Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 15 - Sections 15.1-15.12 - Exercises - Problems by Topic - Page 747: 87c

Answer

$CH_3NH_2(aq) + H_2O(l) \lt -- \gt CH_3N{H_3}^+(aq) + OH^-(aq)$ $K_b = \frac{[OH^-][CH_3N{H_3}^+]}{[CH_3NH_2]}$

Work Step by Step

1. Write the ionization chemical equation: - Write the reaction where $CH_3NH_2$ takes a proton from a water molecule: $CH_3NH_2(aq) + H_2O(l) \lt -- \gt CH_3N{H_3}^+(aq) + OH^-(aq)$ 2. Now, write the $K_b$ expression: - The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants: $K_b = \frac{[Products]}{[Reactants]}$ $K_b = \frac{[OH^-][CH_3N{H_3}^+]}{[CH_3NH_2]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.
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