If the ph at one half the first and second equivilance points of a dibasic acid is 4.2 and 7.34 respectivly, what are the values of pka and pka2

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

You need to divided twenty four by a dozen them by the temperature it is getting cook by