Explanation Always start by defining the equilibrium conditions. Start with the first dissociation, calculate all data, then continue with the second proton. In the end, relate all species.. Step 1: Hydrolysis of the Acid H 2 C 2 O 4 ( a q ) + H 2 O ( l ) → H C 2 O 4 − ( a q ) + H 3 O + ( a q ) Step 2: Get the equilibrium expression for this reaction K a 1 = [ H C 2 O 4 − ] [ H 3 O + ] [ H 2 C 2 O 4 ] Step 3: Substitute equilibrium conditions in constant expression In equilibrium, assuming “x” is the dissociated amount of the first protonation: [ H 2 C 2 O 4 ] = 0.20 − x [ H C 2 O 4 ] = 0 + x [ H 3 O + ] = 0 + x Substitute in K data, solve for x K a 1 = [ H C 2 O 4 − ] [ H 3 O + ] [ H 2 C 2 O 4 ] = ( x ) ( x ) ( 0.20 − x ) 5.9 x 10 − 2 = x 2 0.2 − x x = 0.083 M Step 4: Get equilibrium concentrations [ H 2 C 2 O 4 ] = 0.20 − x = 0.2 − 0.083 = 0.117 M [ H C 2 O 4 − ] = 0 + x = 0.083 M [ H 3 O + ] = 0 + x = 0.083 M Step 5: Second dissociation H C 2 O 4 − ( a q ) + H 2 O ( l ) → C 2 O 4 2 − ( a q ) + H 3 O + ( a q ) Step 6: Get K expression for second dissociation K a 2 = [ C 2 O 4 2 − ] [ H 3 O + ] [ H C 2 O 4 − ] Step 7: Get the equilibrium concentrations initially [ H 2 C 2 O 4 ] = 0

7th Edition

ISBN: 9780321943170

Author: John E. McMurry, Robert C. Fay, Jill Kirsten Robinson

Publisher: PEARSON

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Chapter 15, Problem 15.98SP

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Chapter 15 Solutions

Chemistry (7th Edition)

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2A Ch. 15 - Conceptual PRACTICE 15.3 For the following...Ch. 15 - Conceptual APPLY 15.4 For the following reactions...Ch. 15 - Prob. 15.5P Ch. 15 - Conceptual APPLY 15.6 The following pictures...Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8A Ch. 15 - PRACTICE 15.9 The concentration of H3O+ ions in...Ch. 15 - Prob. 15.10A

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To calculate: The pH and concentrations

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To calculate:

The pH and concentrations