O ACIDS AND BASES Calculating the pH of a weak base titrated with a strong acid

Question

O ACIDS AND BASES Calculating the pH of a weak base titrated with a strong acid An analytical chemist is titrating 78.6 ml. of a 0.1400 M solution of dimethylamine (CH3),NH) with a 0.3200 M solution of HIO, The pK, of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 28.6 mL of the HIO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO3 solution added Round your answer to 2 decimal places.

Explanation / Answer

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

                                              =   0.14*0.0786

                                               = 0.011 moles

no of moles of HIO3                 = molarity * volume in L

                                                 = 0.32*0.0286   = 0.009152 moles

                  (CH3)2NH(aq) + HIO3 ------------------> (CH3)2NH2IO3(aq)

I                   0.011                  0.009152                         0

C                  -0.009152        -0.009152                         0.009152

E                0.001848                0                                     0.009152

             [(CH3)2NH]   = 0.001848M

            [(CH3)2NH2IO3]   = 0.009152M

                        POH   = Pkb + log[(CH3)2NH2IO3]/[(CH3)2NH]

                                   = 3.27 + log0.009152/0.001848

                                    = 3.27 + 0.6948

                                      = 3.9648

                            PH    = 14-POH

                                      = 14-3.9648    = 10.0352 >>>>>answer

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