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: 16.60b

Answer

pH = 10.03

Work Step by Step

1. Drawing the ICE table, we get: -$[OH^-] = [B^+] = x$ -$[Base] = [Base]_{initial} - x$ For approximation, we consider: $[Base] = [Base]_{initial}$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][B^+]}{ [Base]}$ $Kb = 1.3 \times 10^{- 6}= \frac{x * x}{ 8.5 \times 10^{- 3}}$ $Kb = 1.3 \times 10^{- 6}= \frac{x^2}{ 8.5 \times 10^{- 3}}$ $ 1.1 \times 10^{- 8} = x^2$ $x = 1.05 \times 10^{- 4}$ 5% test: $\frac{ 1.05 \times 10^{- 4}}{ 8.5 \times 10^{- 3}} \times 100\% = 1.23\%$ %ionization < 5% : Right approximation. $[OH^-] = x = 1.05 \times 10^{-4}$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 1.05 \times 10^{- 4})$ $pOH = 3.97$ $pH + pOH = 14$ $pH + 3.97 = 14$ $pH = 10.03$
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