Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575d: 77

Answer

$Ka$ for trichloroacetic acid: 0.16

Work Step by Step

1. Since they have the same pH, their $[H_3O^+]$ should be the same too. And, $HClO_4$ is a strong acid, therefore :$[H_3O^+] = [HClO_4] = 0.040M$ - Now, we know that, the hydronium concentration in the trichloroacetic solution is 0.040M too. 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CCl_3C{O_2}^-] = x$ -$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][CCl_3C{O_2}^-]}{ [CCl_3CO_2H]}$ $Ka = \frac{x^2}{[InitialCCl_3CO_2H] - x}$ $Ka = \frac{( 0.04)^2}{ 0.05- 0.04}$ $Ka = \frac{ 1.6\times 10^{- 3}}{ 0.01}$ $Ka = 0.16$
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