General Chemistry 10th Edition

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

Chapter 15 - Acids and Bases - Questions and Problems - Page 656: 15.9

Answer

Please see the work below.

Work Step by Step

As $PH=3.16$ so $log[H_3O^+]=-3.16$ and $H_3O^+=antilog(-3.16)$ $[H_3O^+]=10^{-3.16}=6.91\times 10^{-4}M$
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