Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 17 - Additional Aspects of Aqueous Equilibria - Exercises - Page 768: 17.27b

Answer

$pH \approx 5.018$

Work Step by Step

Using the Henderson-Hasselbalch equation: $pH = pKa +log\frac{[Base]}{[Acid]}$ 1. Find the pKa: $pKa = -log(Ka) = -log(1.8\times 10^{-5}) = 4.745$ 2. Find [Base]: $1CH_3COOH(aq) + 1KOH(aq) --> 1CH_3COO^-(aq) + K^+(aq) + H_2O(l)$ *Since the proportion is 1 to 1, 0.02 M of $CH_3COO^-$ were added. $[Base] = Initial + 0.02 = 0.13 + 0.02 = 0.15M$ 3. Find [Acid]: *Since the proportion is 1 to 1, 0.02 M of $CH_3COOH$ reacted, so: $[Acid] = Inital - 0.02 = 0.10 - 0.02 = 0.08M$ 4. Now, find the pH: $pH = 4.745 + log(\frac{0.15}{0.08})$ $pH = 4.745 + log(1.875)$ $pH \approx 4.745 + 0.273$ $pH \approx 5.018$
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