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[References) Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC,H;O2 (K = 6.4 × 10-5) titrated by 0.10 M NaOH pH at the halfway point %3D pH at the equivalence point = b. 100.0 mL of 0.40 M C2H5NH2 (K = 5.6 x 10-) titrated by 0.10 M HNO3 %3D pH at the halfway point %3! pH at the equivalence point c. 100.0 mL of 0.70 M HCl titrated by 0.15 M NaOH pH at the halfway point pH at the equivalence point %3D

[References) Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC,H;O2 (K = 6.4 × 10-5) titrated by 0.10 M NaOH pH at the halfway point %3D pH at the equivalence point = b. 100.0 mL of 0.40 M C2H5NH2 (K = 5.6 x 10-) titrated by 0.10 M HNO3 %3D pH at the halfway point %3! pH at the equivalence point c. 100.0 mL of 0.70 M HCl titrated by 0.15 M NaOH pH at the halfway point pH at the equivalence point %3D

Chemistry
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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ng resourc.. (27) Could've Been - [References] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC7H502 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.40 M C2H;NH2 (K½ = 5.6 × 10¬4) titrated by 0.10 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.70 M HC1 titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =
Transcribed Image Text:ng resourc.. (27) Could've Been - [References] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC7H502 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.40 M C2H;NH2 (K½ = 5.6 × 10¬4) titrated by 0.10 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.70 M HC1 titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =
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