(b) Explanation The equilibrium reaction of glycine is, + H 3 NCH 2 COO − ( a q ) + H 2 O ( l ) ⇌ H 2 NCH 2 COO − ( a q ) + H 3 O + ( a q ) The change in the concentration of + H 3 NCH 2 COO − is assumed to be x . The ICE table is formed for the given reaction. + H 3 NCH 2 COO − ( a q ) ⇌ H 2 NCH 2 COO − ( a q ) + H 3 O + ( a q ) Initial : 0.050 0 Change : − x + x + x Equilibrium: 0.050 − x x x The liquid component does not affect the value of equilibrium constant. The equilibrium concentration of + H 3 NCH 2 COO − is ( 0.050 − x ) M . The equilibrium concentration of H 2 NCH 2 COO − is x M . The equilibrium concentration of H 3 O + is x M . At equilibrium, the equilibrium constant expression is expressed by the formula, K a = Concentration of products Concentration of reactants Where, K a is acid dissociation constant (c) Interpretation Introduction To determine: The predominant form of glycine in a solution with pH 13 .with pH 1 .

Pearson eText Chemistry: The Central Science -- Access Card

14th Edition

ISBN: 9780136848981

Author: Brown, Theodore, Lemay, H., Bursten, Bruce, Murphy, Catherine, Woodward, Patrick, Stoltzfus, Matthew

Publisher: PEARSON

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Chapter 16, Problem 114AE

(b)

Interpretation Introduction

To determine: The pH of a 0.050 M aqueous solution of glycine.

(c)

Interpretation Introduction

To determine: The predominant form of glycine in a solution with pH 13 .with pH 1 .

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Chapter 16, Problem 113AE

Chapter 16, Problem 115AE

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Pearson eText Chemistry: The Central Science -- Access Card

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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