hence no inductive effect takes place. So formic acid is a stronger acid than ethanoic acid.

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    – Martin - マーチン ♦
    Commented Mar 2, 2015 at 6:57
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We may also consider solvation. Formate ion is relatively compact and thus more readily solvated by water molecules than acetate or other carboxylates, which are encumbered by their hydrocarbon groups.

Oxalate ion, $\ce{C2O4^{2-}}$, enjoys a similar solvation advantage, and this table shows that the second dissociation of oxalic acid is stronger than the dissociation of acetic acid.

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    $\begingroup$ Yeah, totally this! I'm pretty sure this "electron donating" explanation is incorrect, and solvation is probably the important thing here. $\endgroup$
    – Mithoron
    Commented Mar 13, 2023 at 21:40
  • $\begingroup$ @Mithoron Formic acid is more acidic than acetic acid in both solution and the gas phase. Gas phase data can be found here and here $\endgroup$
    – ron
    Commented Mar 19, 2023 at 20:47
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This is because of increasing positive inductive effect of alkyl group attached to the $\ce{-COOH}$ group in acid. In ethanoic acid electron releasing group ($\ce{CH3}$) increase the electron density on the carboxylic group $\ce{-COOH}$. But, methanoic acid doesn't contain electron releasing group and there is no inductive effect.The larger the positive inductive effect, the lesser will be the strength of an acid. Hence, methanoic acid (formic acid) is stronger than ethanoic acid.

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