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What is the ph of a solution made from 500 mL of 1.0 M HC2H3O2 and...

What is the ph of a solution made from 500 mL of 1.0 M HC2H3O2 and 1000 mL of .5 M C2H3O- after 80 mL 0.9 M NaOh has been added?

Solutions

Expert Solution

no of moles of HC2H3O2   = molarity * volume in L

                                           = 1*0.5 = 0.5moles

no of moles of C2H3O2^-    = molarity * volume in L

                                             = 0.5*1 = 0.5 moles

PKa of HC2H3O2    = 4.75

PH   = PKa + log[C2H3O2^-]/[HC2H3O2]

      = 4.75+ log0.5/0.5

         = 4.75+ log1

         = 4.75 + 0 = 4.75

By the addition of NaOH

no of moles of NaoH = molarity * volume in L

                                    = 0.9*0.08   = 0.072 moles of NaOH

no of moles of HC2H3O2 by the addition of NaOH = 0.5-0.072   = 0.428 moles

no of moles of C2H3O2^- by the addition of NaOH = 0.5+0.072    = 0.572 moles

    PH   = PKa + log[C2H3O2^-]/[HC2H3O2]

            = 4.75 + log0.572/0.428

            = 4.75+ 0.126   = 4.876 >>>>answer

      


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