Answer
B
Work Step by Step
$K_w=[H_3O^+][OH^-]$
$10^{-14}=[H_3O^+]0.0012$
$[H_3O^+]=8.33\times10^{-12}$
$pH=-log10([H_3O^+])$
$pH=11.08$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - 16-2 Water and the pH Scale - Review & Check for Section 16-2 - Page 592: 1
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