How is the the pK_a of acetic acid found in this experiment?

How is the dissociation constant of acetic acid found in this experiment?

1. Pipette 25.0 cm³ of 0.10 M acetic acid into a conical flask.
2. Titrate with 0.10 M sodium hydroxide solution, using phenolphthalein as indicator, until the solution is just pink.
3. Add a further 25.0 cm³ of the same acetic solution to the flask and mix thoroughly.
4. Determine the pH of the resulting solution.

Sancho Chiang

1. How this experiment works.The basis for the experiment is the Henderson-Hasselbalch equation:

   pH = pKa + log

   [CH3COO-] [CH3COOH]

   where the pH is minus the log of the hydrogen ion molarity, pK_a is minus the log of the acid dissociation constant, and [CH₃COO^-] and [CH₃COOH] are the equilibrium concentrations of acetate ion and acetic acid, respectively.

   1. 25.0 cm³ of 0.10 M acetic acid means that you have 0.00250 moles of acetic acid in the flask.
   2. Titrating to the phenolphthalein endpoint with NaOH converts this to 0.00250 moles of acetate. A little bit of the acetate will react with water to give you back acetic acid, but neglect this for now.
   3. Adding a further 25.0 cm³ of 0.10 M acetic acid will give you about 0.0025 moles of acetic acid. A little bit of the acetic acid will dissociate and give you more acetate, but neglect this for now.
   4. Now you know (approximately) what the fraction after the log is (0.00250/0.00250 = 1.00, right? I can get the fraction without converting moles to molarity, because the acetate and acetic acid are in the same volume of solution.) If I measure the pH at this point, the Henderson Hasselbalch equation can be solved for pK_a:
      pK_a = pH - log 1.0

      [CH₃COOH] = [CH₃COO^-]. The log term vanishes (since the log of one is zero) so pK_a = pH.

      Finally, if you want the K_a, it is of course just 10^-pK_a.

2. Why simple analytical concentrations can be used instead of equilibrium concentrations in the Henderson-Hasselbalch equation.

   We've made an approximation in the calculation above. We kept things simple by neglecting the fact that some of the acetate will react with water to give us acetic acid, and some of the acetic acid will dissociate to give back acetate. Consider the dissociation of acetic acid into acetate and hydrogen ions:

   CH₃COOH = H⁺ + CH₃COO^- All of that acetate you've produced in the titration will suppress this dissociation. Consider the hydrolysis of acetate to form acetic acid and hydroxide ions: CH₃COO^- + H₂O = CH₃COOH + OH^- All of the acetic acid you added after the titration will suppress this reaction. Thus, changes due to BOTH of these reactions will be small compared to the total acetate and acetic acid concentrations themselves. That allowed us to use analytical and not equilibrium concentrations in the calculation of pK_a above.

   Still not convinced? The approximation is easy to justify. We'll need to do a little bit of algebra, though.

   Let the concentrations of CH₃COOH and CH₃COO^- before equilibration be C_CH3COOH and C_CH3COO^-, and let the concentration after equilibration be [CH₃COOH] and [CH₃COO^-]. We want to show that

   C_CH3COOH ~ [CH₃COOH]
   C_CH3COO^- ~ [CH₃COO^-]
   

   We need to look for relationships that hold both before and after equilibration to relate these concentrations. Two things must be unchanged by the dissociation and hydrolysis reactions: the total moles of acetate in all its forms, and the total charge on the solution (which must be zero).

   The total concentration of acetate in all its forms has to be the same before and after equilibration, if acetate isn't created or destroyed in the process. In your experiment, you have only 0.0025 + 0.0020 = 0.0045 moles of acetic acid and acetate to work with. Hydrolysis and dissociation will change the form of the acetate, but NOT the total amount of acetate. This statement can be written as a mass balance equation:

   C_CH3COOH + C_CH3COO^- = [CH₃COOH] + [CH₃COO^-]

   The solution must be electrically neutral; that is, the total molarity of positive charges has to be equal to the total molarity of negative charges. Hydrogen and sodium ions contribute to the positive charge, in this experiment; hydroxide and acetate contribute to the negative charge. This gives us a charge balance equation:

   [H⁺] + [Na⁺] = [OH^-] + [CH₃COO^-] The sodium ions come from the NaOH in the titration. Because one mole of acetate was produced for every mole of NaOH added, [Na⁺] = C_CH3COO^-.

   Combining the charge balance and mass balance equations gives us the relationships we're looking for:

   [CH₃COOH] = C_CH3COOH - [H⁺] + [OH^-]
   [CH₃COO^-] = C_CH3COO^- + [H⁺] - [OH^-]
   

   These equations make sense. The first one says: The equilibrium concentration of acetic acid is the analytical concentration minus the amount that dissociates (equal to [H⁺], plus the amount produced by hydrolysis of acetate (equal to [OH^-].

   The second one says: The equilibrium concentration of acetate is the analytical concentration plus the amount produced by dissociation of acetic acid (equal to [H⁺], minus the amount converted by hydrolysis back to acetic acid (equal to [OH^-].

   Both equations say that the difference between the analytical and equilibrium acetic acid concentrations is equal to the difference between the hydrogen ion and hydroxide ion concentrations. If the difference is small relative to the analytical concentrations, the analytical and equilibrium concentrations are equal, and we've justified using analytical concentrations when we really should have used equilibrium concentrations in the Henderson-Hasselbalch equation.

   Sometimes, the assumption is NOT justified. That will happen when the acid is really dissociates easily (K_a larger than about 10^-3, or when the concentration of acid or its conjugate base are very small. That tells us that your experimental determination of K_a will fail under those conditions!

   Author: Fred Senese senese@antoine.frostburg.edu

Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/pKa-experiment-HOAc.shtml