Answer
$K_a=2.95\times10^{-10}$
Work Step by Step
$pK_a=-log(K_a)$
$K_a=10^{-9.53}$
$K_a=2.95\times10^{-10}$
It would be above $HCO_3^-$ in the table.
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 629b: 27
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