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.
|moles before reaction||0.00250||0.00250||0|
|change in moles||-0.00250||-0.00250||+0.00250|
|moles after reaction||0||0||0.00250|
|molarity after reaction||0||0||0.0500|
|C2H3O2-(aq)||+ H2O(l) =||HC2H3O2(aq)||+||OH-(aq)|
|molarity before equilibrium||0.050||0||~0|
|change in molarity||- x||+ x||+x|
|molarity at equilibrium||0.050 - x||x||x|
The equilibrium constant expression is
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 firstname.lastname@example.org
Copyright © 1997-2010 by Fred Senese
Comments & questions to email@example.com
Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/print-equivalence-point-HOAc-NaOH.shtml