Question

Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2 with 0.145 M HBr. Determine each of the following.

A) the pH at one-half of the equivalence point

B) the pH at the equivalence point

C) the pH after adding 6.0 mL of acid beyond the equivalence point

Answer 1

A.   no of moles of CH3NH2   = molarity * volume in L

                                                = 0.175*0.026   = 0.00455 moles

no of moles of HBr                  = molarity * volume in L

                                                 = 0.145*0.026   = 0.00377 moles

A) the pH at one-half of the equivalence point

Kb   = 4.4*10^-4

PKb   = -logKb

         = -log4.4*10^-4

         = 3.3565

at one-half of the equivalence point [ salt ] = [base]

    POH   = Pkb + log[Salt]/[Base]

    POH   = 3.3565 + log1/1

POH   = 3.3565

   PH     = 14-POH

              = 14-3.3565   = 10.6435

B.      molarity of CH3NH3^+   = no of moles/total volume in L

                                                = 0.00455/0.052   = 0.0875 M

              CH3NH3^+ + H2O --------------> CH3NH2 + H3O^+

I                0.0875                                        0                0

C               -x                                                + x            +x

E             0.0875-x                                        +x             +x

           Ka    = Kw/Kb

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

                     = 2.272*10^-11

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

             2.272*10^-11    = x*x/0.0875-x

            2.272*10^-11*(0.0875-x) = x^2

                 x = 1.4*10^-6

           [H3O^+]    =   x   = 1.4*10^-6 M

             PH   = -log[H3O^+]

                    = -log1.4*10^-6

                     = 5.8538