Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.20 M HC7H5O₂ (K₂ = 6.4 × 10-5) titrated by 0.10 M NaOH 2 pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.60 M C₂H³NH₂ (K₁ = 5.6 × 10-4) titrated by 0.60 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.40 M HCl titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.20 M HC7H5O₂ (K₂ = 6.4 × 10-5) titrated by 0.10 M NaOH 2 pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.60 M C₂H³NH₂ (K₁ = 5.6 × 10-4) titrated by 0.60 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.40 M HCl titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.
a. 100.0 mL of 0.20 M HC7H₂O₂ (K₂ = 6.4 × 1
: 10-5)
titrated by 0.10 M NaOH
pH at the halfway point =
pH at the equivalence point =
b. 100.0 mL of 0.60 M C₂H5NH₂ (K₁ = 5.6 × 10-
pH at the halfway point =
pH at the equivalence point =
c. 100.0 mL of 0.40 M HCl titrated by 0.15 M NaOH
pH at the halfway point =
pH at the equivalence point =
titrated by 0.60 M HNO3

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