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: 19.20

Answer

pH of buffer = 10.57

Work Step by Step

1 (A) Given : 1. [conjugate acid] = [CH3NH3+] = 0.60 M 2. [base] = [CH3NH2] = 0.50 M 3. pKb of Base CH3NH2 = 3.35 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [base] / [acid] 2. pKa + pKb = pKw = 14 3 (C) Solution : 1. pKa = pKw - pKb = 14 - 3.35 = 10.65 .... (1) 2. pH = pKa + log [base] / [acid] = 10.65 + log (0.50 / 0.60) = 10.65 + (-0.08) = 10.57 Hence, pH of the buffer is 10.57 4 (D) Answer : pH of buffer = 10.57
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