Problem: 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?

FREE Expert Solution

We’re being asked to calculate the Ka of the given monoprotic acid



Recall that at the equivalence point of a titration:


moles acid=moles base


Also, recall that moles = molarity × volume

This means:


MVacid=MVbase


The monoprotic acid will have 1 equivalence point:

HA H+ + A-                                 

 (MA)acid = (MV)base

We are given:

[HA]=0.0950 M

Volume of HA = 25 mL

Volume of Ato reach equivalence point=30 mL

Recall: A-   

1 mL = 10-3 L

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Problem Details

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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