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 717: 16.28b


$K_w = [H_3O^+][OH^-]$

Work Step by Step

$K_w = \frac{[Products]}{[Reactants]}$ $H_2O(l) H_3O^+(aq) + OH^-(aq)$ $K_w = \frac{[H_3O^+][OH^-]}{[H_2O]}$ Since water is a pure liquid, we do not consider it in the constant. $K_w = [H_3O^+][OH^-]$
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