Frank titrated 45.00 mL of 0.30 M formic acid. It required 54.0 mL of 0.25 M LiOH to reach the equivalence point. What is the pH when 27.0 mL of 0.25 M LIOH have been added to the acid? K₂ = 1.8 x 10-4 pH = [?]

Frank titrated 45.00 mL of 0.30 M formic acid. It required 54.0 mL of 0.25 M LiOH to reach the equivalence point. What is the pH when 27.0 mL of 0.25 M LIOH have been added to the acid? K₂ = 1.8 x 10-4 pH = [?]

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter15: Additional Aqueous Equilibria

Section: Chapter Questions

Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

Frank titrated 45.00 mL of 0.30 M formic
acid. It required 54.0 mL of 0.25 M LiOH
to reach the equivalence point. What is
the pH when 27.0 mL of 0.25 M LIOH
have been added to the acid?
K₂ = 1.8 x 10-4
pH = [?]
pH
Enter

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