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.
HC2H3O2(aq) | + | NaOH(aq) | = | C2H3O2-(aq) | + H2O(l) | |
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.
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/print-equivalence-point-HOAc-NaOH.shtml