Question

A chemist titrates 170.0 mL of a 0.4930 M dimethylamine (CH3)NH) solution with 0.2420 M HNO, solution at 25 °C. Calculate the pH at equivelence. The p K, of dimethylamine is 3.27 Round your answer to 2 decimal places Note for advanced stu dents: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added dt Save For LaterSubmit Assignment inn ights Reserved. Teis of

Answer 1

(CH3)2NH(aq) + HNO3(aq) ---------------> (CH3)2NH2NO3

1 mole               1 mole

(CH3)2NH                                           HNO3

M1   = 0.493M                                   M2 = 0.242M

V1    = 170ml                                     V2   =

n1   =1                                               n2 = 1

         M1V1/n1   =   M2V2/n2

             V2         = M1V1n2/M2n1

                           = 0.493*170*1/0.242*1   =346.32ml

Total volume of solution   = 170 + 346.32   = 516.32ml   = 0.51632L

no of moles of (CH3)2NH   = molarity * volume in L

                                        = 0.493*0.17   = 0.08381 moles

        (CH3)2NH(aq) + HNO3(aq) ---------------> (CH3)2NH2NO3

I 0.08381moles      0.08381 moles                 0

C      -.0.08381            -0.08381                          0.08381

E          0                         0                                  0.08381

molarity of (CH3)2NH2NO3    = no of moles/volume in L

                                            =   0.08381/0.51632   = 0.162M

    PKb    = 3.27

PKa    = 14-PKb

          = 14-3.27

PKa    = 10.73

-logKa   = 10.73

       Ka    = 10^-10.73   = 1.86*10^-11

                  (CH3)2NH2^+ (aq) + H2O (l) ----------------> (CH3)2NH(aq)    + H3O^+ (aq)

I                  0.162                                                            0                            0

C                  -x                                                                +x                          +x

E                0.162-x                                                          +x                          +x

                   Ka     =   [(CH3)2NH][H3O^+]/[(CH3)2NH2^+]

                  1.86*10^-11    = x*x/0.162-x

                  1.86*10^-11*(0.162-x)   = x^2

                    x   = 1.735*10^-6

             [H3O^+]     =x    = 1.735*10^-6M

                  PH   = -log[H3O^+]

                         = -log1.735*10^-6

                         = 5.76>>>>>answer