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              150    Modern Analytical Chemistry


                                              where the subscript “eq” is included for clarification. If a portion of this solution
                                              is diluted with an equal volume of water, each of the concentration terms in equa-
                                              tion 6.30 is cut in half. Thus, the reaction quotient becomes
                                                                          ( . )[Ag (NH 3 2    +
                                                                          05
                                                                                     ) ] eq
                                                                   Q =
                                                                                     2
                                                                                           2
                                                                        05
                                                                       (. )[Ag + ] ( . ) [NH 3  ] eq
                                                                                  05
                                                                               eq

                                              which we can rewrite as
                                                                   æ 05.  ö æ [Ag (NH 3 2   +  ö
                                                                                   ) ] eq
                                                               Q =  ç   3  ÷ ç  +    2  ÷ = 4  ´ 2 b
                                                                    è05(. )  ø è [Ag  ][NH 3 ] eq  ø
                                              Since Q is greater than b 2, equilibrium must be reestablished by shifting the reac-
                                                                                             +
                                              tion to the left, decreasing the concentration of Ag(NH 3) 2 . Furthermore, this new
                                              equilibrium position lies toward the side of the equilibrium reaction with the
                                                                           +
                                              greatest number of solutes (one Ag ion and two molecules of NH 3 versus the sin-
                                                                                           +
                                              gle metal–ligand complex). If the solution of Ag(NH 3) 2 is concentrated, by evapo-
                                              rating some of the solvent, equilibrium is reestablished in the opposite direction.
                                              This is a general conclusion that can be applied to any reaction, whether gas-phase,
                                              liquid-phase, or solid-phase. Increasing volume always favors the direction pro-
                                              ducing the greatest number of particles, and decreasing volume always favors the
                                              direction producing the fewest particles. If the number of particles is the same on
                                              both sides of the equilibrium, then the equilibrium position is unaffected by a
                                              change in volume.
                                               6F Ladder Diagrams
                                              When developing or evaluating an analytical method, we often need to under-
                                              stand how the chemistry taking place affects our results. We have already seen,
                                                                                       +
                                              for example, that adding NH 3 to a solution of Ag is a poor idea if we intend to
                                                          +
                                              isolate the Ag as a precipitate of AgCl (reaction 6.29). One of the primary
                                              sources of determinate method errors is a failure to account for potential chemi-
                                              cal interferences.
               ladder diagram                     In this section we introduce the ladder diagram as a simple graphical tool
               A visual tool for evaluating systems at  for evaluating the chemistry taking place during an analysis. Using ladder dia-
                                                                                                 1
               equilibrium.
                                              grams, we will be able to determine what reactions occur when several reagents
                                              are combined, estimate the approximate composition of a system at equilibrium,
                                              and evaluate how a change in solution conditions might affect our results.

                                              6F.1 Ladder Diagrams for Acid–Base Equilibria
                                              To see how a ladder diagram is constructed, we will use the acid–base equilibrium
                                              between HF and F –

                                                                                    +
                                                                                            –
                                                                HF(aq)+H 2 O(l) t H 3 O (aq)+F (aq)
                                              for which the acid dissociation constant is
                                                                                      –
                                                                                   +
                                                                              [ HO ][ F ]
                                                                                 3
                                                                            =
                                                                       K aHF,
                                                                                 [ HF]
                                              Taking the log of both sides and multiplying through by –1 gives
                                                                                           [F – ]
                                                                – log K aHF =  – log [HO + ] – log
                                                                       ,
                                                                                  3
                                                                                           [HF ]
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