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: 19.19

Answer

pH of buffer = 9.47

Work Step by Step

1 (A) Given : 1. [conjugate acid] = [NH4+] = 0.15 M 2. [base] = [NH3] = 0.25 M 3. pKb of Base NH3 = 4.75 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 - 4.75 = 9.25 .... (1) 2. pH = pKa + log [base] / [acid] = 9.25 + log (0.25 / 0.15) = 9.25 + 0.22 = 9.47 Hence, pH of the buffer is 9.47. 4 (D) Answer : pH of buffer = 9.47
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