Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 17 - Questions and Problems - Page 824: 17.12

Answer

The pH of this buffer is equal to $6.5$

Work Step by Step

1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 2 \times 10^{- 7})$ $pKa = 6.699$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.1}{0.15}$ - 0.6667: It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $ \frac{0.1}{2 \times 10^{-7}} = 5\times 10^{5}$ - $ \frac{0.15}{2 \times 10^{-7}} = 7.5\times 10^{5}$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 6.699 + log(\frac{0.1}{0.15})$ $pH = 6.699 + log(0.6667)$ $pH = 6.699 + (-0.176)$ $pH = 6.523$
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