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 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

2. The C₂H₃O₂^-(aq) hydrolyzes.

   	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

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

and solving this for x with K_b = K_w/K_a = 1.0x10^-14/1.8x10^-5 = 5.6x10^-10 gives x = [OH^-(aq)] = 5.27x10^-6. Then [H₃O⁺(aq)] = K_w/[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

Copyright © 1997-2001 by Fred Senese
Comments & questions to senese@antoine.frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101-hidden/acidbase/faq/equivalence-point-HOAc-NaOH.shtml