Answer
Please see the work below.
Work Step by Step
We know that
$POH=-log[OH]^-$
$POH=-log(0.025)=1.602$
Now $PH=14.00-1.602=12.40$
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 - Exercises - Page 651: 15.8
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