My friend sent me this for his homework, and I haven't done a titration calculation in years.

A student titrated $\pu{10.00 mL}$ aliquots of her unknown amino acid solution with standard $\pu{0.1521 M}$ $\ce{NaOH}$ and with $\pu{0.0986 M}$ $\ce{HCl}$. No inflection points were observed with the $\ce{HCl}$ titrations but two were observed in the titration with $\ce{NaOH}$. One occured after titrant had been added and another after a total of $\ce{18.61 mL}$ of $\ce{NaOH}$ had been added. Calculate the amino acid concentration. HINT: Assume the $\pu{10 mL}$ unknown solution contains a mixture of the $\ce{A}$ and $\ce{A+}$ forms of amino acid.

So based on the information, I have a diprotic amino acid. I know at the second equivalence point, mol $\ce{H2A}$ = mol $\ce{A^2-}$, so if I can find mol $\ce{A^2-}$, I'm basically done.

At the second equivalence point I have added

$$\pu{0.1521 M}\, \ce{NaOH} \times \pu{18.61 mL}\, \ce{NaOH} = \pu{2.83 mmol}\, \ce{NaOH}$$

If mol $\ce{A^2-}$ = mol $\ce{NaOH}$, then my concentration of $\ce{H2A}$ would be: $\pu{2.83 mmol/10 mL = .283 M}$

I don't think that's right because I haven't even used the first equivalent point for anything, and I don't think it's necessarily true that at the 2nd equivalent point, mol $\ce{A^2-}$ = mol $\ce{NaOH}$, so I'm not sure how to proceed.

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    $\begingroup$ Welcome to ChemSE. Take a moment to familiarize yourself with the formatting this page allows (chemistry.meta.stackexchange.com/questions/86/…). It is a bit worrying to read "One occured after titrant [i.e. $\ce{NaOH}$?] had been added" that actually does not say how much (i.e., mL of the lye) was used. Otherwise, I would assume all $\pu{18.61 mL}$ of $\pu{0.1521 mol/L}$ $\ce{NaOH}$ was needed to reach two deprotonations, i.e. $\pu{1.42 mmol}$ of amino acid present in the aliquot. $\endgroup$ – Buttonwood Jul 5 '17 at 0:51
  • $\begingroup$ @Buttonwood Yea that's how I read it too. Unfortunately, I don't have any clarifying information since that how it was written on my friend's assignment $\endgroup$ – Brenton Jul 5 '17 at 4:43

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