Answer
A
Work Step by Step
$pH=-log[H_3O^+]$
$[H_3O^+]=10^{-3.02}$
$[H_3O^+]=9.55\times10^{-4}\ M=[A^-], [HA]=0.039\ M$
$K_a=[H_3O^+][A^-]/[HA]$
$K_a=(9.55\times10^{-4})^2/0.039$
$K_a=2.34\times10^{-5}$
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 - 16-7 Calculations with Equilibrium Constants - Review & Check for Section 16-7 - Page 615: 2
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