Answer
Percent ionization $ \approx 4.9 \%$
Work Step by Step
Percent ionization :
$\frac{x}{[C_6H_5COOH]_{initial}}\times 100\% = \frac{ 2.449 \times 10^{- 3}}{ 0.05} \times 100\% = 4.899\%$
** The "x" value was calculated in "59a".
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 59b
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