Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 16 - Acid-Base Equilibria - Exercises - Page 719: 16.64c

Answer

The percent ionization for 0.02M of propionic acid is 2.54%

Work Step by Step

Ka (Propionic acid) = $1.3 \times 10^{-5}$ 1. Drawing the ICE table we get: -$[H^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x$ For approximation, we consider: $[HA] = [HA]_{initial}$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H^+][A^-]}{ [HA]}$ $Ka = 1.3 \times 10^{- 5}= \frac{x * x}{ 2 \times 10^{- 2}}$ $Ka = 1.3 \times 10^{- 5}= \frac{x^2}{ 2 \times 10^{- 2}}$ $ 2.6 \times 10^{- 7} = x^2$ $x = 5.09 \times 10^{- 4}$ Percent ionization: $\frac{ 5.09 \times 10^{- 4}}{ 2 \times 10^{- 2}} \times 100\% = 2.54\%$ %ionization < 5% : Right approximation.
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.