A sample of 0.2505 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M . The acid required 30.0 mL of base to reach the equivalence point.
After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?
We’re being asked to calculate the Ka of the given monoprotic acid.
Recall that at the equivalence point of a titration:
Also, recall that moles = molarity × volume.
The monoprotic acid will have 1 equivalence point:
HA → H+ + A-
(MA)acid = (MV)base
We are given:
Volume of HA = 25 mL
Volume of A- to reach equivalence point=30 mL
1 mL = 10-3 L