Inorganic and Applied Chemistry

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Example 5- J: Titration of a divalent acid

We look at the general divalent acid H 2 A which we assume have the following two Ka value:

Ka1 = 1.0 · 10-4 M and Ka2 = 1.0 · 10-9 M => pKa1 = 4.0 and pKa2 = 9.0

We are to titrate 20.0 mL 0.100 M solution of H 2 A with a 0.100 M solution of the strong base NaOH and we wish to determine pH during titration.

At the start of titration we have a 0,100 M solution of the acid H 2 A which is a weak acid what may be seen from the first Ka value.

Analogously earlier calculations we have the following equilibrium with corresponding expression of equilibrium:

H 2 A (aq) + H 2 O HA-(aq) + H 3 O+ (aq)

H A

K M HO HA a 2

1 1.^01043

As earlier we now look at start and equilibrium conditions. The initial concentrations are:

[H 2 A] 0 = 0.100 M [HA-] 0 = 0 M [H 3 O+] 0 * 0 M (the autoprotolysis of water is neglected)

and the end concentrations are thereby:

[H 2 A] = (0.100 – x) M [HA-] = x M [H 3 O+]* x M

which is why we get the following equation:

x M x

x x H A

K M HO HA a

3 2

(^43)
1 1.^0100. 100 3.^110

(^) (
Thereby [H 3 O+] = 3.1 · 10-3 M and the pH value becomes:
Acids and bases