Answer
$[HNO_3] = 3.02 \times 10^{-4}$
Work Step by Step
1. Find $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.52}$
$[H_3O^+] = 3.02 \times 10^{- 4}M$
2. Since this is a strong acid: $[HNO_3] = [H_3O^+] = 3.02 \times 10^{-4}$
Chemistry 10th Edition
by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.
Published by
Brooks/Cole Publishing Co.
ISBN 10:
1133610668
ISBN 13:
978-1-13361-066-3
Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 743: 24
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