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: 15.74


Please see the work below.

Work Step by Step

We know that $H_3O^+=\frac{K_w}{OH^-}$ $H_3O^+=\frac{1.0\times 10^{-14}}{0.050}=2\times 10^{-3}M$ $PH=-log(H_3O^+)$ $PH=-log (2\times 10^{-13})=12.698$
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