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 - Page 869: 19.27

Answer

See explanation below.

Work Step by Step

1 (A) Given : 1. [conjugate base] = [A-] = 0.1500 M 2. [acid] = [HA] = 0.2000 M 3. pH of buffer = 3.35 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid] 3 (C) Solution : 1. pH = pKa + log [conjugate base] / [acid] So, 3.35 = pKa + log (0.1500 / 0.2000) pKa = 3.35 - (-.012) pKa = 3.47 ...... (1) 2. [NaOH] added = Moles added / Volume [NaOH] = 0.0015 moles / 0.500 L = 0.003 M NaOH 3. Now, on addition of 0.003 M NaOH, the acid concentration will increase by 0.003 M and base concentration will decrease by 0.003 M. So, new [A-] = 0.1500 + 0.003 = 0.1530 M new [HA] = 0.2000 - 0.003 = 0.1970 M new pH = pKa + log([A-] / [HA]) = 3.47 + log(0.1530 / 0.1970) = 3.47 + (-0.06) = 3.37 Hence, the new pH of the buffer is 3.37.
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