Chemistry: Atoms First (2nd Edition)

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

Chapter 13 - Exercises - Page 575d: 78

Answer

$Ka \approx 2\times 10^{- 9}$

Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CCl_3C{O_2}^-] = x$ -$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$ 2. Using the pH, calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.95}$ $[H_3O^+] = 1.122 \times 10^{- 5}$ So: $x = 1.122 \times 10^{-5}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$ $Ka = \frac{x^2}{[InitialHOBr] - x}$ $Ka = \frac{( 1.122\times 10^{- 5})^2}{ 0.063- 1.122\times 10^{- 5}}$ $Ka = \frac{ 1.259\times 10^{- 10}}{ 0.06299}$ $Ka = 1.999\times 10^{- 9}$
Small 1531587707
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.