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Determination Of An Unknown Amino Acid From Titration - College Paper

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 ...

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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 ...

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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 ...

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PAPER DETAILS
Added: 3/1/2004 08:49:52 PM
Category: Science & Nature
Type: Free Paper
Words: 1634
Pages: 6

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