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Determination of Amino Acid

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data

    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?


    2. Relevant equations

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

    3. 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
     
  2. jcsd
  3. Sep 12, 2013 #2

    epenguin

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    Does any amino acid have only one titratable group?
     
  4. Sep 14, 2013 #3

    chemisttree

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    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.
     
  5. Sep 15, 2013 #4
    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

    No. It should have at least two titratable groups
     
  6. Sep 15, 2013 #5

    Borek

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    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?
     
  7. Sep 15, 2013 #6

    epenguin

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    Your calculation appears implicitly based on 1
     
  8. Sep 15, 2013 #7
    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:

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

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

    Does this appear to be reasonable?
     
  9. Sep 15, 2013 #8
    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
     
  10. Sep 15, 2013 #9

    epenguin

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    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-
     
  11. Sep 15, 2013 #10

    epenguin

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    :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?
     
  12. Sep 15, 2013 #11
    Ah, yes. I apologize for that. I'll be more specific next time.

    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.
     
  13. Sep 16, 2013 #12

    Borek

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    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.
     
  14. Sep 16, 2013 #13

    epenguin

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

    That would normally be a good question, however both what you are told and your experimental data say diprotic.
     
  15. Sep 16, 2013 #14

    chemisttree

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    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"?
     
  16. Sep 16, 2013 #15

    epenguin

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    :blushing: Shome confushion, sorry.
     
  17. Sep 17, 2013 #16

    chemisttree

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    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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