Determine the equilibrium constant for the following reaction, CH3COOH(aq) +F(aq) CH3COO (aq)+ HF(aq) given Ka (CH3COOH) = 1.8 x 10 and Ka (HF) = 6.8 x 104. -5

Chemistry for Engineering Students
4th Edition
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Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.78PAE
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Determine the equilibrium constant for the following reaction,
CH3COOH(aq) + F (aq) CH3COO (aq) + HF(aq)
given Ka (CH3COOH) = 1.8 x 10P and Ka (HF) = 6.8 x 104.
O a. 8.2 x 107
O b.12x 108
OC70x 104
O d. 3.8 x 10
O e. 26 x 10-2
Transcribed Image Text:Determine the equilibrium constant for the following reaction, CH3COOH(aq) + F (aq) CH3COO (aq) + HF(aq) given Ka (CH3COOH) = 1.8 x 10P and Ka (HF) = 6.8 x 104. O a. 8.2 x 107 O b.12x 108 OC70x 104 O d. 3.8 x 10 O e. 26 x 10-2
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