Answer
See the answer below.
Work Step by Step
$pH=-log([H_3O^+])$
$[H_3O^+]=10^{-pH}$
$[H_3O^+]=0.005\ M=[F^-]$
$K_a=[H_3O^+][F^-]/[HF]$
$7.2\cdot10^{-4}=0.005^2/(x-0.005)$
$x=0.0397\ M=[HF]_0$
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 - Study Questions - Page 629d: 58
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