Answer
Please see the work below.
Work Step by Step
We know that
$log[H_3O^+]=-PH=-5.12$
$log[H_3O^{+}]=-5.12$
$[H_3O^+]=anti log(-5.12)$
$[H_3O^+]=10^{-5.12}=7.58\times 10^{-6}M$
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
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: 15.75
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