Interpretation Introduction To determine: The given compounds are the neutral base or in the protonated form. Explanation Given: The pH of the stomach is 2.5 The value of equilibrium constant for nicotine is 7 × 10 − 7 . The value of equilibrium constant for caffeine is 4 × 10 − 14 . The value of equilibrium constant for strychnine is 1 × 10 − 6 . The value of equilibrium constant for quinine is 1.1 × 10 − 6 Formula used: The relationship between pOH and pH is, pOH + pH = 14 Calculation: Rearrange the equation to calculate the value of pOH , pOH = 14 − pH Substitute the value of pH in the above equation, pOH = 14 − 2 .5 =11 .5 [ OH − ] =10 − 11.5 = 3.16 × 10 − 12 M The general ionization reaction of each base is, B ( a q ) + H 2 O ( l ) ⇌ BH + ( a q ) + OH − ( a q ) The equilibrium constant expression for the above reaction is, K b = [ BH + ] [ OH − ] [ B ] Where, K b is the base ionization constant. [ BH + ] represents the protonated form of a base. [ B ] represents the base. Rearrange the above equation in the form of ratio of [ BH + ] [ B ] [ BH + ] [ B ] = K b [ OH − ] For nicotine, [ BH + ] [ B ] = 7 × 10 − 7 3

Chemistry: The Central Science (14th Edition)

14th Edition

ISBN: 9780134414232

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 16, Problem 113AE

Interpretation Introduction

To determine: The given compounds are the neutral base or in the protonated form.

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Chapter 16, Problem 112AE

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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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