Answer
See the answer below.
Work Step by Step
a) The first because it has a higher Kb
b) $K_b=[BH^+][OH^-]/[B]$
$4.3\cdot10^{-4}=x\cdot\ x/(0.10-x)$
$x=[OH^-]=0.0063\ M$
$pH=14+log\ [OH^-]=11.80$
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 629f: 95
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