Question

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of NaOH.

Answer 1

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.