What is the pH at the equivalence point

Question

Calculate the pH at the equivalence point during the titration of 25.0 mL of 0.75 M acetic acid with 0.75 M sodium hydroxide. (Ka for acetic acid is 1.8 * 10^-5)

Answer 1

Titration of a weak acid (acetic acid, HAc) with a strong base (NaOH) - creates a buffer and pH at equivalence will be greater than 7.

HAc + NaOH ==> NaAc + H₂O

At equivalence all HAc is converted to Ac^- which acts as the conjugate base and undergoes hydrolysis:

Ac^- + H₂O ==> HAc + OH^-

25 ml x 1 L/1000 ml x 0.75 mol/L = 0.01875 moles in a total volume of 50 ml (25 ml HAc + 25 ml NaOH)

[Ac^-] = 0.01875 mol/0.050 L = 0.375 M

Kb for Ac^- = Kw/Ka = 1x10^-14/1.8x10^-5 = 5.5x10^-10

5.5x10^-10 = [HAc][OH^-]/[Ac^-] = (x)(x)/0.375

x² = 2.06x10^-10

x = 1.44x10^-5 M = [OH^-]

pOH = -log [OH^-] = 4.84

pH = 14 - pOH

pH = 9.16 (greater than 7 as predicted)