6:30 PM Mon Mar 25 ED Calculate the pH at the equivalence point for the following titrations. 4A. Benzoic 50. mL of 0.12 M benzoic acid titrated by 0.10 M potassium hydroxide = ka C6H5COOH 6.3 x 10-5 4B. 50. mL of 0.15 M ethylamine titrated by 0.20 M chloric acid Ethylamine CH3CH2NH2 6.4 x 104 K
6:30 PM Mon Mar 25 ED Calculate the pH at the equivalence point for the following titrations. 4A. Benzoic 50. mL of 0.12 M benzoic acid titrated by 0.10 M potassium hydroxide = ka C6H5COOH 6.3 x 10-5 4B. 50. mL of 0.15 M ethylamine titrated by 0.20 M chloric acid Ethylamine CH3CH2NH2 6.4 x 104 K
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter14: Acid- Base Equilibria
Section: Chapter Questions
Problem 65E
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Question
![6:30 PM Mon Mar 25
ED
Calculate the pH at the equivalence point for the following titrations.
4A.
Benzoic
50. mL of 0.12 M benzoic acid titrated by 0.10 M potassium hydroxide
= ka
C6H5COOH
6.3 x 10-5
4B.
50. mL of 0.15 M ethylamine titrated by 0.20 M chloric acid
Ethylamine
CH3CH2NH2
6.4 x 104 K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c4fb4c2-0fc5-420f-aa03-c19db8aa74de%2F0d582236-db4d-4812-986d-f54262b07474%2Fsyb157k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6:30 PM Mon Mar 25
ED
Calculate the pH at the equivalence point for the following titrations.
4A.
Benzoic
50. mL of 0.12 M benzoic acid titrated by 0.10 M potassium hydroxide
= ka
C6H5COOH
6.3 x 10-5
4B.
50. mL of 0.15 M ethylamine titrated by 0.20 M chloric acid
Ethylamine
CH3CH2NH2
6.4 x 104 K
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