Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575e: 102

Answer

The percentage of pyridine that forms pyridium ion is $0.01304 \%$.

Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [C_5H_5NH^+] = x$ -$[C_5H_5N] = [C_5H_5N]_{initial} - x = 0.1 - x$ For approximation, we consider: $[C_5H_5N] = 0.1M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$ $Ka = 1.7 \times 10^{- 9}= \frac{x * x}{ 0.1}$ $Ka = 1.7 \times 10^{- 9}= \frac{x^2}{ 0.1}$ $ 1.7 \times 10^{- 10} = x^2$ $x = 1.304 \times 10^{- 5}$ Percent ionization: $\frac{ 1.304 \times 10^{- 5}}{ 0.1} \times 100\% = 0.01304\%$ %ionization < 5% : Right approximation.
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