Question

Assuming that K_a = 1.85×10^-5 for acetic acid, calculate the pH at the equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH.

Answer 1

To reach equivalence point, volume of NaOH required = 50 mL = 0.050 L

Given:
M(CH3COOH) = 0.10 M
V(CH3COOH) = 0.050 L
M(NaOH) = 0.10 M
V(NaOH) = 0.050 L

mol(CH3COOH) = 0.005 mol
mol(NaOH) = 0.00500 mol

0.005 mol of both will react to form CH3COO- and H2O
CH3COO- here is strong base
CH3COO- formed = 0.005 mol
Volume of Solution = 0.050 + 0.050 = 0.100 L
Kb of CH3COO- = Kw/Ka = 1.0E-14/1.85E-5 = 5.41E-10
concentration ofCH3COO-,c = 0.005/0.100 = 0.050 M

for simplicity lets write weak base as A-

A-        + H2O   ----->     AH   +   OH-
0.050                        0         0
0.050-x                      x         x

Kb = [AH][OH-]/[B-]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.405405405405406E-10)*0.050) = 5.20E-6

[OH-] = x = 5.20E-6M

pOH = -log [OH-] = -log (5.20E-6) = 5.28

PH = 14 - pOH = 14 - 5.28 = 8.72
PH = 8.72