Chemistry: Atoms First (2nd Edition)

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

Chapter 13 - Exercises - Page 575e: 103

Answer

$Kb = 9.908\times 10^{- 10}$

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. -$[OH^-] = [CH_3C_6H_4N{H_3}^+] = x$ -$[CH_3C_6H_4NH_2] = [CH_3C_6H_4NH_2]_{initial} - x$ 2. Calculate the [OH^-] pH + pOH = 14 8.6 + pOH = 14 pOH = 5.4 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 5.4}$ $[OH^-] = 3.981 \times 10^{- 6}$ 3. Write the Kb equation, and find its value: $Kb = \frac{[OH^-][CH_3C_6H_4N{H_3}^+]}{ [CH_3C_6H_4NH_2]}$ $Kb = \frac{x^2}{[Initial CH_3C_6H_4NH_2] - x}$ $Kb = \frac{( 3.981\times 10^{- 6})^2}{ 0.016- 3.981\times 10^{- 6}}$ $Kb = \frac{ 1.585\times 10^{- 11}}{ 0.016}$ $Kb = 9.908\times 10^{- 10}$
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