Page 69 - Modul Asam Basa
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Contoh Soal


                                                 -1
                                  +
                               [H ]      = 1 x 10 M
                                                   +
                               pH        = - log [H ]
                               pH        = - log 10 M
                                                   -1

                               pH        = 1


                           2.  Tentukan pH larutan sebelum dan sesudah diencerkan dari 200 mL

                               larutan Mg(OH)2 0,1 M ditambah 200 mL air!

                               Jawab:
                                                                                -
                                                                    2+
                               Reaksi ionisasi: Mg(OH)2 (aq) → Mg (aq) + 2OH (aq)

                               Sebelum dilakukan pengenceran
                               [OH ]     = x. Mb
                                   -

                                   -
                                                       -1
                               [OH ]     = 2 x 0,1 x 10 M
                                   -             -2
                               [OH ]     = 2 x 10 M
                               pOH       = - log [OH ]
                                                     -

                                                      -2
                               pOH       = - log 2 x 10 M

                               pOH       = 2 – log 2
                               pH        = 14 – (2 – log 2) = 12 + log 2

                               Setelah dilakukan pengenceran
                               V₁ × M₁ = V₂ × M₂

                               200 mL x 0,1 M = 400 mL x M2

                               20 M = 400 x M2
                               M2 = 0,05 M

                                   -
                               [OH ]     = x. Mb
                                   -                 -2
                               [OH ]     = 2 x 5 x 10  M
                               [OH ]     = 1 x 10  M
                                                 -1
                                   -

                                                   +
                               pOH       = - log [H ]
                                                   -1
                               pOH       = - log 10 M
                               pOH       = 1

                               pH        = 14 – 1 = 13



                        E-MODUL KIMIA ASAM BASA                                                      57
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