Determination of An Unknown Amino Acid From Titration


Abstract


Experiment 11 used a titration curve to determine the identity of an
unknown amino acid. The initial pH of the solution was 1.96, and the pKa’s
found experimentally were 2.0, 4.0, and 9.85. The accepted pKa values were
found to be 2.10, 4.07, and 9.47. The molecular weight was calculated to be
176.3 while the accepted value was found to be 183.5. The identity of the
unknown amino acid was established to be glutamic acid, hydrochloride.

Introduction

Amino acids are simple monomers which are strung together to form polymers
(also called proteins). These monomers are characterized by the general
structure shown in figure 1.


Fig. 1

Although the general structure of all amino acids follows figure 1, the presence
of a zwitterion is made possible due to the basic properties of the NH2 group
and the acidic properties of the COOH group. The amine group (NH2) is Lewis
base because it has a lone electron pair which makes it susceptible to a
coordinate covalent bond with a hydrogen ion. Also, the carboxylic group is a
Lewis acidic because it is able to donate a hydrogen ion (Kotz et al., 1996).
Other forms of amino acids also exist. Amino acids may exists as acidic or
basic salts. For example, if the glycine reacted with HCl, the resulting amino
acid would be glycine hydrochloride (see fig. 2). Glycine hydrochloride is an
example of an acidic salt form of the amino acid. Likewise, if NaOH were added,
the resulting amino acid would be sodium glycinate (see fig. 3), an example of a
basic salt form.

Fig. 2

Fig. 3

Due to the nature of amino acids, a titration curve can be employed to identify
an unknown amino acid. A titration curve is the plot of the pH versus the volume
of titrant used. In the case of amino acids, the titrant will be both an acid
and a base. The acid is a useful tool because it is able to add a proton to the
amine group (see fig. 1). Likewise the base allows for removal of the proton
from the carboxyl group by the addition of hydroxide. The addition of the
strong acid or base does not necessarily yield a drastic jump in pH. The acid
or base added is unable to contribute to the pH of the solution because the
protons and hydroxide ions donated in solution are busy adding protons to the
amine group and removing protons from the carboxyl group, respectively. However,
near the equivalence point the pH of the solution may increase or decrease
drastically with the addition of only a fraction of a mL of titrant. This is
due to the fact that at the equivalence point the number of moles of titrant
equals the number of moles of acid or base originally present (dependent on if
the amino acid is in an acidic or basic salt form). Another point of interest
on a titration curve is the half-equivalence point. The half-equivalence point
corresponds to the point in which the concentration of weak acid is equal to the
concentration of its conjugate base. The region near the half-equivalence point
also establishes a buffer region (Jicha, et al., 1991). (see figure 4).

Fig. 4

The half-equivalence point easily allows for the finding of the pKa values
of an amino acid. A set pKa values can be extremely helpful in identifying an
amino acid. Through a manipulation of the Henderson-Hasselbalch equation, the
pH at the half-equivalence point equals the pKa. This is reasoned because at
the half-equivalence point the concentration of the conjugate base and the acid
are equal. Therefore the pH equals the pKa at the half-equivalence point (see
figure 5.)

Fig. 5 [base]
pKa= pH - log -------
[acid]

[base]
log -------- = log 1 = 0
[acid]

therefore, pH = pKa



However, many substances characteristically have more than one pKa value.
For each value, the molecule is able to give up a proton or accept a proton.
For example H3PO4 has three pKa values. This is due to the fact that it is able
to donate three protons while in solution. However, it is much more difficult
to remove the second proton than the first. This is due to the fact that it is
more difficult to remove a proton from a anion. Furthermore, the more negative
the anion, the more difficult to remove the proton.

The trapezoidal method can be employed to find the equivalence points as
shown if figure 6. The volume of titrant between two equivalence points is
helpful in the determination of the molecular weight of the amino