A student dissolved 0.1319g of an unknown diprotic acid in 100mL of distilled water. The acid was then titrated with 0.1005M NaOH solution. The second equivalence point showed the sharpest change in pH, and so it was used to determine the molecular weight of the unknown acid.

a. The NaOH volume just before the pH jump was 24.9mL and the NaOH volume just after the pH jump was 25.6mL.Calculate the volume of NaOH added at the equivalence point.

b. Calculate the number of moled of NaOH used in the titration to reach the second equivalence point.

c. Calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.

d. Calculate the molar mass of diprotic acid

e. What is the volume of NaOH used at the first equivalence point

f. What is the volume of NaOH used at the first half-titration

g- What is the volume of NaOH used at the second half-titration

h. Calculate Ka1 and Ka2 if the pH at the first half-titration is 1.95 and the pH at the second half-titration is 6.10