Question

The Ka for C₂H₅COOH (propanoic acid) is 1.30 x 10^-5 the reaction below

C₂H₅COOH ÜÞ H⁺ + C₂H₅COO^-

Determine the ph of a buffer system that has 2.5 M C₂H₅COOH and 2.5 M C₂H₅COONa in one liter of solution. Determine the ph after 25ml of 3M NaOH has been added and give the ratio of base/acid for a ph of 5.20

Please step by step work so I can understand. Thank you

Answer 1

no of moles of C₂H₅COOH   = molarity * volume in L

                                                = 2.5*1    = 2.5 moles

no of moles of C₂H₅COONa   = molarity * volume in L

                                                = 2.5*1    = 2.5 moles

no of moles of NaOH              = molarity * volume in L

                                                = 3*0.025   = 0.075 moles

no of moles of C₂H₅COOH after the addition of 0.075 moles of NaOH = 2.5 -0.075   = 2.425 moles

no of moles of C₂H₅COONa after the addition of 0.075 moles of NaOH = 2.5+0.075   = 2.575 moles

        Ka   = 1.3*10^-5

               PKa = -logKa

                         = -log1.3*10^-5

                          = 4.886

         PH    = Pka + log[ C₂H₅COONa]/[ C₂H₅COOH]

                = 4.886 + log2.575/2.425

                  = 4.886 + 0.026

                   = 4.912

     2.

          PH    = Pka + log[ C₂H₅COONa]/[ C₂H₅COOH]

             5.2 = 4.886 + log[ C₂H₅COONa]/[ C₂H₅COOH]

log[ C₂H₅COONa]/[ C₂H₅COOH]   = 5.2-4.886

log[ C₂H₅COONa]/[ C₂H₅COOH]   = 0.314

   [ C₂H₅COONa]/[ C₂H₅COOH]     = 10^0.314

                                                          = 2.0606

   [ C₂H₅COONa]/[ C₂H₅COOH]   = 2.0606/1 >>>>>answer