Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 19 - Ionic Equilibria II: Buffers and Titration Curves - Exercises - The Common Ion Effect and Buffer Solutions - Page 773: 10

Answer

(a) $pH = 3.27$ (b) $pH = 4.84$

Work Step by Step

(a) 1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 7.2 \times 10^{- 4})$ $pKa = 3.143$ 2. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 3.143 + log(\frac{0.2}{0.15})$ $pH = 3.143 + log(1.333)$ $pH = 3.143 + 0.1249$ $pH = 3.268$ (b) 1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.745$ 2. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ ** $0.025M Ba(CH_3COO)_2 = 0.050M CH_3COO^-$ $pH = 4.745 + log(\frac{0.05}{0.04})$ $pH = 4.745 + log(1.25)$ $pH = 4.745 + 0.09691$ $pH = 4.842$
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