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liz.cancio liz.cancio
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11 years ago
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!
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#1
11 years ago
The equilibrium is:

HAc Leftwards Arrow-> 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
wrote...
#2
11 years ago
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
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