Answer
Please see the work below.
Work Step by Step
We know that
$H_3O^+=\frac{K_w}{OH^-}$
$H_3O^+=\frac{1.0\times 10^{-14}}{0.050}=2\times 10^{-3}M$
$PH=-log(H_3O^+)$
$PH=-log (2\times 10^{-13})=12.698$
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: 15.74
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