Answer
$pH = 12.290$
Work Step by Step
1. Since $KOH$ is a strong base:
$[OH^-] = [KOH] = 0.0195M$
2. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.0195)$
$pOH = 1.710$
$pH + pOH = 14$
$pH + 1.710 = 14$
$pH = 12.290$
Chemistry: A Molecular Approach (3rd Edition)
by Tro, Nivaldo J.
Published by
Prentice Hall
ISBN 10:
0321809246
ISBN 13:
978-0-32180-924-7
Chapter 15 - Sections 15.1-15.12 - Exercises - Cumulative Problems - Page 750: 138c
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