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In the wikipedia article "Step-growth Polymerization/Kinetics", it is stated that:
The simple esterification is an acid-catalyzed process in which the protonation of the acid is followed by interaction with the alcohol to produce an ester and water
My question is: How are acids even protonated? Aren't they the ones supposed to be giving out protons? How are they the ones accepting the protons?
A general acid-base reaction is one of the kind $(1)$.
$$\ce{\underset{\text{acid}}{HA} + \underset{\text{base}}{B} <=> \underset{\text{conj. acid}}{HB+} + \underset{\text{conj. base}}{A-}}\tag{1}$$
All that the acid needs to react as an acid is some hydrogen atom bonded to something, and all that the base needs to react as a base (although it is not shown here) is a lone pair which can accept the acid’s proton. Depending on the relative strength of the acid and the conjugate acid, this equilibrium will be shifted. If the acid is stronger, it will be shifted towards the product; if the conjugate acid is stronger, towards the reactant. If the acidity are within reasonable range of each other, you can always expect a catalytic amount of the stronger acid to be protonated.
The above was the general introduction; now how does this apply to your question?
The ‘acid’ you are talking about is a carboxylic acid. These have the functional group $\ce{-COOH}$ and each of the oxygens in there has two lone pairs. Thus technically, a carboxylic acid could be protonated up to four times.[1] We only need one further protonation, so we can define the carboxylic acid to be compound $\ce{B}$ above — the base in the acid-base reaction.
In fact, acid-catalysed esterification is typically catalysed by the addition of a small drop of sulfuric acid. An average carboxylic acid has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 5$, sulfuric acid $-3$. The hydronium ion $\ce{H3O+}$, that forms if water takes the place of the base in $(1)$, has a $\mathrm{p}K_\mathrm{a}$ of $-1.7$ or $0$, depending on who you ask. Other alcoholic oxonium ions have similar $\mathrm{p}K_\mathrm{a}$ values. The lower the $\mathrm{p}K_\mathrm{a}$, the stronger the acid. Thus, we can assume that most carboxylic acid molecules remain as $\ce{-COOH}$, while the catalytic amounts of sulfuric acid (practically completely deprotonated once) protonated a few solvent molecules at equilibrium. The stronger acid — sulfuric acid — practically governs that none of the weaker acid — the carboxylic acid — can lose its proton.[2]
Even though it is equilibrium, a few of the acid molecules — the ‘next-weakest links in the chain’ — will also be protonated at a given point in time, even though we can expect the $\mathrm{p}K_\mathrm{a}$ value of a protonated carboxylic acid to be rather low. This small proportion is enough to get the reaction going.
Notes:
[1]: However, each successive protonation introduces a positive charge which always makes further protonation less favourable. A positively charged compound is, generally, also more inclined to transfer hydrogen to something else, i.e. is more acidic.
[2]: This can actually be extended. There are a number of superacids with even lower $\mathrm{p}K_\mathrm{a}$ values — in fact, those values are so low that they can no longer accurately be calculated and instead a different value, the Hammett acidity is used. Some of the strongest superacids are even able to protonate methane which is unsuspiscious of providing a lone pair an acidic proton could interact with. Hence Ivan’s comment of:
Everything can be protonated, given enough effort.
