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: 19.28

Answer

See explanation below.

Work Step by Step

1 (A) Given : 1. [base] = [B] = 0.40 M 2. [conjugate acid] = [BH+] = 0.25 M 3. pH of buffer = 8.88 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pOH = pKb + log [conjugate acid] / [base] 2. pH = 14 - pOH 3 (C) Solution : 1. pH = 14 - pOH So, pOH = 14 - 8.88 = 5.12 2. pOH = pKb + log [conjugate acid] / [base] So, 5.12 = pKb + log (0.25 / 0.40) pKb = 5.12 - (-0.20) = 5.32 3. [HCl] added = Moles added / Volume [HCl] = 0.0020 moles / 0.25 L = 0.008 M HCl 4. Now, on addition of 0.008 M HCl, the acid concentration will increase by 0.008 M and base concentration will decrease by 0.008 M. So, new [B] = 0.40 - 0.008 = 0.392 M new [HB+] = 0.25 + 0.003 = 0.258 M new pOH = pKb + log [conjugate acid] / [base] = 5.32 + log(0.258 / 0.392) = 5.32 + (-0.182) = 5.14 pH = 14 - pOH = 14 - 5.14 = 8.86 Hence, the new pH of the buffer is 8.86.
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