Chemistry Reference
In-Depth Information
we can deduce that
[RCO
2
−
]
[RCO
2
H]
=
log
pH
−
p
K
a
=
7
.
0
−
2
.
0
=
5
.
0
Thus
[RCO
2
−
]
[RCO
2
H]
=
10
5
=
10 000 : 1
Therefore, the carboxylic acid group of an amino acid can be considered to be completely ionized in solution at
pH 7.0.
Now let us consider the
amino group
in
α
-amino acids. The p
K
a
values of the conjugate acids are found to
range from about 8.8 to 10.8. We shall consider a typical group with p
K
a
10.0. From
[RNH
2
]
[RNH
3
+
]
pH
=
p
K
a
+
log
[RNH
2
]
[RNH
3
+
]
=
pH
−
p
K
a
=
−
=−
log
7
.
0
10
.
0
3
.
0
Thus
[RNH
2
]
[RNH
3
+
]
=
10
−
3
=
1 : 1000
Therefore, as with the carboxylic acid group, we find that the amino group of an amino acid is effectively ionized
completely, i.e. fully protonated, in solution at pH 7.0.
Therefore, we can deduce that
α
-amino acids in solution at pH 7.0 exist as dipolar ions; these are called
zwitterions
(German;
zwitter
=
hybrid) (see Section 13.1).
pH 7.0
R
CO
2
R
CO
2
H
NH
3
NH
2
α-amino acid
zwitterion
Some amino acids have additional ionizable groups in their side-chains. These may be acidic or potentially acidic
(aspartic acid, glutamic acid, tyrosine, cysteine), or basic (lysine, arginine, histidine). We use the term 'potentially
acidic' to describe the phenol and thiol groups of tyrosine and cysteine respectively; under physiological
conditions, these groups are unlikely to be ionized. It is relatively easy to calculate the amount of ionization
at a particular pH, and to justify that latter statement.
Similar calculations as above for the basic side-chain groups of arginine (p
K
a
12.48) and lysine (p
K
a
10.52),
and the acidic side-chains of aspartic acid (p
K
a
3.65) and glutamic acid (p
K
a
4.25) show essentially complete
ionization at pH 7.0. However, for cysteine (p
K
a
of the thiol group 10.29) and for tyrosine (p
K
a
of the phenol
group 10.06) there will be negligible ionization at pH 7.0.
For
cysteine
at pH 7.0, the Henderson - Hasselbalch equation leads to
[A
−
]
[HA]
=
log
pH
−
p
K
a
=
7
.
0
−
10
.
29
=−
3
.
29
and
[A
−
]
[HA]
=
10
−
3
.
29
10
−
4
=
5
.
1
×
i.e. no significant ionization.
