D. Harvey - Modern Analytical Chemistry (794078), страница 46
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Decreasingammonia’s concentration causes reaction 6.29 to move from products toreactants, decreasing the solubility of AgCl.Increasing or decreasing the partial pressure of a gas is the same as increasingor decreasing its concentration.† The effect on a reaction’s equilibrium position canbe analyzed as described in the preceding example for aqueous solutes. Since theconcentration of a gas depends on its partial pressure, and not on the total pressureof the system, adding or removing an inert gas has no effect on the equilibrium position of a gas-phase reaction.Most reactions involve reactants and products that are dispersed in a solvent.If the amount of solvent is changed, either by diluting or concentrating the solution, the concentrations of all reactants and products either decrease or increase.The effect of these changes in concentration is not as intuitively obvious as whenthe concentration of a single reactant or product is changed.
As an example, let’sconsider how dilution affects the equilibrium position for the formation of theaqueous silver-amine complex (reaction 6.28). The equilibrium constant for thisreaction isβ2 =[Ag(NH 3 )2+ ]eq2[Ag + ]eq[NH 3 ]eq6.30*Adding AgNO3 decreases the solubility of AgCl.†The relationship between pressure and concentration can be deduced from the ideal gas law. Starting with PV = nRT, wesolve for the molar concentrationnPMolar concentration ==VRTOf course, this assumes an ideal gas (which is usually a reasonable assumption under normal laboratory conditions).1491400-CH06 9/9/99 7:40 AM Page 150150Modern Analytical Chemistrywhere the subscript “eq” is included for clarification.
If a portion of this solutionis diluted with an equal volume of water, each of the concentration terms in equation 6.30 is cut in half. Thus, the reaction quotient becomesQ =(0.5)[Ag(NH 3 )2+ ]eq2(0.5)[Ag + ]eq (0.5)2 [NH 3 ]eqwhich we can rewrite as 0.5 [Ag(NH 3 )2+ ]eq = 4 × β2Q =2 (0.5)3 [Ag + ][NH 3 ]eqSince Q is greater than β2, equilibrium must be reestablished by shifting the reaction to the left, decreasing the concentration of Ag(NH3)2+. Furthermore, this newequilibrium position lies toward the side of the equilibrium reaction with thegreatest number of solutes (one Ag+ ion and two molecules of NH3 versus the single metal–ligand complex).
If the solution of Ag(NH3)2+ is concentrated, by evaporating 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 producing the greatest number of particles, and decreasing volume always favors thedirection producing the fewest particles. If the number of particles is the same onboth sides of the equilibrium, then the equilibrium position is unaffected by achange in volume.6F Ladder Diagramsladder diagramA visual tool for evaluating systems atequilibrium.When developing or evaluating an analytical method, we often need to understand how the chemistry taking place affects our results.
We have already seen,for example, that adding NH3 to a solution of Ag+ is a poor idea if we intend toisolate the Ag + as a precipitate of AgCl (reaction 6.29). One of the primarysources of determinate method errors is a failure to account for potential chemical interferences.In this section we introduce the ladder diagram as a simple graphical toolfor evaluating the chemistry taking place during an analysis.1 Using ladder diagrams, we will be able to determine what reactions occur when several reagentsare 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 EquilibriaTo see how a ladder diagram is constructed, we will use the acid–base equilibriumbetween HF and F–HF(aq) + H2O(l)t H3O+(aq) + F–(aq)for which the acid dissociation constant isKa, HF =[H 3O + ][F – ][HF]Taking the log of both sides and multiplying through by –1 gives– log Ka, HF = – log [H 3O + ] – log[F – ][HF]1400-CH06 9/9/99 7:40 AM Page 151Chapter 6 Equilibrium Chemistry151Finally, replacing the negative log terms with p-functions and rearranging leaves uswithpH = pKa + log[F – ][HF]6.31F–Examining equation 6.31 tells us a great deal about the relationship betweenpH and the relative amounts of F– and HF at equilibrium.
If the concentrations ofF– and HF are equal, then equation 6.31 reduces topHpH = pKa,HF = –log(Ka,HF) = –log(6.8 × 10–4) = 3.17pH = pKa,HF = 3.17For concentrations of F– greater than that of HF, the log term in equation 6.31 ispositive andHFpH > pKa,HForpH > 3.17This is a reasonable result since we expect the concentration of hydrofluoric acid’sconjugate base, F–, to increase as the pH increases. Similar reasoning shows that theconcentration of HF exceeds that of F– whenpH < pKa,HForpH < 3.17Figure 6.4Now we are ready to construct the ladder diagram for HF (Figure 6.4).The ladder diagram consists of a vertical scale of pH values oriented so thatsmaller (more acidic) pH levels are at the bottom and larger (more basic) pHlevels are at the top. A horizontal line is drawn at a pH equal to pKa,HF. This line,or step, separates the solution into regions where each of the two conjugate formsof HF predominate.
By referring to the ladder diagram, we see that at a pHof 2.5 hydrofluoric acid will exist predominately as HF. If we add sufficient baseto the solution such that the pH increases to 4.5, the predominate form becomes F–.Figure 6.5 shows a second ladder diagram containing information aboutHF/F– and NH4+/NH3. From this ladder diagram we see that if the pH is lessthan 3.17, the predominate species are HF and NH4+. For pH’s between 3.17and 9.24 the predominate species are F– and NH4+, whereas above a pH of9.24 the predominate species are F– and NH3.Ladder diagrams are particularly useful for evaluating the reactivity ofacids and bases.
An acid and a base cannot coexist if their respective areas ofpredominance do not overlap. If we mix together solutions of NH3 and HF,the reactionHF(aq) + NH3(aq)t NH4+(aq) + F–(aq)6.32Ladder diagram for HF, showing areas ofpredominance for HF and F–.NH3pH = pKa,NH3 = 9.24NH4+pHoccurs because the predominance areas for HF and NH3 do not overlap. Before continuing, let us show that this conclusion is reasonable by calculatingthe equilibrium constant for reaction 6.32.
To do so we need the followingthree reactions and their equilibrium constants.HF(aq) + H 2 O(l)NH 3 (aq) + H 2 O(l)t H3O + (aq) + F – (aq)t OH – (aq) + NH4+ (aq)H 3O + (aq) + OH – (aq)t 2H2 O(l)K =Ka = 6.8 × 10 –4F–pH = pKa,HF = 3.17HFFigure 6.5Ladder diagram for HF and NH3.K b = 1.75 × 10 –511=Kw1.00 × 10 –141400-CH06 9/9/99 7:40 AM Page 152152Modern Analytical ChemistryAdding together these reactions gives us reaction 6.32, for which the equilibriumconstant isK =Ka K b(6.8 × 10 –4 )(1.75 × 10 –5 )= 1.19 × 106=Kw(1.00 × 10 –14 )Since the equilibrium constant is significantly greater than 1, the reaction’s equilibrium position lies far to the right.
This conclusion is general and applies to all ladder diagrams. The following example shows how we can use the ladder diagram inFigure 6.5 to evaluate the composition of any solution prepared by mixing togethersolutions of HF and NH3.EXAMPLE6.7Predict the pH and composition of a solution prepared by adding 0.090 mol ofHF to 0.040 mol of NH3.SOLUTIONSince HF is present in excess and the reaction between HF and NH 3 isfavorable, the NH3 will react to form NH4+. At equilibrium, essentially no NH3remains andMoles NH4+ = 0.040 molConverting NH3 to NH4+ consumes 0.040 mol of HF; thusMoles HF = 0.090 – 0.040 = 0.050 molMoles F– = 0.040 molAccording to the ladder diagram for this system (see Figure 6.5), a pH of 3.17results when there is an equal amount of HF and F–.
Since we have more HFthan F–, the pH will be slightly less than 3.17. Similar reasoning will show youthat mixing together 0.090 mol of NH3 and 0.040 mol of HF will result in asolution whose pH is slightly larger than 9.24.If the areas of predominance for an acid and a base overlap each other, thenpractically no reaction occurs. For example, if we mix together solutions of NaF andNH4Cl, we expect that there will be no significant change in the moles of F– andNH4+.
Furthermore, the pH of the mixture must be between 3.17 and 9.24. BecauseF– and NH4+ can coexist over a range of pHs we cannot be more specific in estimating the solution’s pH.The ladder diagram for HF/F – also can be used to evaluate the effect ofpH on other equilibria that include either HF or F–. For example, the solubility ofCaF2CaF2(s)t Ca2+(aq) + 2F–(aq)is affected by pH because F– is a weak base. Using Le Châtelier’s principle, if F– isconverted to HF, the solubility of CaF2 will increase.
To minimize the solubility ofCaF2 we want to control the solution’s pH so that F– is the predominate species.From the ladder diagram we see that maintaining a pH of more than 3.17 ensuresthat solubility losses are minimal.1400-CH06 9/9/99 7:40 AM Page 153Chapter 6 Equilibrium Chemistry1536F.2 Ladder Diagrams for Complexation EquilibriaThe same principles used in constructing and interpreting ladder diagrams foracid–base equilibria can be applied to equilibria involving metal–ligand complexes. For complexation reactions the ladder diagram’s scale is defined by theconcentration of uncomplexed, or free ligand, pL.
Using the formation ofCd(NH3)2+ as an exampleCd2+(aq) + NH3(aq)t Cd(NH3)2+(aq)we can easily show that the dividing line between the predominance regions forCd2+ and Cd(NH3)2+ is log(K1).Cd2+log K1 = 2.55[Cd(NH 3 )2 + ]K1 =[Cd2 + ][NH 3 ]log(K1 ) = logCd(NH3)2+[Cd(NH 3 )2 + ]– log[NH 3 ][Cd2 + ]log K2 = 2.01Cd(NH3)22+p NH3log(K1 ) = log[Cd(NH 3 )2 + ]+ pNH3[Cd2 + ]log K3 = 1.34Cd(NH3)32+pNH3 = log(K1 ) + log[Cd2 + ]log K4 = 0.84[Cd(NH 3 )2 + ]Cd(NH3)42+Since K1 for Cd(NH3)2+ is 3.55 × 102, log(K1) is 2.55. Thus, for a pNH3 greater than2.55 (concentrations of NH3 less than 2.8 × 10–3 M), Cd2+ is the predominatespecies. A complete ladder diagram for the metal–ligand complexes of Cd 2+ andNH3 is shown in Figure 6.6.EXAMPLE 6.8Using the ladder diagram in Figure 6.7, predict the result of adding 0.080 molof Ca2+ to 0.060 mol of Mg(EDTA)2–.
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