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 - Review Questions - Page 745: 23

Answer

$K_a \times K_b = K_w$

Work Step by Step

- Let's create a imaginary acid, and its conjugate base: $HA$ and $A^-$. - The $K_a$ expression for $HA$ will be: $K_a = \frac{[H_3O^+][A^-]}{[HA]}$ - The $K_b$ expression for $A^-$ will be: $K_b = \frac{[OH^-][HA]}{[A^-]}$ - So, if we multiply $K_a$ and $K_b$: $K_a * K_b$ $\frac{[H_3O^+][A^-]}{[HA]} * \frac{[OH^-][HA]}{[A^-]}$ - We can cancel $[H_3O^+]$ and $[OH^-]$ using algebra. $[H_3O^+] * [OH^-]$, which is equal to $K_w$ - This is the expression for $K_w$ - Therefore: $K_a * K_b = K_w$
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