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Title: Calculate the acetate ion and acetic acid concentration at equilibrium, as well as the dissociation constant.? Post by: liz.cancio on Sep 7, 2012 You have 10 mL of 1.0 M acetic acid in a 50 mL beaker. The observed pH is 2.51. What are the concentrations of acetate and acetic acid? Also, what is the dissociation constant, Ka?
Thank you both so much! Title: Calculate the acetate ion and acetic acid concentration at equilibrium, as well as the dissociation constant.? Post by: lewywong on Sep 7, 2012 The equilibrium is:
HAc <---> H+ + Ac- (Ac- = CH3COO-, HAc = CH3COOH) The expression for Ka is: Ka = [H+][Ac-]/[HAc] Since pH = 2.51, [H+] = 10^-2.51 = 3.09 X 10^-3 M. Since you have just a solution of acetic acid, [H+] must equal [Ac-]. So, [Ac-] = 3.09 X 10^-3 M. The concentration of HAc = 1.0 - 3.09 X 10^-3 = 0.997 So, Ka = (3.09X10^-3)^2 / 0.997 = 9.58 X 10^-6 Title: Calculate the acetate ion and acetic acid concentration at equilibrium, as well as the dissociation constant.? Post by: sodaff on Sep 7, 2012 If pH = 2.51
[H+] = 10^-pH [H+] = 10^-2.51 [H+] = 3.09*10^-3M The [CH3COO-] ion is the same as [H+] Because the dissociation is small, it is common to take the [acid] as equal to the molarity of the solution, But for accuracy [CH3COOH] = 1.0 - 3.09*10^-3 = 0.997M Solve for Ka dissociation constant: Ka = [H+] [CH3COO-] / [CH3COOH] Ka = (3.09*10^-3)² / 0.997 Ka = 9.57*10^-6 The accepted value for Ka CH3COOH is 1.8*10^-5 |