Question

23. Consider the titration of a 88.6 mL sample of 0.355 M CH3NH2 , methylamine, with 0.432 M HBr. The Kb of methylamine is 4.4x10−4. Determine the pH at the equivalence point.

Answer 1

no of moles of CH3NH2    = molarity* volume in L

                                       = 0.355*0.0886 = 0.031453moles

at equivalent point

no of moles of CH3NH2   = no of moles of NaOH

no of moles of NaOH   = 0.031453moles

no of moles fo NaOH   = molarity * volume in L

0.031453                    = 0.432*volume in L

volume in L               = 0.031453/0.432 = 0.0728L    = 72.8ml

              CH3NH2(aq) + HBr(aq) ---------------> CH3NH3^+ (aq) + Br^-(aq)

               0.031453moles   0.031453moles

total volume = 88.6+72.8   = 161.4ml   = 0.1614

             molarity of CH3NH3^+    = no of moles /total volume

                                                  = 0.031453/0.1614   = 0.195M

Ka   = Kw/Kb

         = 1*10^-14/4.4*10^-4   = 2.27*10^-11

                        CH3NH3^+ (aq) + H2O -----------> CH3NH2 + H3O^+

   I                      0.195                                              0               0

   C                     -x                                                     +x            +x

   E                 0.195-x                                                 +x            +x

                  Ka   = [CH3NH2][H3O^+]/[CH3NH3^+]

                 2.27*10^-11   = x*x/0.195-x

                 2.27*10^-11*(0.195-x) = x^2

                 x   = 2.1*10^-6

                [H3O^+]   = x   = 2.1*10^-6 M

                PH   = -log[H3O^+]

                         = -log2.1*10^-6

                           = 5.6778>>>>answer

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