Answer
See the answer below.
Work Step by Step
$K_a=[H_3O^+][CN^-]/[HCN]$
$[H_3O^+]=[CN^-]=x$
$[HCN]=0.025-x$
$4\cdot10^{-10}=x^2/(0.025-x)$
$x=3.16\cdot10^{-6}\ M=[H_3O^+]=[CN^-]$
$[HCN]\approx0.025\ M$
$pH=-log(H_3O^+)$
$pH=5.50$
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 629c: 51
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