Why Histidine is triprotic under biological condition?

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rorsara

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what would be the correct choice?

Under most biological conditions, however, histidine is considered to be a triprotic acid. This is due to the pKa of the _________ .

A. Other nitrogen in the imidazole ring, that possesses a hydrogen but no charge being much lower than the pH values normally found in a cell.

B. Other nitrogen in the imidazole ring, that possesses a hydrogen but no charge being much higher than the pH values normally found in a cell.

C. Carboxylic acid being much higher than the pH values normally found in a cell.

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It potentially has three acidic protons. The carboxylic acid group, and two protons on the side chain. The pH of the cell is about 7.4. The pka of COOH is about 2, pka of NH3 is about 9 and the side chain's imidazole pka is about 6 with two nitrogens. So at physiological pH the side chain imidazole nitrogen is protonated.
 
I am really sorry to post what may be perceived as rude, but unless you know a topic really well, you should refrain from posting what might be incorrect information. With the rush to print materials for the 2015 MCAT, there are more errors floating around than anytime I've seen in the past ten years. So it is imperative that this site be as perfect as possible, so that posters can come here for great insights.

The term triprotic refers to the number of protons that can dissociate. The addition of "under most biological conditions" is there to simply emphasize that it is an aqueous environment (and thus limited by the presence of water). Histidine has three protons with pKas in the 1.5 to 10.5 range. They are the carboxyl terminal (just over 2), the sidechain proton on the cationic nitrogen (around 6), and the amino terminal (about 9.5).

The point of this question is that you CANNOT remove the other proton on the imidazole ring (the one found on the neutral nitrogen), because it is just not acidic enough to be removed (in the presence of water). In essence, it has a pKa so high that the pH of the aqueous solution could never be raised high enough to remove it. If you added a base strong enough to remove it, then that base would remove a proton from water before touching the H on the neutral N of the imidazole ring.

I really hope this didn't come across as rude, but I have been reading factually incorrect (or at the least partially incorrect) information to many questions in this section and figured this was one I could respond to. Etiquette dictates that if it's a question from a source besides BR, I refrain from commenting. I'm not sure where this question comes from, but wirth only three choices I assume it si not an MCAT question. It is a very good question though, and fits the MCAT topic list.
 
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