Question

Calculate the pH of a titration mixture of acetic acid with NaOH when 45.00 mL of NaOH has been added, considering that the equivalence point of the mixture is 22.98 mL.

Concentration of acetic acid = 0.1081mol/L

Concentration of NaOH = 0.1185 mol/L

Volume of Acetic Acid sample = 25.00 mL

DO NOT use henderson hasselbalch equation

Answer 1

NaOH + HCl ---------> NaCl + H2O

Molarity of Sodium hydroxide = 0.1185 mol /L

Volume of NaOH soln added = 45ml

No of mol of NaOH = (0.1185mol/1000ml)×45ml = 0.0053325

Molarity of acetic acid = 0.1081mol/L

Volume of acetic acid solution = 25ml

No of mole of Acetic acid = 0.0027025

0.0027025mole acetic acid react with 0.0027025mole of NaOH

Remaining mole of NaOH = 0.0053325 - 0.0027025 = 0.00263

Total volume = 25ml + 45ml = 70ml

[ NaOH ] =(0.00263mol/70ml)×1000ml = 0.03757M

[ OH-] = 0.03757M

pOH = 1.43

pH = 14 - pOH

= 14 - 1.43

= 12.57