Not sure if this is correct, but is the answer pH = 4.921

Question

A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

Answer 1

4.91 is not the answer I get. Here is what I did.

mols acetic acid = 0.05 L x 0.150 mol/L = 0.0075 mols

mols NaOH added = 0.02 L x 0.150 mol/L = 0.003 mols

CH3COOH + NaOH ==> CH3COONa + H2O

0.0075.........0.003...............0...........Initial

-0,003,.........-0.003............+0.003....Change

0.0045..............0.................0.003....Equilibrium

Final volume = 50 ml + 20 ml = 70 ml = 0.070 L

Final concentrations:

[CH3COOH] = 0.0045 mol/0.070 L = 0.0643 M

[CH3COO-] = 0.003 mol/0.070 L = 0.0429 M

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

pKa = - log Ka = 4.74

pH = 4.74 + log (0.0429/0.0643)

pH = 4.74 - 0.176

pH = 4.56