How do you get pKa of a diprotic acid?

Question

A diprotic acid, H2A, has the following constants:
Ka1 = 0.0011
Ka2 = 0.0000025
In order to make a solution of pH 5.8 what is the pKa of the acid component?

Answer 1

Roughly, the Ka1 informs the first buffering point around pH = 3, and the Ka2 informs the second buffering point around pH < 6.
Remember, Ka1 is defined for:
H2A <==> H+ + HA-, and
Ka1 = [H+]*[HA-]/ [H2A]
and Ka2 is defined for:
HA- <==> H+ + A(2-), and
Ka2 = [H+]*[A--]/ [HA-]
Hence if you have a mixture of salts KHA and K2A with right ratio, you can get pH = 5.8 exactly.
Remember the definition of pH:
pH = - log(base 10) [H+]
We have:
pH = 5.8 = -log(Ka2) + log([A--]/ [HA-])
or: [A--]/ [HA-] = 1.58 = [K2A] / [KHA]