Pka = -logka
          =
-log3.5*10-8
        Â
=Â Â 7.4559
no of moles of HOCl = molarity * volume in L
                                  Â
= 0.5*0.1 = 0.05 moles
no of moles of NaOCl = 0.4*0.1 = 0.04 moles
                        Â
PHÂ Â = PKa + log[NaOCl]/[HOCl]
                                Â
= 7.4559 + log0.04/0.05
                                Â
= 7.4559-0.0969 = 7.359
By the addition of NaOH
no of moles of NaOH = 1*0.01 = 0.01 moles
no of moles of HOCl  = 0.05-0.01 = 0.04 moles
no of moles of NaOCl  = 0.04+0.01 = 0.05 moles
  PH = Pka + log[NaOCl]/[HOCl]
        = 7.4559 +
log0.05/0.04
      = 7.4559 + 0.09691 =
7.5528
