Acid and Base Titration Curves Video Lessons

Concept:

# 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:

Also, recall that moles = molarity × volume

This means:

$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

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?