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 659: 15.76


Please see the work below.

Work Step by Step

We know that $log[H_3O^+]=-PH=-3.85$ $log[H_3O^{+}]=-3.85$ $[H_3O^+]=anti log(-3.85)$ $[H_3O^+]=10^{-3.85}=1.41\times 10^{-4}M$
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