Answer
$[OH^-]=0.022 M$
$pH=12.34$
Work Step by Step
$[OH^-]=x$
$K_b=6.4\times10^{-4}=\frac{x\times x}{0.075-x}$
We can assume that
$x\lt\lt0.075$
Therefore, $x=\sqrt {6.4\times 10^-4\times 0.075}$
$x=[OH^-]=0.022 M$
$pOH=1.66$
$pH=14-pOH=12.34$
Chemistry: The Central Science (13th Edition)
by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.
Published by
Prentice Hall
ISBN 10:
0321910419
ISBN 13:
978-0-32191-041-7
Chapter 16 - Acid-Base Equilibria - Exercises - Page 719: 16.71
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