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: 19.30

Answer

See explanation below.

Work Step by Step

1 (A) Given : 1. [base] = [B] = 1.05 M 2. [conjugate acid] = [BH+] = 0.75 M 3. pH of buffer = 9.50 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [base] / [acid] 3 (C) Solution : 1. pH = pKa + log [conjugate base] / [acid] So, 9.50 = pKa + log (1.05 / 0.75) pKa = 9.50 - 0.15 pKa = 9.35 ...... (1) 2. [HCl] added = Moles added / Volume [HCl] = 0.0050 moles / 0.500 L = 0.010 M HCl 3. Now, on addition of 0.01 M HCl, the acid concentration will increase by 0.01 M and base concentration will decrease by 0.01 M. So, new [B] = 1.05 - 0.01 = 1.04 M new [BH+] = 0.75 + 0.01 = 0.76 M new pH = pKa + log([B] / [BH+]) = 9.35 + log(1.04 / 0.76) = 9.35 + 0.14 = 9.49 Hence, the new pH of the buffer is 9.49
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