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 4 O 6 ( a q ) + H 2 O ( l ) → H C 4 O 6 − ( 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 A ] = 0.50 − x [ H A − ] = 0 + x [ H 3 O + ] = 0 + x Substitute in K data, solve for x K a 1 = [ H A − ] [ H 3 O + ] [ H 2 A ] = ( x ) ( x ) ( 0.50 − x ) 10 − 2.89 = x 2 0.5 − x x = 0 .02474 M Step 4: Get equilibrium concentrations [ H 2 A ] = 0.50 − x = 0.5 − 0.025 = 0.475 M [ H A − ] = 0 + x = 0.025 M [ H 3 O + ] = 0 + x = 0.025 M Step 5: Second dissociation H C 4 O 6 − ( a q ) + H 2 O ( l ) → C 4 O 6 2 − ( a q ) + H 3 O + ( a q ) Step 6: Get K expression for second dissociation K a 2 = [ A 2 − ] [ H 3 O + ] [ H A − ] Step 7: Get the equilibrium concentrations initially [ H 2 A ] = 0

7th Edition

ISBN: 9780321943170

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

Publisher: PEARSON

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Chapter 15, Problem 15.99SP

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