Answer
$E. pH = pK_a$
Work Step by Step
- Analyzing the Henderson-Hasselbalch equation:
$pH = pK_a + log(\frac{[Base]}{[Acid]})$
If $[Base] = [Acid]$, then: $\frac{[Base]}{[Acid]} = 1$
$pH = pK_a + log(1)$
$pH = pK_a + 0$
$pH = pK_a$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Applying Concepts - Page 693f: 109g
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