Question

Trying to determine the molar weight and PKa values for an unknown diprodic acid.

I have .9010 molar NAOH titrant I have a mass of Unknown acid being .1101 g. Equivalence point of Steepest point on graph equals 22.5mL NAOH. Equivalence OH- is 2.027×10^ -2

moles.

question---I need help to convert moles of OH ion added to initial moles diprotic acid H2A PH 12Approximately pka1 is approximately 3.5 at 6.25 mL OH.

data-

NaOHmL / PH 2.5 3 3.50 7.50 3.75 10.00 44 – Bottom line 12.50 4.15€ Eq pnt 1 15 16.50€ topline 22.00 8 € Bottomline 22.50 / 11.5 E q ent 2 24 [12&top line

Answer 1

C.B. TAI At a.s canivalence paint [H₂A) = THAI the corresponding pH = pka, = 3.5 C. From Henderson. Handbalth eanation we get i PH - pkat log I conjugate base here, c at u.s equivalence A = Acidl point Than) = [HA] where, ha is L.B. of the So, PH = pka tlaga apka apkay=35 . pkai =35 Similarly at ls equivalence point [HA-) = [A-] then corresponding pt = pkan here, at is equivalence point PH Z 8.5 Pkaz= 6.5