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1400-CH06  9/9/99  7:41 AM  Page 167






                                                                                 Chapter 6 Equilibrium Chemistry  167

                                                        –
                                                +
                                            [NH 4 ] = [OH ]
                     We continue by multiplying together the equations for K sp and b 2 , giving
                                                      –
                                             (
                                          [ Ag NH 3 2   +  I ]   9 –
                                                 ) ][
                                  K sp b 2 =       2     = .   10
                                                          14 ´
                                             [ NH 3 ]
                 Substituting in the new mass balance equations for NH 3 and I –
                                            – 2
                                            I
                                           []             –9
                                                 =  . 14 ´ 10
                                          C   ) 2
                                          ( NH 3
                                  –
                 and solving for the [I ] gives
                                           – 2
                                           I
                                          []             –9
                                                 =  . 14 ´ 10
                                        (. 010  M ) 2
                                               –
                                             [] =   . 37 ´ 10 –6  M
                                              I

                     Before accepting this answer, we first check our assumptions. Using the K sp
                                         +
                 equation we calculate the [Ag ] to be
                                               . ´10
                                        K sp  83     –17         –11
                                                           . ´10
                                 [Ag  + ] =  –  =    –6  = 22      M
                                         I
                                               . ´10
                                        []     37
                                                       –
                 From the simplified mass balance equation for I , we have
                                                   –
                                                             –6
                                              +
                                     [Ag(NH 3 ) 2 ]=[I ] = 3.7 ´10 M
                                                                                  +
                                             +
                 Our first assumption that the [Ag ] is significantly smaller than the [Ag(NH 3 ) 2 ],
                                                                                 +
                 therefore, is reasonable. Furthermore, our third assumption that the [Ag(NH 3 ) 2 ] is
                 significantly less than the [NH 3 ] also is reasonable. Our second assumption was
                                                               +
                                          +
                                     [NH 4 ] << [NH 3 ] + [Ag(NH 3 ) 2 ]
                                                                  +
                 To verify this assumption, we solve the K b equation for [NH 4 ]
                                   [NH  + ][OH –  ]  [NH  + ] 2
                                      4        =    4    =  . 175 ´ 10 –5
                                          ]       . 010 M
                                      [NH 3
                 giving
                                                        –3
                                              +
                                          [NH 4 ] = 1.3 ´10 M
                                 +
                 Although the [NH 4 ] is not significantly smaller than the combined concentrations
                                   +
                 of NH 3 and Ag(NH 3 ) 2 , the error is only about 1%. Since this is not an excessively
                 large error, we will accept this approximation as reasonable.
                                                          –
                     Since one mole of AgI produces one mole of I , the solubility of AgI is the same
                                                   –6
                 as the concentration of iodide, or 3.7 ´10 mol/L.
                  6H Buffer Solutions
                 Adding as little as 0.1 mL of concentrated HCl to a liter of H 2 O shifts the pH from
                 7.0 to 3.0. The same addition of HCl to a liter solution that is 0.1 M in both a weak
                                                                                         buffer
                 acid and its conjugate weak base, however, results in only a negligible change in pH.
                                                                                         A solution containing a conjugate weak
                 Such solutions are called buffers, and their buffering action is a consequence of the  acid/weak base pair that is resistant to a
                 relationship between pH and the relative concentrations of the conjugate weak  change in pH when a strong acid or
                 acid/weak base pair.                                                    strong base is added.
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