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 - Problems by Topic - Page 747: 93

Answer

$Kb (Morphine) = 6.7\times 10^{- 7}$

Work Step by Step

1. We have these concentrations at the equilibrium: -$[OH^-] = [Conj. Acid] = 0 + x = x$ -$[Morphine] = [Morphine]_{initial} - x$ 2. Calculate the [OH^-] pH + pOH = 14 10.5 + pOH = 14 pOH = 3.5 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 3.5}$ $[OH^-] = 3.2 \times 10^{- 4}$ Therefore : $x = 3.2 \times 10^{- 4}$ 3. Write the Kb equation, and find its value: $Kb = \frac{[OH^-][Conj. Acid]}{ [Morphine]}$ $Kb = \frac{x^2}{[Initial Morphine] - x}$ $Kb = \frac{( 3.2 \times 10^{- 4})^2}{ 0.15- 3.2\times 10^{- 4}}$ $Kb = \frac{ 1.0 \times 10^{- 7}}{ 0.15}$ $Kb = 6.7\times 10^{- 7}$
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