Question

A 83.0 mL sample of 0.0500 M HNO₃ is titrated with 0.100 M KOH solution. Calculate the pH after the following volumes of base have been added.

(a) 11.2 mL pH = _________	(b) 39.8 mL pH = __________	(c) 41.5 mL pH = ___________
(d) 41.9 mL pH = ____________	(e) 79.3 mL pH = _____________

Answer 1

a.

    HNO3                                             KOH

MA =0.05                                                MB = 0.1

VA   = 83ml                                              VB   = 11.2

       M    =   MAVA-MBVB/VA+VB

               = 0.05*83-.1*11.2/83+11.2   = 3.03/94.2 = 0.032M

    M = [H+] = 0.032M

   PH = -log[H+]

        = -log0.032   = 1.4948

b.

HNO3                                             KOH

MA =0.05                                                MB = 0.1

VA   = 83ml                                              VB   = 39.8ml

       M    =   MAVA-MBVB/VA+VB

                = 0.05*83-0.1*39.8/83+39.8   = 0.17/122.8   = 0.00138M

     M      =   [H+] = 0.00138M

    PH   = -log[H+]

             = -log0.00138 = 2.86

c.

HNO3                                             KOH

MA =0.05                                                MB = 0.1

VA   = 83ml                                              VB   = 41.5ml

       M    =   MAVA-MBVB/VA+VB

              = 0.05*83-0.1*41.5/83+41.5   = 0

no of moles of HNO3 = no of moles of KOH

      PH   = 7

d.

HNO3                                             KOH

MA =0.05                                                MB = 0.1

VA   = 83ml                                              VB   = 41.9ml

       M    =   MBVB-MAVA/VA+VB

               = 0.1*41.9-0.05*83/83+41.9   = 0.04/124.9   = 0.00032M

        M = [OH-]   = 0.00032M

      POH   = -log0.00032   = 3.4948

      PH   = 14-POH

               = 14-3.4948   = 10.5052