Answer
$pH = 1.187$
Work Step by Step
1. Since $HNO_3$ is a strong acid:
$[H_3O^+] = [HNO_3]_{initial}$
2. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.0650)$
$pH = 1.187$
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: 138a
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