Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

Chapter 19 - Ionic Equilibria II: Buffers and Titration Curves - Exercises - The Common Ion Effect and Buffer Solutions - Page 773: 11

Answer

$pH = 7.75$

Work Step by Step

1. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 7.45 + log(\frac{0.0444}{0.0222})$ $pH = 7.45 + log(2)$ $pH = 7.45 + 0.301$ $pH = 7.75$

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