Explanation The dissociation reaction of tartaric acid ( H 2 C 4 H 4 O 6 ) is, H 2 C 4 H 4 O 6 ( a q ) + H 2 O ( l ) ⇌ 2H + ( a q ) + C 4 H 4 O 6 2 − ( a q ) The change in the concentration of H 2 C 4 H 4 O 6 is assumed to be x . The ICE table is formed for the given reaction. H 2 C 4 H 4 O 6 ( a q ) + H 2 O ( l ) ⇌ C 4 H 4 O 6 2 − ( a q ) + 2H + ( a q ) Initial : 0.250 0 0 Change : − x + x + 2 x Equilibrium: 0.250 − x x 2 x The liquid component does not affect the value of equilibrium constant. The equilibrium concentration of H 2 C 4 H 4 O 6 is ( 0.250 − x ) M . The equilibrium concentration of H 2 C 4 H 4 O 6 − is x M . The equilibrium concentration of H + is 2 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 the acid dissociation constant

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 66E

Interpretation Introduction

To determine: The pH and the concentration of tartrate ion for a 0.250 M solution of tartaric acid and if any assumptions or approximation had to be made to calculate the concentration of the tartrate ion.

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Chapter 16, Problem 65E

Chapter 16, Problem 67E

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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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