What is the identity of the monoamine monocarboxylic amino acid?

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The discussion revolves around identifying a monoamine monocarboxylic amino acid based on titration data. The titration involved 7.39 mmol of base to titrate 432 mg of the amino acid from pH 0.8 to 12.0, leading to a calculated molecular weight of 117 mg/mmol, which corresponds to valine. Participants debated the implications of the initial pH and the titration of a hydrochloride salt versus the free base form of the amino acid. The conversation highlights the need for clarity in experimental data and the importance of understanding titration curves and inflection points in determining the correct identity of the amino acid. Overall, the amino acid in question is likely valine, although discrepancies in the provided data raise questions about the accuracy of the initial conditions.
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Homework Statement



7.39 mmol of base was used when titrating 432 mg of a monoamine monocarboxylic amino acid from pH 0.8 to 12.0. What is the name of the amino acid?


Homework Equations



None I suppose. Henderson-Hasselbalch perhaps, but I don't see how that helps me

The Attempt at a Solution



Well I decided to start with the simplest calculation first - Determining a molecular mass by dividing the given mass by the given amount of mmols. This provided me with 58.46 g/mol, which doesn't match any known amino acid weights.

So now I'm thinking that the pH range provided during the titration is supposed to be a hint, but I can't quite figure out its relevance. I'm really just looking for a tip in the right direction here
 
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Youngster said:

Homework Statement



7.39 mmol of base was used when titrating 432 mg of a monoamine monocarboxylic amino acid from pH 0.8 to 12.0. What is the name of the amino acid?


Homework Equations



None I suppose. Henderson-Hasselbalch perhaps, but I don't see how that helps me No it doesn't.

The Attempt at a Solution



Well I decided to start with the simplest calculation first - Determining a molecular mass by dividing the given mass by the given amount of mmols. This provided me with 58.46 g/mol, which doesn't match any known amino acid weights.

So now I'm thinking that the pH range provided during the titration is supposed to be a hint, but I can't quite figure out its relevance. I'm really just looking for a tip in the right direction here

Does any amino acid have only one titratable group?
 
That's a very unusual amino acid to have a native pH of 0.8! Are you sure this problem isn't a titration of a hydrochloride salt of an amino acid? If that's the case your analysis needs to account for the FW of the hydrochloride salt of the amino acid rather than it's free base.
 
chemisttree said:
That's a very unusual amino acid to have a native pH of 0.8! Are you sure this problem isn't a titration of a hydrochloride salt of an amino acid? If that's the case your analysis needs to account for the FW of the hydrochloride salt of the amino acid rather than it's free base.

I probably should have specified that. The amino acid was dissolved in 0.2 N HCl. So the amine group is probably protonated, and the overall charge is +1

epenguin said:
Does any amino acid have only one titratable group?

No. It should have at least two titratable groups
 
Youngster said:
I probably should have specified that. The amino acid was dissolved in 0.2 N HCl. So the amine group is probably protonated, and the overall charge is +1

Then you should take into account amount of base required to titrate HCl.

Something doesn't add up. Have you listed all the information you are given?
 
Youngster said:
No. It should have at least two titratable groups

Your calculation appears implicitly based on 1
 
Alright well, I've tried something else. Starting with the equation for the reaction:

H2X + 2NaOH → Na2X + 2H2O

Where X is the amino acid. The equation states that 2 moles of base titrates 1 mole of amino acid.

Assuming the change from pH 0.8 to pH 12.0 is the full titration (meaning the both the carboxyl group and the amino group are deprotonated), 7.39 mmol was used in the ionization of the entire amino acid.

So 3.70 mmol of amino acid was titrated by the 7.39 mmol of base. A molecular weight can be obtained by dividing the mass of the amino acid by the moles titrated:

\frac{432 mg amino acid}{3.70 mmol amino acid} = 117 mg/mmol

This molecular weight corresponds with valine, which also fits the description of a monoamine monocarboxylic amino acid.

Does this appear to be reasonable?
 
Borek said:
Then you should take into account amount of base required to titrate HCl.

Something doesn't add up. Have you listed all the information you are given?

Sorry, it seems the 0.2 N HCl information only applies to my actual lab data. The question above is what I assume to be a practice calculation to be done with my lab data
 
Youngster said:
:

\frac{432 mg amino acid}{3.70 mmol amino acid} = 117 mg/mmol

This molecular weight corresponds with valine, which also fits the description of a monoamine monocarboxylic amino acid.

Does this appear to be reasonable?

Yes it does. And I have just noticed that the question itself gives you the suggestion of the point you previously missed since it specifies "monoamine monocarboxylic amino acid".

I am not very comfortable with your chemical symbology and think you would be showing more understanding if you wrote the ionic forms, that you are going from RCHNH3+COOH to RCHNH2COO-
 
  • #10
Borek said:
Then you should take into account amount of base required to titrate HCl.

Something doesn't add up. Have you listed all the information you are given?

Youngster said:
Sorry, it seems the 0.2 N HCl information only applies to my actual lab data. The question above is what I assume to be a practice calculation to be done with my lab data

:confused: we can't work out what you don't tell us. Is this lab data, or is it a made up problem as I have assumed?
 
  • #11
epenguin said:
I am not very comfortable with your chemical symbology and think you would be showing more understanding if you wrote the ionic forms, that you are going from RCHNH3+COOH to RCHNH2COO-

Ah, yes. I apologize for that. I'll be more specific next time.

epenguin said:
:confused: we can't work out what you don't tell us. Is this lab data, or is it a made up problem as I have assumed?

This is a made up problem that goes along with a lab assignment.

To elaborate, the 0.2 N HCl information goes along with my titration curve below:

wPyKMmR.png


As far as I know, the actual amino acid unknown solution was prepared by dissolving 4.5g of the unknown amino acid in 300mL of 0.2 N HCl.

The titration, however, only used 20 mL of the solution.

Currently, I'm at a loss, because the technically one equivalent of OH should be used to titrate one of the ionizable groups of the amino acid. At 2 equivalents, however, there doesn't appear to be any rise in pH. I'm beginning to wonder if the amino acid above is actually triprotic.
 
  • #12
Youngster said:
As far as I know, the actual amino acid unknown solution was prepared by dissolving 4.5g of the unknown amino acid in 300mL of 0.2 N HCl.

The titration, however, only used 20 mL of the solution.

If so, you titrated 20 mL of 0.2 N HCl and 20/300*4.5 g of the amino acid. But 20/300*4.5 is 0.3 g, not 0.432 g. Plus, you have not used 7.39 mmol of NaOH to neutralize the aminoacid, but 7.39-20*0.2=3.39 mmol.

Perhaps - if the question is made up - information about initial pH is simply wrong and put there without a second thought. Generally speaking valine looks reasonable, it just doesn't fit rest of the information.
 
  • #13
Youngster said:
At 2 equivalents, however, there doesn't appear to be any rise in pH. I'm beginning to wonder if the amino acid above is actually triprotic.

Yes I think I know what you mean. The opposite of what you say :biggrin: - there is a very steep rise in pH there.

Youngster said:
I'm beginning to wonder if the amino acid above is actually triprotic.
That would normally be a good question, however both what you are told and your experimental data say diprotic.
 
  • #14
Borek said:
If so, you titrated 20 mL of 0.2 N HCl and 20/300*4.5 g of the amino acid. But 20/300*4.5 is 0.3 g, not 0.432 g. Plus, you have not used 7.39 mmol of NaOH to neutralize the aminoacid, but 7.39-20*0.2=3.39 mmol.

Perhaps - if the question is made up - information about initial pH is simply wrong and put there without a second thought. Generally speaking valine looks reasonable, it just doesn't fit rest of the information.

I don't agree. There are two inflection points. The first one is the HCl titration at ~1 "Equivalents NaOH", whatever that means in this case. Is it 40 grams? I don't think so. The second inflection point is the ammonium proton. The difference between these two peaks should be the amount of base corresponding to the monoamine monocarboxylic acid, right?

What is presented isn't data. It is partially-reduced data without an explanation. Raw data would be presented as pH vs volume. Equivalents of base means nothing without an understanding of the underlying assumptions.

What is meant by "Equivalents NaOH"?
 
  • #15
epenguin said:
Yes I think I know what you mean. The opposite of what you say :biggrin: - there is a very steep rise in pH there.

:blushing: Shome confushion, sorry.
 
  • #16
This is a monoprotic amino acid. The amount of base used is that amount between the inflection points NOT the amount required to go from either pH 0.8 to 12 or from pH 7 to 12. pH 12 is an arbitrary endpoint for this experiment designed only to fully describe the inflection point at pH 11. You still need to determine mmol of NaOH used between the inflection points.
 
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