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Topic: 4 - aminobenzoic acid, charges in water?  (Read 1765 times)

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Offline Sporkatronic

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4 - aminobenzoic acid, charges in water?
« on: October 29, 2013, 01:18:47 PM »
Hello everyone, I have a chemisty problem and would be grateful if somebody could point me in the right direction. With 4 aminobenzoic acid

http://en.wikipedia.org/wiki/4-Aminobenzoic_acid

in pH neutral water, will the NH2 be NH3, the COOH become COO-, or both (zwitterion)? Or more correctly, what will the ratios of ionised to unionised be for each group? I have the pKa's for the various protonations and deprotonations but it is more complex than I thought as the Pka for each group depends on whether or not the other one is ionised.

Thankyou

Offline Sporkatronic

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Re: 4 - aminobenzoic acid, charges in water?
« Reply #1 on: October 29, 2013, 02:13:09 PM »
Based on the Henderson Hasslebach equation

http://en.wikipedia.org/wiki/Henderson_hasselbach

and a pKa of neutral going to anion + H+ (ie the COOH becoming COO-) of 4.83, and this paper

http://pubs.acs.org/doi/pdf/10.1021/jo00317a002

calculating that 10 percent of them are zwitterions, I come to the conclusion that a vast majority of the molecules will have their COOH groups go to COO-, and 10 percent of them will have the NH2 go to NH3 as well.

Is this correct?

Cheers

Offline Borek

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Re: 4 - aminobenzoic acid, charges in water?
« Reply #2 on: October 29, 2013, 04:52:42 PM »
Technically it is

H2A+ ::equil:: HA ::equil:: A-

with pKa1 = 2.4 and pKa2 = 4.9 (more or less, according to the paper you listed, 0.1 error is not that important here).

At pH 7 we are far to the right with the second equilibrium, and [itex]\frac {[A^-]} {[HA]} \approx 10^{7-4.9}[/itex] (concentration of H2A+ is negligible).
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