Question

In: Chemistry

Given 100.0 ml of a buffer that is 0.50 M in HOCl and 0.40 M in...

Given 100.0 ml of a buffer that is 0.50 M in HOCl and 0.40 M in NaOCl, what is the pH after 10.0 ml of 1.0 M NaOH has been added?

(Ka for HOCl =3.5 x 10-8)

Solutions

Expert Solution

   Pka = -logka

           = -log3.5*10-8

          =   7.4559

no of moles of HOCl = molarity * volume in L

                                    = 0.5*0.1 = 0.05 moles

no of moles of NaOCl = 0.4*0.1 = 0.04 moles

                          PH   = PKa + log[NaOCl]/[HOCl]

                                  = 7.4559 + log0.04/0.05

                                  = 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

queen_honey_blossom answered 2 years ago
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