a) Explanation Given The given pH is 9 .95 . The value of pOH can be calculated by the following formula, pH + pOH = 14 pOH = 14 − pH Substitute the value of pH in the above reaction, pOH = 14 − 9 .95 pOH = 4 .05 The value of pOH is calculated by the formula, pOH = − log [ OH − ] Rearrange the above expression to calculate the value of [ OH − ] , pOH = − log [ OH − ] [ OH − ] = 10 − pOH Substitute the value of pOH in the above expression. [ OH − ] = 10 − pOH [ OH − ] = 10 − 4 .05 [ OH − ] = 8 .91 × 10 − 5 M The dominant equilibrium reaction that takes place in the given case is, C 18 H 21 O 3 N ( a q ) + H 2 O ( l ) ⇌ C 18 H 21 NO 3 H + ( a q ) + OH − The change in the concentration of C 18 H 21 O 3 N is assumed to be x . The ICE table is formed for the given reaction. C 18 H 21 NO 3 ( a q ) + H 2 O ( l ) ⇌ C 18 H 21 NO 3 H + ( a q ) + OH − ( a q ) Initial : 5.0 × 10 − 3 0 0 Change : − 8.91 × 10 − 5 + 8

Chemistry: The Central Science (14th Edition)

14th Edition

ISBN: 9780134554563

Author: Brown

Publisher: PEARSON

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Chapter 16, Problem 74E

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To determine: The equilibrium constant Kb and pKb for Codeine C18H21NO3 .

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Chapter 16, Problem 73E

Chapter 16, Problem 75E

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Chapter 16 Solutions

Chemistry: The Central Science (14th Edition)

Ch. 16.2 - Practice Exercise 1 Consider the following...Ch. 16.2 - Prob. 16.1.2PE Ch. 16.2 - Prob. 16.2.1PE Ch. 16.2 - Practice Exercise 2 When lithium oxide (Li2O) is...Ch. 16.2 - Based on information in Figure 16.4, place the...Ch. 16.2 - Practice Exercise 2 For each reaction, use Figure...Ch. 16.3 - Practice Exercise 1 In a certain acidic solution...Ch. 16.3 - Practice Exercise 2 Indicate whether solutions...Ch. 16.3 - Prob. 16.5.1PE Ch. 16.3 - Prob. 16.5.2PE

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The equilibrium constant K b and p K b for Codeine C 18 H 21 NO 3 .

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To determine: The equilibrium constant Kb and pKb for Codeine C18H21NO3 .

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Chapter 16 Solutions