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Is it because the $\ce{OH}$ group next to $\ce{COOH}$ in carbonic acid donates electron density to the $\ce{COO-}$ by resonance, making it a stronger base, compared to a weaker electron donating inductive effect of $\ce{CH3}$ and thus its conjugate acid is weaker?

What does the resonance contribution look like in this case?

What makes resonance a stronger factor than induction?

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Before you ask "why", ask "if".

Carbonic acid is not weaker than acetic. In fact, it is stronger, just as one might expect by looking at that extra electron-withdrawing substituent. The trouble is, carbonic acid is also pretty unstable. At any given moment, most of it exists as $\ce{CO2}$. This leads to much lower apparent acidity.

Think of it this way: you add some acetic acid to a carbonate salt. An equilibrium sets in, much closer to the starting compounds than to the products. In other words, you'll have just a very tiny amount of $\ce{H2CO3}$. But as soon as you have it, it starts decomposing, thus shifting the equilibrium more and more to the right.

Look again at that Wikipedia page, it says:

3.6 ($\rm pK_{a1}$ for $\ce{H2CO3}$ only), 6.3 ($\rm pK_{a1}$ including $\ce{CO2(aq)}$)...

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  • $\begingroup$ This is something new! Thank you for this information. Although I'd like to ask, doesn't OH DONATE electron density by resonance, or is that not possible in this case? $\endgroup$ – Diracc Jul 7 '17 at 13:57
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    $\begingroup$ Even if it does, that doesn't outweigh the electron-withdrawing effect. $\endgroup$ – Ivan Neretin Jul 7 '17 at 14:04

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