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: 16.69a

Answer

$(CH_3)_2NH + H_2O -> (CH_3)_2NH_2^+ + OH^- $ $Kb = \frac{[CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}$

Work Step by Step

1. Write and balance the reaction with water: $(CH_3)_2NH + H_2O -> (CH_3)_2NH_2^+ + OH^- $ $1(CH_3)_2NH + 1H_2O -> 1(CH_3)_2NH_2^+ + 1OH^- $ 2. Now, write the Kb formula: $Kb = \frac{[Conjugate. acid]*[OH^-]}{[Base]}$ $Kb = \frac{[CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}$
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