A 0.338 g sample of an unknown triprotic acid dissolved in water is titrated to the second equivalence point using 22.2 mL of 0.163 M NaOH. Calculate the molar mass of this acid. The correct answer is 187g/mol. a) Explain in detail how to arrive at the answer of 187 g/mol and why certain equations are used. b) Explain why is the answer 187 g/mol and not 280 g/mol?
A 0.338 g sample of an unknown triprotic acid dissolved in water is titrated to the second equivalence point using 22.2 mL of 0.163 M NaOH. Calculate the molar mass of this acid. The correct answer is 187g/mol. a) Explain in detail how to arrive at the answer of 187 g/mol and why certain equations are used. b) Explain why is the answer 187 g/mol and not 280 g/mol?
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.118QP: A 0.239-g sample of unknown organic base is dissolved in water and titrated with a 0.135 M...
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A 0.338 g sample of an unknown triprotic acid dissolved in water is titrated to the second equivalence point using 22.2 mL of 0.163 M NaOH. Calculate the molar mass of this acid.
The correct answer is 187g/mol.
a) Explain in detail how to arrive at the answer of 187 g/mol and why certain equations are used.
b) Explain why is the answer 187 g/mol and not 280 g/mol?
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