4.60 g of formic acid (molar mass 46.0 g/mol) was added to a flask and dissolved in water to make 100.00 mL of solution. It required 40.00 mL of 2.5 M NaOH to reach the equivalence point. What is the pH at the equivalence point of the titration? K₂ = 1.77 x 10-4 a pH = [?] pH Enter

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 52QAP: A solution is prepared by dissolving 0.350 g of benzoic acid, HC7H5O2, in water to make 100.0 mL of...
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4.60 g of formic acid (molar mass
46.0 g/mol) was added to a flask and
dissolved in water to make 100.00 mL
of solution. It required 40.00 mL of 2.5 M
NaOH to reach the equivalence
point. What is the pH at the equivalence
point of the titration?
K₂ = 1.77 x 10-4
a
pH = [?]
pH
Enter
Transcribed Image Text:4.60 g of formic acid (molar mass 46.0 g/mol) was added to a flask and dissolved in water to make 100.00 mL of solution. It required 40.00 mL of 2.5 M NaOH to reach the equivalence point. What is the pH at the equivalence point of the titration? K₂ = 1.77 x 10-4 a pH = [?] pH Enter
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