Determine the pH when 250 mL of 015 M acetic acid HC2H3O2 is

Determine the pH when 25.0 mL of 0.15 M acetic acid (HC2H3O2) is titrated with 35.0 mL of 0.15 M KOH.

Solution

pka for acetic acid = 4.74,

acetic acid + KOH <----------> CH3COOK + H2O ,

after neutralisation we have excess KOH ,

excess OH- moles = ( 0.035x0.15-0.025x0.15) = 0.0015,

[OH-] = (0.0015)/(0.025+0.035) = 0.025 M,

pOH = -log(0.025) = 1.6,

pH = 14-1.6 = 12.4