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
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?
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What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Acid and Base Titration Curves concept. You can view video lessons to learn Acid and Base Titration Curves. Or if you need more Acid and Base Titration Curves practice, you can also practice Acid and Base Titration Curves practice problems.
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