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: 15.73


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 $PH=-log (H_3O^+)=-log(2.5\times 10^{-12})=11.602$
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