Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 772: 16.62

Answer

$Ka = 3.982\times 10^{- 11}$

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. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x$ 2. Now, convert the pH into hydronium ion concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 6.2}$ $[H_3O^+] = 6.3 \times 10^{- 7}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{( 6.31\times 10^{- 7})^2}{ 0.01- 6.31\times 10^{- 7}}$ $Ka = \frac{ 3.982\times 10^{- 13}}{ 0.01}$ $Ka = 3.982\times 10^{- 11}$
Small 1531317675
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.