Question

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. (25 points)

Answer 1

no of mole ofCH3COOH taken = 25*0.1 = 2.5 mmole

a) initial

   pH = 1/2(pka-logC)

   pka of cH3COOH = 4.76

    C = concentration of CH3COOH = 0.1 M

   pH = 1/2(4.76-log0.1)

      = 2.88

b) after 5 ml KOH added

   no of mole ofCH3COOH taken = 25*0.1 = 2.5 mmole

   no of mole of KOH added = 5*0.2= 1 mmole

pH = pka +log(base/acid)

     = 4.76+log(1/(2.5-1))

     = 4.584

c) after 10 ml KOH added

   no of mole ofCH3COOH taken = 25*0.1 = 2.5 mmole

   no of mole of KOH added = 10*0.2= 2 mmole

pH = pka +log(base/acid)

     = 4.76+log(2/(2.5-2))

     = 5.36

   d) after 12.5 ml KOH added

   no of mole ofCH3COOH taken = 25*0.1 = 2.5 mmole

   no of mole of KOH added = 12.5*0.2= 2.5 mmole

   EQUIVALENCE POINT

   pH = 7+1/2(pka+logC)

    C = concentration of salt = n/V = 2.5/(37.5) = 0.067 M

     pka of CH3COOH = 4.76

    pH = 7+1/2(4.76+log0.067)

        = 8.8

   e) after 15 ml KOH added

   no of mole ofCH3COOH taken = 25*0.1 = 2.5 mmole

   no of mole of KOH added = 15*0.2= 3 mmole

   concentration of excess KOH = (3-2.5)/40 = 0.0125 M

   pH = 14 - (-log(OH-))

      = 14 - (-log0.0125)

       = 12.1