General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 16 - Acid-Base Equilibria - Questions and Problems - Page 704: 16.63

Answer

Please see the work below.

Work Step by Step

We know that A. $K_b=\frac{K_w}{K_a}$ $K_b=\frac{1.0\times 10^{-14}}{4.5\times 10^{-4}}=2.22\times 10^{-11}$ B . We know that $K_a=\frac{K_w}{K_a}$ $K_a=\frac{1.0\times 10^{-14}}{1.4\times 10^{-9}}=7.14\times 10^{-6}$
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.