Why assume a neutral amino acid is given for acid-base reaction?

Question

The questions is:

$1.812\ \mathrm{g}$ of a crystallized α-amino acid ($\mathrm{p}K_\mathrm{a1} = 2.4, \mathrm{p}K_\mathrm{a2} = 9.7$) has a $\mathrm{pH}$ of $10.4$ when dissolved in $100\ \mathrm{mL}$ of $0.1~\mathrm{M}\ \ce{NaOH}$. Calculate the molecular mass of this amino acid.

As usual we can obtain the ratio of the conjugate base and acid around $\mathrm{p}K_\mathrm{a2}$ by using the Henderson-Hasselbalch equation. However, once we get the ratio, the solutions key says that we know $c(\text{Base}) = 0.1$ because all of the $\ce{OH-}$ reacted with our acid. While I totally agree that all the $\ce{OH-}$ reacted with the acid, since it is a strong base, how do we know there was no $c(\text{Base})$ before the addition of $\ce{NaOH}$? Thus making the concentration $c_0(\text{Base}) + 0.1$. Is this because it came from a solid crystal where all of the amino acid was in the acid form?

Answer 1

There are a few observations to make.

1. The amino acid only has two $\mathrm{p}K_\mathrm{a}$ values. Thus, it cannot have a side chain that is in any way acidic or basic. We only have three different possibilities: $\ce{H2AA+, HAA, AA-}$.

2. It was crystalline when added. Therefore, it must have been in its neutral $\ce{HAA}$ state. Otherwise, it would have been given as ‘the $\ce{HCl}$ salt …’ or ‘the $\ce{NaOH}$ salt …’

If we had any excess base in there, it would have to have been given in the question as the sodium hydroxide crystal adduct or something.