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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 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

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?