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.

KA2 KA1

[H3O+] [A2-]

[HA-]

[H3O+] [HA-]

[H2A]

It can be shown that if > 103, then

the effects of the second dissociation is negligible during the titration of H2A.

The same test can be used for other weak acids with more than two titratible acid groups and bases.

You must test K values in sequence. To compare KA1 to KA3 has no meaning.

KA1 KA2

[H3O+][H2PO4-]

[H3PO4]

[H3O+][HPO42-]

[H2PO4-]

[H3O+][PO43-]

[HPO42-]

H3PO4 H2PO4- HPO4

2- PO43-

KA3 KA2 KA1

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 is not greater than 1000 does not mean it can’t be titrated.

It is still possible to determine equilibrium concentrations at any point during a titration.

However, when KA1/KA2 < 1000, you don’t get a sharp ‘first’ endpoint. You can

Titrate to the ‘final endpoint’ Titrate to a form where Kx/KY > 1000.

We’ll now review the calculations involved for the titration of a ‘well behaved’ diprotic acid.

This really is nothing more that doing two separate sets of calculations.

Triprotic acids (or higher) simply amount to repeating the process as many times as necessary.

Construct a curve for the titration of 25.00 ml of 0.1000 M sodium carbonate (Na2CO3) with 0.1000 M HCl.

KB1 = 2.00 x 10-4 =

KB2 = 2.51 x 10-8 =

KB1 / KB2 = 8.00 x 103 so our base is ‘well behaved’ and we can treat each equilibrium separately.

[ OH- ] [HCO3- ]

[ H2CO3 ]

[ OH- ] [ CO32- ]

[HCO3- ] KB1 = 2.00 x 10-4 =

[ OH- ] = [ CO32- ]

[ HCO3- ] = 0.1000M - [ CO3

2- ]

[ OH- ] [ CO32- ]

[HCO3- ]

[OH- ]2 0.1000 - [OH-]

[CO32]

[HCO3-]

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 15.0 60 10.12 17.5 70 9.93 20.0 80 9.70 22.5 90 9.35

0

2

4

6

8

10

12

0 5 10 15 20 25

pH

ml HCl pH = pKA1 + pKA2

2

0

2

4

6

8

10

12

0 5 10 15 20 25

pH

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

0

2

4

6

8

10

12

0 10 20 30 40 50

pH

ml HCl

[H3O+] [HCO3-]

[H2CO3]

[H3O+ ]2 0.03333

KA1 =

pH = 3.94 0

2

4

6

8

10

12

0 10 20 30 40 50

pH

ml HCl

0

2

4

6

8

10

12

0 20 40 60 80

pH

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.

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

bicarbonate only

mixed system

What happens when the K values do not differ by 1000 or more?

Can you still titrate them?

What would the curves look like?

First, lets outline how to predict what a titration curve would look.

A few points are relatively easy to estimate.

At 50% conversion of any specific form, pH will be relatively close to the pKA value.

The equivalence point pH of intermediate forms can be estimated based on the average of the pKA values.

You should also be able to calculate the pH when over titrating the sample.

Let’s take a look at the following two acids:

pKA1 pKA2 Sulfurous 1.9 7.2 Tartaric 3.0 4.4

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 titrant.

Sulfurous Tartaric 1st buffer region 1.9 3.0 1st eq. point 4.6 3.7 2nd buffer region 7.2 4.4

The basic shape of a titration curve should still be more or less the same.

A change of about 2 pH units from 10 to 90% titration - assuming no overlap of the curves.

0 100 200 300

1

3

5

7

9

11

13

You can see that we still have a usable 2nd equivalence point in both cases.

For tartaric acid, it would be difficult to detect the 1st eq. point.

sulfurous

tartaric

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.