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 705: 16.100


Please see the work below.

Work Step by Step

We know that $K_b=\frac{K_w}{K_a}$ $K_b=\frac{1.0\times 10^{-4}}{4.8\times 10^{-13}}=2.08\times 10^{-2}$ As $K_b=\frac{K_w}{K_a}$ $K_b=\frac{1.0\times 10^{-14}}{1.1\times 10^{-12}}=9.09\times 10^{-13}$
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