[yolop]

I have a problem I'm stuck on figuring out. The equations just aren't coming to me so maybe another brain will help.

A solution of 0.1M glutamic acid is at its isoelectric point (pI = 3.22)
What is the concentration of the major isoelectric form and what is the concentration of the protonated form?

Someone told me there were 8 forms of this throughout the titration and all I could think of was D and L conformations. Is that the reason there are 8 or are there really 6 forms?

[Yggdrasil]

Here the different forms are not isomers of eachother (as in D- and L-) but they are conjugate acids/bases.  Remember that glutamic acid has three acidic protons, the alpha amino group, the main carboxylic acid, and the side-chain carboxylic acid.  Since each of these can exist in acid (protonated) and base (deprotonated) form, there are theoretically 6 different forms (however, in reality there are less because, for example, you can't have a protonated carboxyl group in the presence of an unprotonated amine with a higher pKa).

So, one hint would be to first draw out the different forms of glutamic acid going from low pH to high pH.  Next, figure out which form(s) are present at the isoelectric point and use the Henderson-Hasselbalch equation to determine their concentrations.

[yolop]

Here the different forms are not isomers of eachother (as in D- and L-) but they are conjugate acids/bases.  Remember that glutamic acid has three acidic protons, the alpha amino group, the main carboxylic acid, and the side-chain carboxylic acid.  Since each of these can exist in acid (protonated) and base (deprotonated) form, there are theoretically 6 different forms (however, in reality there are less because, for example, you can't have a protonated carboxyl group in the presence of an unprotonated amine with a higher pKa).

so there are only 6 forms not 8? That was my original thought but D/L conf. made it equal to 8

So, one hint would be to first draw out the different forms of glutamic acid going from low pH to high pH.  Next, figure out which form(s) are present at the isoelectric point

that would be the +1 form from the beginning of the titration where the alpha carboxyl, amine group and R chain are all protonated, and the neutral form with the unprotonated alpha carboxyl.

Total I have 4 forms which was why I believed D/L to make the total of 8 forms.
I have:
COOH-CH2(NH3)CH2CH2COOH    [AA+]
COO-CH2(NH3)CH2CH2COOH      [AA0]
COO-CH2(NH3)CH2CH2COO-       [AA-]
COO-CH2(NH2)CH2CH2COO-       [AA-2]

use the Henderson-Hasselbalch equation to determine their concentrations.

This is where I am stuck.  I calculated the pI to be 3.22 but how do I use [H][A]/[HA] to calculate the concentration of the major form and how do I know what the major form is?

[yolop]

Am I to assume I'm doing a titration with say NaOH for this problem and that would make my c.a./c.b. pairs?

[yolop]

let me post what I have done so far.

1. Found the pI using 2.19 and 4.25
pI = (pKa1+pKa2)/2 = 3.22

2. Using Hen.Has equation
ph = pKa+log[A-]/[HA]
3.22 = 4.25+log[A-]/[HA]

this comes to 1.03. The antilog of 1.03 = 0.093

I have tried a couple different things from here which I will list below. None of them seem correct.
First i tried to say pH = -log[H+]
3.22=-log[H+]
10^-3.22 = 6.02x10^-4
0.1M-6.02x10^-4= 0.999   This is way too concentrated

Next I tried using 2.19 instead of 4.25
This got me an antilog equal to 10.715
[A-]/[HA] = 10.72
[A+]=10.72/11.72 = 0.914
[HA]=1/11.72=0.0853

Neither of these seem to fit anything else. I tried subtracting HA from A then that number from 0.1M but still didn't get an answer that seems logical.

[enahs]

Maybe I am just reading the question wrong? but it says it is at it's isoelectric point. You know what this is. You are not adding an acid or base, you are just given a concentration at its isoelectric point. Thus, half of it is in the undisassociated form, the other half is in its disassociated.

I am tired, so maybe I am missing something in the question, but I do not see where it asks you to actually calculate anything with the Henderson-Hasslebach equation?
But you know because it is at its isoelectric point the ratio in the Henderson-Hasslebach equation is 1:1 so that is 0, so the pH = isoelectric point.


Again, maybe I am totally missing some part of the question?

[Yggdrasil]

enahs, the problem here is that there are multiple acid/base groups in the amino acid.  This complicates things so that the pI does not fall on the pKa of one of the groups and you cannot say that the ratio of acidic to basic form is 1:1.

Anyway, at the pI the predominant form will be the neutral amino acid (AA⁰ in your notation).  However, there will be small amounts of the AA⁺ form and the AA^- form due to the fact that the pKas of the two carboxylic acids are fairly close to the pI (you can ignore the presence of AA²⁺ form because the pKa of the amino group is very far away from the pI).  In summary, you need to worry about two acid/base equilibria:

H⁺ + AA^- <--> AA⁰  pKa = 2.19
H⁺ + AA⁰ <--> AA⁺ pKa = 4.25

From the Henderson-Hasselbalch equation, you can calculate the ratios [AA^-]/[AA⁰] and [AA⁰]/[AA⁺].  Furthermore, since these three are essentially the only forms of your amino acid present you know that [AA^-] + [AA⁰] + [AA⁺] = 0.1M.  Now you have three equations and three unknowns.

[yolop]

That's what I was doing wrong. I was only using ONE equation with the protonated and the neutral form. I'll do the math in the morning.



This was a problem from my brother's class and I thought I would try to tackle it. Looked hard from the beginning. Well...not hard just like it put together equations I hadn't put together. Something I hadn't done.

When would something like this come in hand?

[yolop]

I found one more using the van't Hoff plot but instead of being a straight line it is concave down. Not sure what to do with this.

[enahs]

enahs, the problem here is that there are multiple acid/base groups in the amino acid.  This complicates things so that the pI does not fall on the pKa of one of the groups and you cannot say that the ratio of acidic to basic form is 1:1.

I understand that. But the pKa's are more then 1pKa unit apart. Thus you can make the general assumption that the buffer action does not overlap to much extent (all though, they are just slightly more then 1 pKa unit apart, so there will be some minor buffer action). You can therefor conclude that they are roughly 1:1, and so it would be 0.05M to 0.05M in the appropriate form (again, rough estimates). And when you do the math, you get 0.048M, 0.004M, 0.048M. No?

[yolop]

See now here is the problem I'm having. I did the math and got 0.47 for [AA+], .047 for {AA-} and .00445 for [AA0]

The answer for the approximate concentration for the major isoelectric for m is 0.0843M and the concentration of the completly protonated form is 0.0079M

The answer also says there are 8 forms but as previously discussed I only found 6 with C.A and C.B. unless I'm suppose to count some twice.

[enahs]

See now here is the problem I'm having. I did the math and got 0.47 for [AA+], .047 for {AA-} and .00445 for [AA0]

You did your math wrong (or typed it here wrong). Those do not add up to 1.

[yolop]

The concentration is 0.1M not 1M. If they added up to 1 it would be too concentrated.

[enahs]

The concentration is 0.1M not 1M. If they added up to 1 it would be too concentrated.

That is just a mathematical expression. They do not add up to 0.1, either. They do not add up to the whole (which in mathematics is 1).

[chiralic]

Graphic of Acid-Base Chemistry of Glutamic Acid....


Edit: picture changed to png format