1. It is still possible to develop analytical
methods for species that contain
multiple acid or base groups.
This unit will review the types of
calculations and titration curves
involved.
Some specific examples will also be
included.
KA1 KA2 [H3O+] [HA-]
[H2A]
[H3O+] [A2-]
[HA-]
K
It can be shown that if KA1 > 103, then KA1 KA2 KA3
A2
H3PO4 H2PO4- HPO42- PO43-
the effects of the second dissociation is
negligible during the titration of H2A.
[H3O+][H2PO4-]
The same test can be used for other weak [H3PO4]
acids with more than two titratible acid groups
[H3O+][HPO42-]
and bases.
[H2PO4-]
You must test K values in sequence. To [H3O+][PO43-]
compare KA1 to KA3 has no meaning. [HPO42-]

2. KA1 / KA2 = 7.5x10-3 / 6.0x10-8
= 1.25 x 105
KA2 / KA3 = 6.0x10-8 / 4.8x10-13
= 1.25 x 105
This indicates that it is possible to
treat each step separately during
a titration.
Just because you have a material where KA1/KA2 We’ll now review the calculations involved for
is not greater than 1000 does not mean it the titration of a ‘well behaved’ diprotic acid.
can’t be titrated.
It is still possible to determine equilibrium This really is nothing more that doing two
concentrations at any point during a titration. separate sets of calculations.
However, when KA1/KA2 < 1000, you don’t get
a sharp ‘first’ endpoint. You can Triprotic acids (or higher) simply amount to
repeating the process as many times as
Titrate to the ‘final endpoint’
necessary.
Titrate to a form where Kx/KY > 1000.
Construct a curve for the titration of 25.00 ml of
0.1000 M sodium carbonate (Na2CO3) with
0.1000 M HCl.
[ OH- ] [ CO32- ] [ OH- ] [ CO32- ]
KB1 = 2.00 x 10-4 = [HCO3- ] KB1 = 2.00 x 10-4 = [HCO3- ]
- -
KB2 = 2.51 x 10-8 = [ OH ] [HCO3 ]
[ H2CO3 ]
[ OH- ] = [ CO32- ]
KB1 / KB2 = 8.00 x 103
so our base is ‘well behaved’ and we can treat [ HCO3- ] = 0.1000M - [ CO32- ]
each equilibrium separately.

3. [OH- ]2
0.1000 - [OH-]
mlHCl % titration1st eq.pt. pH
0.0 0 11.65
2.5 10 11.25
5.0 20 10.90
7.5 30 10.67
10.0 40 10.48
12.5 50 10.30
[CO32] 15.0 60 10.12
[HCO3-] 17.5 70 9.93
20.0 80 9.70
22.5 90 9.35
12
10
8
pH
6
4
2
0
0 5 10 15 20 25 pKA1 + pKA2
pH =
ml HCl 2

4. 12
10
8
pH
6
4
2
0
0 5 10 15 20 25
ml HCl
mlHCl % titration1st eq.pt. pH
25.0 0 8.35
27.5 10 7.35
30.0 20 7.00
32.5 30 6.77
35.0 40 6.58
37.5 50 6.40
40.0 60 6.22
42.5 70 6.03
45.0 80 5.80
47.5 90 5.45
12
10
8
[H3O+] [HCO3-]
pH
6
[H2CO3]
4
2
0
0 10 20 30 40 50
ml HCl

5. 12
10
8
pH
6
4
[H3O+ ]2 2
KA1 =
0.03333 0
0 10 20 30 40 50
pH = 3.94 ml HCl
12
10
8
pH
6
4
2
0
0 20 40 60 80
ml HCl
Having used carbonate as an example, we
should point out some of the specifics of
bicarbonate titrations.
Carbonates and bicarbonate/carbonate
mixtures are commonly measured. They are
relatively common in nature.
The recommended titrant is HCl.
Indicator used is based on the type of sample.

6. Neither the phenolphthalein or methyl orange
endpoint detection is stellar.
An alternate approach is to use methyl red
This indicator can only be used to detect the
second endpoint.
The transition is from yellow to red and is still
gradual unless some additional steps are
taken.
carbonate only
mixed system
bicarbonate only
What happens when the K values do not differ A few points are relatively easy to estimate.
by 1000 or more?
At 50% conversion of any specific form, pH
will be relatively close to the pKA value.
Can you still titrate them?
The equivalence point pH of intermediate
What would the curves look like? forms can be estimated based on the average
of the pKA values.
First, lets outline how to predict what a titration You should also be able to calculate the pH
curve would look. when over titrating the sample.

7. Let’s take a look at the following two acids: Sulfurous Tartaric
1st buffer region 1.9 3.0
1st eq. point 4.6 3.7
pKA1 pKA2
2nd buffer region 7.2 4.4
Sulfurous 1.9 7.2
Tartaric 3.0 4.4
The basic shape of a titration curve should still
be more or less the same.
We’ll assume that we have 100 ml of a 0.10 M
for each and we are using 0.10 M NaOH as our A change of about 2 pH units from 10 to 90%
titrant. titration - assuming no overlap of the curves.
13
11
You can see that
9 sulfurous we still have a usable
2nd equivalence point
7 in both cases.
5
For tartaric acid, it
3 would be difficult to
tartaric detect the 1st eq. point.
1
0 100 200 300
So when working with diprotic acids (or bases)
where K2 is < 1000 K1, you can only work with
the final equivalence point.
Usable results can still be obtained but you can’t
work with mixed systems.
The same is true if you have more than 2
titratible groups.
However, you can obtain a good endpoint when
any adjacent K values differ by 1000.