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¼
H þ
½
HA 3
A 4
ðA1:3Þ
K 4
2
K 3 K 4 A 4
¼
H þ
¼
H þ
½
½
H 2 A 2
HA 3
ðA1:4Þ
K 3
3
K 2 K 3 K 4 A 4
H þ
¼
H þ
½
½
H 3 A
H 2 A 2
½
¼
ðA1:5Þ
K 2
4
H þ
H þ
½
½
H 3 A
A 4
½
H 4 A
¼
½
¼
ðA1:6Þ
K 1
K 1 K 2 K 3 K 4
By using these four relations, the expression for the total concentration can be
written in terms of [A 4 ]:
4
3
2
H þ
þ
H þ
þ
H þ
þ
H þ
þ
½
½
½
½
A 4
A 4
A 4
A 4
A 4
C 0 ¼
K 1 K 2 K 3 K 4
K 2 K 3 K 4
K 3 K 4
K 4
ðA1:7Þ
Dividing through by [A 4 ] leads to
4
K 1 K 2 K 3 K 4 þ
3
K 2 K 3 K 4 þ
2
K 3 K 4 þ
H þ
H þ
H þ
H ½
K 4 þ
C 0
A 4
1
a 4 ¼
½
½
½
¼
ðA1:8Þ
1
Writing the right side of Eq. (A1.8) as a fraction with a common denominator and
then inverting the fraction gives the desired final expression
K 1 K 2 K 3 K 4
a 4 ¼
K 1 K 2 K 3 þ K 1 K 2 K 3 K 4 :
ðA1:9Þ
4
3
2
H þ
H þ
H þ
H þ
½
þ
½
K 1 þ
½
K 1 K 2 þ
½
An important feature of Eq. (A1.9) to notice is that each term in the expression
actually represents the contribution of a particular protonated form, thus:
A 4
K 1 K 2 K 3 K 4 ,
H þ
HA 3
½
K 1 K 2 K 3 ,
2
H þ
H 2 A 2
½
K 1 K 2 ,
3
H þ
H 3 A
½
K 1 ,
4
H þ
H 4 A
This insight makes it easy to write the fraction of the polybasic acid that is in a
particular form: the term representing the particular protonated form appears in
the numerator, while the denominator is simply the sum of all the possible terms.
For example, the fraction existing as HA 3 is
½
,
H þ
½
K 1 K 2 K 3
a 3 ¼
K 1 K 2 K 3 þ K 1 K 2 K 3 K 4 ;
ðA1:10Þ
4
3
2
H þ
H þ
H þ
H þ
½
þ
½
K 1 þ
½
K 1 K 2 þ
½
and the fraction existing in the doubly deprotonated H 2 A 2 form is
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