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: 19.11

Answer

pH of buffer = 5.25 Hydronium ion concentration = 5.5 x 10^-6 M

Work Step by Step

1 (A) Given : 1. [conjugate base] = [CH3CH2COO-] = 0.35 M 2. [acid] = [CH3CH2COOH] = 0.15 M 3. Ka of propanoic acid = 1.3 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 (1.3 x 10^-5) = 4.89 .... (1) 2. pH = pKa + log [conjugate base] / [acid] = 4.89 + log (0.35 / 0.15) = 5.25 Hence, pH of the buffer is 5.26. 3. pH of the solution = -log [H+] So, [H+] = 10^(-pH) = 10^(-5.25) = 5.6 X 10^-6 M 4 (D) Answer : pH of buffer = 5.25 Hydronium ion concentration = 5.5 x 10^-6 M
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