Page 165 - Modern Analytical Chemistry
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1400-CH06  9/9/99  7:40 AM  Page 148





              148    Modern Analytical Chemistry


                                                  Solving for K gives the equilibrium constant as
                                                                           log K = 40.6558

                                                                           K = 4.527 ´10 40
                                                                            +
                                                                                                   2+
                                                  (c) The potential when the [Ag ] is 0.020 M and the [Cd ] is 0.050 M is
                                                    calculated using equation 6.24 employing the appropriate relationship for
                                                    the reaction quotient Q.
                                                                             . 0 05916  [Cd 2 + ]
                                                                         °
                                                                    E =  E –       log
                                                                               n      [Ag + 2
                                                                                          ]
                                                                                . 0 05916  (. 0 050 )
                                                                      = . 1 2026  –   log
                                                                                  2      (. 0 020 ) 2

                                                                      = . 114V





                                               6E Le Châtelier’s Principle
                                              The equilibrium position for any reaction is defined by a fixed equilibrium con-
                                              stant, not by a fixed combination of concentrations for the reactants and products.
                                              This is easily appreciated by examining the equilibrium constant expression for the
                                              dissociation of acetic acid.
                                                                   [H O + ][CH COO – ]
                                                                             3
                                                                      3
                                                              K a =                  = . 175  ´10 –5          6.26
                                                                      [CH COOH  ]
                                                                          3
                                              As a single equation with three variables, equation 6.26 does not have a unique so-
                                                                                                     +
                                                                                            –
                                              lution for the concentrations of CH 3 COOH, CH 3 COO , and H 3 O . At constant
                                                                                                               +
                                              temperature, different solutions of acetic acid may have different values for [H 3 O ],
                                                       –
                                              [CH 3 COO ] and [CH 3 COOH], but will always have the same value of K a .
                                                  If a solution of acetic acid at equilibrium is disturbed by adding sodium acetate,
                                                          –
                                              the [CH 3 COO ] increases, suggesting an apparent increase in the value of K a . Since
                                              K a must remain constant, however, the concentration of all three species in equa-
                                              tion 6.26 must change in a fashion that restores K a to its original value. In this case,
                                                                                                          +
                                                                                                 –
                                              equilibrium is reestablished by the partial reaction of CH 3COO and H 3O to pro-
                                              duce additional CH 3 COOH.
                                                  The observation that a system at equilibrium responds to a stress by reequili-
               Le Châtelier’s principle       brating in a manner that diminishes the stress, is formalized as Le Châtelier’s prin-
               When stressed, a system that was at  ciple. One of the most common stresses that we can apply to a reaction at equilib-
               equilibrium returns to its equilibrium  rium is to change the concentration of a reactant or product. We already have seen,
               state by reacting in a manner that
               relieves the stress.           in the case of sodium acetate and acetic acid, that adding a product to a reaction
                                              mixture at equilibrium converts a portion of the products to reactants. In this in-
                                              stance, we disturb the equilibrium by adding a product, and the stress is diminished
                                              by partially reacting the excess product. Adding acetic acid has the opposite effect,
                                              partially converting the excess acetic acid to acetate.
                                                  In our first example, the stress to the equilibrium was applied directly. It is also
                                              possible to apply a concentration stress indirectly. Consider, for example, the fol-
                                              lowing solubility equilibrium involving AgCl
                                                                                       –
                                                                               +
                                                                    AgCl(s) t Ag (aq)+Cl (aq)                 6.27
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