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How do I compute the pH at the equivalence point in the titration of acetic acid with NaOH?


Let's consider a specific example. Suppose we're titrating 25.0 mL of 0.1000 M HC2H3O2(aq) with 0.1000 M NaOH(aq). The equivalence point will occur at 25.0 mL added NaOH(aq) solution. Picture this as a two step process. The first step involves stoichiometric reaction between the NaOH and the acetic acid; the second step involves hydrolysis of the acetate formed by the first step.

  1. The NaOH(aq) completely reacts with the HC2H3O2(aq). Mixing the two solutions causes the following reaction to occur:
    HC2H3O2(aq)+NaOH(aq)=C2H3O2-(aq) + H2O(l)
    moles before reaction0.002500.002500
    change in moles-0.00250-0.00250+0.00250
    moles after reaction000.00250
    molarity after reaction000.0500

  2. The C2H3O2-(aq) hydrolyzes.
    C2H3O2-(aq)+ H2O(l) =HC2H3O2(aq)+ OH-(aq)
    molarity before equilibrium0.0500~0
    change in molarity- x+ x+x
    molarity at equilibrium0.050 - xxx

The equilibrium constant expression is

Kb = [OH-(aq)][HC2H3O2(aq)]/[C2H3O2-(aq)] = x2/(0.050 - x)

and solving this for x with Kb = Kw/Ka = 1.0x10-14/1.8x10-5 = 5.6x10-10 gives x = [OH-(aq)] = 5.27x10-6. Then [H3O+(aq)] = Kw/[OH-(aq)] = 1.0x10-14/5.27x10-6 = 1.9x10-9, and pH ~ -log [H3O+(aq)] = 8.72.

Author: Fred Senese senese@antoine.frostburg.edu



General Chemistry Online! How do I compute the pH at the equivalence point in the titration of acetic acid with NaOH?

Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/equivalence-point-HOAc-NaOH.shtml