your question is missing that we need Ka1 & Ka2 from the values of Pka1 & Pka2
we are going to use H-H equation:
PH = Pka + ㏒[A-/HA]
when at one half the equivalence, the acid and conjugate base are equal:
so, [A-] = [HA]
a) so to get Pka1:
PH = Pka1 + ㏒1
∴PH = Pka1
= 4.2
and when Pka1 = -㏒ Ka1
by substitution:
4.2 = -㏒ Ka1
∴ Ka1 = 6.3 x 10^-5
b) to get Pka2:
when [A-] = [HA]
so, PH = Pka2 + ㏒1
PH = Pka2
∴Pka2 = 7.34
when Pka2 = - ㏒ Ka2
7.34 = -㏒ Ka2
∴ Ka2 = 4.57 x 10^-8
