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Calculate the pH of a titration with 25 mL of 0.1 M NH3 with 0.1 MHCl...

Calculate the pH of a titration with 25 mL of 0.1 M NH3 with 0.1 MHCl at the following volume points: 0, 10, 24.9, 25, 25.1, 40,50 mL

Solutions

Expert Solution

    no of moles of NH3 = molarity * volume in L

                                    = 0.1*0.025 = 0.0025 moles

      NH3 + H2O -------> NH4+ + OH-

I                   0.0025                       0              0

C            -x                             +x              +x

E           0.0025-x                      +x              +x

          Kb    = [NH4+][OH-]/[NH3]

        1.8*10-5    =   x*x/(0.0025-x)

       1.8*10-5 *(0.0025-x) = x2

             x    = 0.000203

         [OH-]   = x   = 0.000203M

        POH   = -log[OH-]

                  = -log0.000203   = 3.6925

        no of moles of HCl   = molarity * volume in L

                                          = 0.1* 0.01   = 0.001 moles

             NH3 + HCl ---------> NH4Cl

I           0.0025 0.001               0

C         -0.001    -0.001             0.001

E          0.0015      0                 0.001

      POH   = PKb + log[NH4Cl]/[NH3]

                = 4.75 + log0.001/0.0015

                 = 4.75-0.176   = 4.574

        PH    = 14-POH

                  = 14-4.574   = 9.426

        no of moles of HCl   = 0.1*0.0249 = 0.00249moles

   NH3 + HCl ---------> NH4Cl

I           0.0025    0.00249               0

C         -0.00249    -0.00249             0.00249

E          0.00001      0                  0.00249

      POH   = PKb + log[NH4Cl]/[NH3]

                = 4.75 + log0.00249/0.00001

                 = 4.75+2.3961 = 7.1461

    PH      = 14-POH

               = 14-7.1461 = 6.8539

   at 25ml

   NH4+   = 0.0025 moles

                 NH4+ + H2O ---------> NH3 + H3O+

I             0.0025                           0          0

C              -x                                 +x         +x

E             0.0025-x                        +x         +x

                  ka = [NH3][H3O+]/[NH4+]

                 5.6*10-10   = x*x/0.0025-x

                 5.6*10-10 *(0.0025-x) = x2

                    x   = 1.2*10-6

                  [H3O+] = x = 1.2*10-6 M

                PH   = -log[H3O+]

                        = -log1.2*10-6

                       = 5.92

     at 25.1 ml

                 [H3O+]   = no of moles of HCl - no of moles of NH3/total volume

                               = 0.00251-0.0025/50.1    = 0.00001/50.1   = 2*10-7 M

                PH   = -log2*10-7

                         = 6.6989

              

       

queen_honey_blossom answered 1 year ago
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