Answer
See the answer below.
Work Step by Step
$K_a=[H_3O^+][X^-]/[HX]$
$1.3\cdot10^{-3}=x\cdot x/(0.010-x)$
$x=0.003\ M=[H_3O^+],\ [HX]=0.007\ M$
$pH=-log([H_3O^+)]=2.52$
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 629e: 89
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