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Very basic question, but I'm rather confused.

Say I want to separate a carboxylic acid $(\ce{RCOOH},$ $\mathrm{p}K_\mathrm{a}~5)$ from a phenol $(\ce{ArOH},$ $\mathrm{p}K_\mathrm{a}~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(\ce{NaHCO3}),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).

The question is:

Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?

If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?

enter image description here

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    $\begingroup$ The picture is a bit confusing: Phenol should $\ce{PhOH}$ and nothing else. Sometimes, $\ce{Ar}$ is used to denote an aromatic compound, but that does not seem to be the case here. $\endgroup$ – TAR86 May 19 at 16:17
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    $\begingroup$ @TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why. $\endgroup$ – chompion May 19 at 16:29
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The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $\ce{CO2}$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $\ce{NaOH}$ (a strong base) or $\ce{NaHCO3}$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $\ce{OH-}$, because any stronger base will just deprotonate a readily available water molecule.

To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $\ce{RCOOH}$ and $\ce{HCO3-}$ do not have to come close to each other to effectively transfer a proton between them.

Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.

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  • $\begingroup$ I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...? $\endgroup$ – chompion May 19 at 16:50
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    $\begingroup$ @chompion That's basically correct. $\endgroup$ – TAR86 May 19 at 17:06

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