Question

an aqueous solution contains 0.361 M acetic acid (HOAc). If you add 83.0mL of 0.278 M potassium hydroxide to 150 mL of this solution, what is the final pH? (Ka of HOAc = 1.8*10-5

Answer 1

The number of mmoles of acetic acid = molarity* volume in mL = 0.361*150 = 54.15 mmoles

The number of mmoles of KOH added          = 0.278*83 = 23.074 mmoles

The reaction between and CH3COOH and KOH is
         
               CH3COOH         +      KOH         <----------------->    CH3COOK      + H2O

initial       54.15 mmoles         23.074 mmoles                           -              -

After the      31.076mmoles         -                                    23.074mmoles
reaction

Now the mixture forms acidic buffer solution and its pH is given by

   pH = pKa + log(conjugate base /acid) = pKa + log(CH3COOK/CH3COOH)

       = 4.74+ log(23.074/31.076) = 4.61

pH of the buffer solution = 4.61