Practice Buffer Problem and Key by xfz11675

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									                        Practice Buffer Problem and Key
 (This was the mandatory problem the year it was given. If you can do this, you can
  do anything they can throw at you! Remember, even if you just get parts a and b,
  you will be ahead of most of the students taking it! Also keep in mind that on the
     real exam, you would have about 18 minutes, more or less, to spend on this.)

Here it is!
                           CH3NH2 + H2O <=> CH3NH3+ + OH-
Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The
value of the ionization constant, Kb, is 5.25x10-4. Methylamine forms salts such as
methylammonium nitrate, (CH3NH3+)(NO3-).
(a) Calculate the hydroxide ion concentration, [OH-], of a 0.225-molar aqueous solution
    of methylamine.
(b) Calculate the pH of a solution made by adding 0.0100 mole of solid
    methylammonium nitrate to 120.0 milliliters of a 0.225-molar solution of
    methylamine. Assume no volume change occurs.
(c) How many moles of either NaOH or HCl (state clearly which you choose) should be
    added to the solution in (b) to produce a solution that has a pH of 11.00? Assume
    that no volume change occurs.
(d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is
    the pH of the solution affected? Explain.

                      (Answer given on next 2 pages)
                                             
               [ C H 3 N H 3 ][ O H               ]
      K b 
                    [ C H 3N H 2 ]
(a)
                     CH3NH2 + H2O <=> CH3NH3+ + OH-
      [ ]i               0.225           0        0
      [ ]                  -X          +X       +X
      [ ]eq           0.225-X            X        X
                                                                              2
                              
                               4               [ X ][ X ]                 X
      K b  5 . 2 5  0
                      1                                           
                                           [ 0 . 2 2 5  X ]            0 .2 2 5

    X = [OH-] = 1.09x10-2 M
    solved using quadratic: X = [OH-] = 1.06x10-2 M
(b) [CH3NH3+] = 0.0100 mol / 0.1200 L = 0.0833 M
    or CH3NH2 = 0.120 L x 0.225 mol/L = 0.0270 mol
                              
                               4           [ 0 . 0 8 3 3  X ][ X ]               0 . 0 8 3 3X
      K b  5 . 2 5  0
                      1                                                     
                                                [ 0 . 2 2 5  X ]                   0 .2 2 5

      X = [OH-] = 1.42x10-3 M; pOH = 2.85; pH = 11.15
      OR (you should really write out the chemicals if you use Henderson Hasselbach
      below instead of what the key says, with "base" and "acid" being too vague, in my
      opinion)
                                   [b as e]
      p H  p K a  lo g
                                   [ a cid ]
                          4
                          1
                1  0
                   1                                        1 1
      K a                   
                               4
                                       1 . 9 1  0
                                                  1             ; p K a 1 0 . 7 2
               5 . 2 5  0
                        1
                                      ( 0 .2 2 5 )
      p H  1 0 . 7 2  lo g                           1 1 . 1 5
                                     ( 0 . 0 8 3 3)

      OR
                                       [ a ci d ]
      p O H  p K b  lo g                          ; p K b  3 . 2 8
                                       [b as e]
                                      ( 0 . 0 8 3 3)
      p O H  3 . 2 8  lo g                             2 . 8 5; p H 1 1 . 1 5
                                          ( 0 . 2 2 5)

(c) HCl must be added.
                              
                               4           [ 0 . 0 8 3 3  X ][ 0 . 0 0 1 0]
      K b  5 . 2 5  0
                      1              
                                                      [ 0 . 2 2 5  X ]

      X = 0.0228 M
      0.0228 mol/L x 0.120 L = 2.74x10-3 mol HCl
      OR
                                       [ b a s e]            [ b a s e]
    1 1 . 0 0  1 0 . 7 2  lo g                  ; lo g                 0 . 2 8
                                       [ a ci d ]            [ a ci d ]
     [ b a s e]                      ( 0 . 2 2 5  X )
                  1 . 9 0 5                            ; X  0 . 0 2 2 7M
     [ a ci d ]                     ( 0 . 0 8 3 3  x )

    0.0227 M x 0.120 L = 2.73x10-3 mol HCl

                            +
            [C H 3 N H      3   ]
            [C H N H
              3   2             ]
(d) The              ratio does not change in this buffer solution with dilution, therefore,
    no effect on pH.

								
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