Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 54

Answer

$Kb = 3.2 \times 10^{-2}$

Work Step by Step

1. Write the Kb equation, and calculate its value: $Kb = \frac{[CH_3N{H_3}^+][OH^-]}{[CH_3NH_2]}$ $Kb = \frac{(0.0080)^2}{0.002}$ $Kb = \frac{6.4 \times 10^{-5}}{0.002}$ $Kb = 3.2 \times 10^{-2}$
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