Answered: A student titrates 20.0 mL of 1.0 M…

A student titrates 20.0 mL of 1.0 M NAOH with 2.0 M formic acid, HCOOH (K, = 1.8 x 10-4). How much formic acid is required to reach the equivalence point? 10.0 mL 25.0 mL 20.0 mL 30.0 mL

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter14: Principles Of Neutralization Titrations

Section: Chapter Questions

Problem 14.39QAP

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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

A student titrates 20.0 mL of 1.0 M NaOH with 2.0 M formic acid, HCOOH
(K, 1.8 x 10-4). How much formic acid is required to reach the
equivalence point?
10.0 mL
25.0 mL
20.0 mL
30.0 mL

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Consider the following four titrations. i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated by 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated by 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated by 0.10 M NaOH Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing pH at the halfway point in equivalence. d. increasing pH at the equivalence point. How would the rankings change if C5H5N replaced CH3NH2 and if HOC6H5 replaced HF?
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