General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 16 - Acid-Base Equilibria - Questions and Problems - Page 704: 16.81



Work Step by Step

We know that $PH=-logK_a+log\frac{[base]}{[acid]}$ $PH=-log(7.14\times 10^{-6})+log \frac{(0.15)}{(0.10)}=5.322$
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