Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 19 - Problems: 19.12

Answer

pH of buffer = 4.13 Hydronium ion concentration = 7.413 x 10^-5 M

Work Step by Step

1 (A) Given : 1. [conjugate base] = [C6H5COO-] = 0.28 M 2. [acid] = [C6H5COOH] = 0.33 M 3. Ka of Benzoic acid = 6.3 X 10^-5 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid] 2. pKa = -log Ka 3. pH = -log [H+] 3 (C) Solution : 1. pKa = -log Ka = - log (6.3 x 10^-5) = 4.20 .... (1) 2. pH = pKa + log [conjugate base] / [acid] = 4.20 + log (0.28 / 0.33) = 4.13 Hence, pH of the buffer is 4.13. 3. pH of the solution = -log [H+] So, [H+] = 10^(-pH) = 10^(-4.13) = 7.413 x 10^-5 M 4 (D) Answer : pH of buffer = 4.13 Hydronium ion concentration = 7.413 x 10^-5 M
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