## Chemistry: The Molecular Science (5th Edition)

$E. pH = pK_a$
- Analyzing the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[Base]}{[Acid]})$ If $[Base] = [Acid]$, then: $\frac{[Base]}{[Acid]} = 1$ $pH = pK_a + log(1)$ $pH = pK_a + 0$ $pH = pK_a$