Answer
Please see the work below.
Work Step by Step
We know that
$H_3O^+=\frac{K_w}{OH^-}$
$H_3O^+=\frac{1.0\times 10^{-14}}{0.0040}=2.5\times 10^{-12}M$
Now $PH=-log (H_3O^+)=-log(2.5\times 10^{-12})=11.602$
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: 15.73
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