A 20ml sample of .125M diprotic acid solution was titrated with .1019M KOH.the constants for the acid are k1=5.2*10 ^-5 and K2=3.4*10 ^-10. What added volume of base does each equivalence point occur?

Question

A 20ml sample of .125M diprotic acid solution was titrated with .1019M KOH.the constants for the acid are k1=5.2*10 ^-5 and K2=3.4*10 ^-10. What added volume of base does each equivalence point occur?
  
i got .0254L for the first one and .0508 for the second. Can anyone check if i did it rightÉ

DrBob222 · Answer

First equivalence point is
M x L = moles
moles/0.1019 = L. Close to 0.0254 but I get 0.0245 L = 24.5 mL. I wonder if that 20 mL sample really is 20.00 an the 0.125 M is really 0.1250 in which case more s.f. are available and the volume for the first equivalence point is 24.53 mL.

For the second one, it is
40.00 x 0.1250/0.1019 = 49.07
How do we use the constants in this problem? The fact that they are 10^5 apart (approximately) means the two equivalence points are distinct and can be titrated separately.