no of moles of base B = molarity * volume in L
                                  Â
= 0.16*0.52 = 0.0832 moles
no of moles of conjugate base = molarity * volume in L
                                                Â
= 0.27*0.52 = 0.1404 moles
     POH = 14-PH
            Â
= 14-8.56 = 5.44
      POH = PKb +
log[B]/[BH+]
       5.44
   = Pkb + log0.1404/0.0832
     Â
5.44Â Â Â Â Â = Pkb +0.2272
     Pkb    Â
= 5.44-0.2272
                 Â
= 5.2128
After addition of 0.02moles of Ba(OH)2
no of moles of base BÂ Â = 0.0832+0.02 =
0.1032moles
no of moles of conjugate acid BH+ = 0.1404-0.02 =
0.1204 moles
  PoH  = Pkb + log[BH+]/[B]
           Â
= 5.2128+ log0.1204/0.1032
            Â
= 5.2128+0.06694Â Â = 5.2797 >>>>answer
