Amino Acids & Titrations

[SaintJude]

I had to share this question, because it's a great MCAT question. I got it wrong the first time, and I wanted to share it because it I don't want anyone else getting a similar question wrong so we can all get a 40+ on this exam.

The titration curve of alanine, an amino acid (structure drawn), is shown below.
Q. Draw the structure of the amino acid at point I, II, III.

View attachment Picture 37.png

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[chiddler]

I had to share this question, because it's a great MCAT question. I got it wrong the first time, and I wanted to share it because it I don't want anyone else getting a similar question wrong so we can all get a 40+ on this exam.

The titration curve of alanine, an amino acid (structure drawn), is shown below.
Q. Draw the structure of the amino acid at point I, II, III.

View attachment 18465

<3

lets see. I is when the carboxylic acid is protonated.

II is when protonated carboxylic acid = deprotonated

III is when protonated amine = deprotonated amine.

amiright?

 

[SaintJude]

Here is the picture of the answer & Kaplan's explanation:
View attachment Picture 39.png

Alanine is an an amino acid. Like all amino acids, alanine is amphoteric--it can act as a base or acid. The amino group is protonated in acidic solutions (stage I here) and the carboxylic acid group is deprotonated in basic solutions (stage III)

At some intermediate pH, unique for every amino acid, the amino group is protonated and the carboxylic acid is deprotonated. This point (stage II here) is called the isolelectric point and the amino acid is in the form of a zwitterion, positively charged on the protonated amino group and negatively charged on the deprotonated COOH group. All amino acids follow a process similar to this and exist in various ionic forms depending on the pH.

 

[mcloaf]

No, sorry! It got me too.

Here is the picture of the answer & Kaplan's explanation:
View attachment 18466

Alanine is an an amino acid. Like all amino acids, alanine is amphoteric--it can act as a base or acid. The amino group is protonated in acidic solutions (stage I here) and the carboxylic acid group is deprotonated in basic solutions (stage III)

At some intermediate pH, unique for every amino acid, the amino group is protonated and the carboxylic acid is deprotonated. This point (stage II here) is called the isolelectric point and the amino acid is in the form of a zwitterion, positively charged on the protonated amino group and negatively charged on the deprotonated COOH group. All amino acids follow a process similar to this and exist in various ionic forms depending on the pH.

That's what chiddler said...

 

[chiddler]

That's what chiddler said...

no if I understand correctly, I was a bit incorrect. I thought those were points of equilibrium, but they're not. When the line remains straight-ish (as opposed to rising sharply), then the mid point of that region is the equilibrium point.

Since they asked for the lines rising sharply, then these are points when one form predominates over the other.

So to correct myself:

I is protonated CO2H. (correct)
II is only deprotonated CO2(-).
III is only deprotonated amine.

for reference:

 

[chiddler]

Which one of those sharp increases is the equivalence point (pI)?

ok i'm not sure if i'm right anymore. would appreciate feedback, thanks.

 

[mcloaf]

Oops, I guess I didn't read your answer very carefully. For some reason I assumed your = signs meant "becomes" or something, don't really know what I was thinking. Yes, you were talking about the buffer regions rather than the equivalence points, sorry.

 

[mcloaf]

wait then where is equivalence point.

ok i'm not sure if i'm right anymore. would appreciate feedback, thanks.

It's a little confusing. The equivalence point means you've added an equivalent amount of (in this case base) titrant to fully protonate or deprotonate whatever it is you're titrating. In this case that means when you've added enough base to fully convert your acid species to the conjugate base (obviously equilibria affect this, hence the curved graph rather than one with sharp angles). The half-equivalence point is when you've added half as much base, which means the species you're titrating is half acid, half conjugate base. This is the flat region of the graph, and this is where buffers exist. The half equivalence point is where pH = pKa.

 

[chiddler]

It's a little confusing. The equivalence point means you've added an equivalent amount of (in this case base) titrant to fully protonate or deprotonate whatever it is you're titrating. In this case that means when you've added enough base to fully convert your acid species to the conjugate base (obviously equilibria affect this, hence the curved graph rather than one with sharp angles). The half-equivalence point is when you've added half as much base, which means the species you're titrating is half acid, half conjugate base. This is the flat region of the graph, and this is where buffers exist. The half equivalence point is where pH = pKa.

i just realized pI is different from equivalence point :x

so point II is the isoelectric point, right? and equivalence point is at the end of the first straight region which indicates when all the carboxylic acid has been deprotonated by base.

 

[mcloaf]

The isoelectric point is where the predominant species is neutral overall. For alanine point II is the isoelectric point, yes. More generally, the pI is the average of the two pKas surrounding the neutral species. For alanine, since there are only two pKas the pI is the average, in this case 6. For amino acids with acidic or basic side chains, you need to figure out which two pKas to average to give you the pI. Take aspartic acid, for example, which has a carboxylic acid R group. The pKa of the carboxyl terminus is 1.95, the pKa of the side chain is 3.71, and the pKa of the amino terminus is 9.66. Looking at aspartic acid, you can see that it will be neutral when the carboxyl terminus is deprotonated, the amino terminus is protonated, and the side chain is protonated. When will this be the case? It must be when the pH is above the pKa of carboxyl terminus but below the pKa of the R group. So the pI will be the average of these two pKas: (1.95 + 3.71) / 2 = 2.77

And we can be more precise about the equivalence point than "the end of the straight region." It is the region where the slope is greatest.

 

[chiddler]

thanks for the very helpful explanation.