# Titration of alpha-amino acid with strong base

Question:

$\mathrm{50~ml}$ of the conjugate acid $\ce{H3N+CH(R)COOH}$ of an $\alpha$-amino acid required $\mathrm{30~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ to completely convert into the dipolar ion $\ce{H3N+CH(R)COO-}$. The pH of the solution was 6.4 during the titration with $\mathrm{15~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ and 7.8 at the isoelectric point. Calculate the dissociation constants for the amino acid.

Attempt:

Reactions involved:

$$\ce{H3N+CH(R)COOH <=> H3N+CH(R)COO- + H+}$$

$$\ce{H3N+CH(R)COOH + OH- -> H3N+CH(R)COO- + H2O}$$

At the isoelectric point, $\mathrm{pH}=\frac{\mathrm pK_{\mathrm a1}+\mathrm pK_{\mathrm a2}}{2}$. Initially, number of milli moles of conjugate acid = number of milli moles of NaOH = $\mathrm{30\times 0.1=3~mmol}$ of salt during titration (neglecting the amount produced by acid) = 1.5

A buffer forms during titration. My doubts is, does the concentration of acid which was present initially, change while adding NaOH? Using Henderson equation will give pH of Buffer. I know the concentration of salt and pH. But I am stuck with the concentration of acid. 1.5 milli moles of acid will react to form salt. If I use 1.5,I get wrong answer. If I use 3, I get right answer.

• I improved the formatting of your post using the \ce command. For more information on how to do this yourself please see here and here. – bon Sep 5 '15 at 18:46