D. Harvey - Modern Analytical Chemistry (794078), страница 47
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EDTA is an abbreviation for the ligandethylenediaminetetraacetic acid.SOLUTIONThe predominance regions for Ca 2+ and Mg(EDTA) 2– do not overlap,therefore, the reactionCa2+ + Mg(EDTA)2–t Mg2+ + Ca(EDTA)2–will take place. Since there is an excess of Ca2+, the composition of the finalsolution is approximatelyMoles Ca2+ = 0.080 – 0.060 = 0.020 molMoles Ca(EDTA)2– = 0.060 molFigure 6.6Ladder diagram for metal–ligand complexesof Cd2+ and NH3.1400-CH06 9/9/99 7:40 AM Page 154154Modern Analytical ChemistryMoles Mg2+ = 0.060 molMoles Mg(EDTA)2– = 0 molCa2+log Kf,Ca(EDTA)2– = 10.69p EDTACa(EDTA)2–Mg2+log Kf,Mg(EDTA)2– = 8.79Mg(EDTA)2–Figure 6.7Ladder diagram for metal–ligand complexes of ethylenediaminetetraacetic acid (EDTA)with Ca2+ and Mg2+.We can also construct ladder diagrams using cumulative formation constantsin place of stepwise formation constants. The first three stepwise formation constants for the reaction of Zn2+ with NH3t Zn(NH3)2+(aq) K1 = 1.6 × 102Zn(NH3)2+(aq) + NH3(aq) t Zn(NH3)22+(aq)K2 = 1.95 × 102Zn(NH3)22+(aq) + NH3(aq) t Zn(NH3)32+(aq)K3 = 2.3 × 102Zn2+(aq) + NH3(aq)show that the formation of Zn(NH3)32+ is more favorable than the formation ofZn(NH3)2+ or Zn(NH3)22+.
The equilibrium, therefore, is best represented by thecumulative formation reactionZn2+(aq) + 3NH3(aq)t Zn(NH3)32+(aq)for whichβ3 =[Zn(NH 3 )32 + ]= 7.2 × 106[Zn2 + ][NH 3 ]3Taking the log of each side giveslog β3 = log[Zn(NH 3 )32 + ]– 3 log [NH 3 ][Zn2 + ]1400-CH06 9/9/99 7:40 AM Page 155Chapter 6 Equilibrium Chemistry155orpNH3 =The concentrations ofZn2+11[Zn2 + ]log β3 + log33[Zn(NH 3 )32 + ]2+,and Zn(NH3)3pNH3 =Zn2+therefore, are equal when3 log β3 = 2.2911log β3 = log(7.2 × 106 ) = 2.2933pNH3A complete ladder diagram for the Zn2+–NH3 system is shown in Figure 6.8.Zn(NH3)32+6F.3 Ladder Diagram for Oxidation–Reduction Equilibrialog K4 = 2.03Ladder diagrams can also be used to evaluate equilibrium reactions in redox systems. Figure 6.9 shows a typical ladder diagram for two half-reactions in whichthe scale is the electrochemical potential, E. Areas of predominance are defined bythe Nernst equation.
Using the Fe3+/Fe2+ half-reaction as an example, we writeE = E ° – 0.05916 log[Fe2 + ][Fe2 + ]0771005916V.–.log=+[Fe3 + ][Fe3 + ]Zn(NH3)42+Figure 6.8Ladder diagram for Zn2+, Zn(NH3)32+, andZn(NH3)42+, showing how cumulativeformation constants are included.For potentials more positive than the standard-state potential, the predominatespecies is Fe3+, whereas Fe2+ predominates for potentials more negative than E°.When coupled with the step for the Sn4+/Sn2+ half-reaction, we see that Sn2+ can beused to reduce Fe3+. If an excess of Sn2+ is added, the potential of the resulting solution will be near +0.154 V.Using standard-state potentials to construct a ladder diagram canpresent problems if solutes are not at their standard-state concentraFe3+tions.
Because the concentrations of the reduced and oxidized speciesare in a logarithmic term, deviations from standard-state concentraE °Fe3+/Fe2+ = +0.771tions can usually be ignored if the steps being compared are separated1bby at least 0.3 V. A trickier problem occurs when a half-reaction’s potential is affected by the concentration of another species.
For example,Fe2+the potential for the following half-reactionUO22+(aq) + 4H3O+(aq) + 2e–t U4+(aq) + 6H2O(l)Sn4+depends on the pH of the solution. To define areas of predominance inthis case, we begin with the Nernst equationE = 0.327 –EE °Sn4+/Sn2+ = +0.1540.05916[U 4 + ]log2[UO22 + ][H 3O + ]4Sn2+and factor out the concentration of H3O+.E = 0.327 +0.059160.05916[U 4 + ]log [H 3O + ]4 –log22[UO22 + ]Figure 6.9Ladder diagram for the Fe3+/Fe2+ and Sn4+/SN2+ halfreactions.From this equation we see that the areas of predominance for UO22+and U4+ are defined by a step whose potential isE = 0.327 +0.05916log [H 3O + ]4 = 0.327 – 0.1183 pH2Figure 6.10 shows how a change in pH affects the step for the UO22+/U4+ half-reaction.1400-CH06 9/9/99 7:40 AM Page 156156Modern Analytical Chemistry6G Solving Equilibrium Problems2+UO2+0.327 V (pH = 0)E+0.209 V (pH = 1)Ladder diagrams are a useful tool for evaluating chemical reactivity, usually providing a reasonable approximation of a chemical system’s composition at equilibrium.When we need a more exact quantitative description of the equilibrium condition, aladder diagram may not be sufficient.
In this case we can find an algebraic solution.Perhaps you recall solving equilibrium problems in your earlier coursework inchemistry. In this section we will learn how to set up and solve equilibrium problems. We will start with a simple problem and work toward more complex ones.6G.1 A Simple Problem: Solubility of Pb(IO3)2 in Water+0.090 V (pH = 2)U4+Figure 6.10Ladder diagram showing the effect of achange in pH on the areas of predominancefor the UO22+/U4+ half-reaction.When an insoluble compound such as Pb(IO3)2 is added to a solution a small portion of the solid dissolves.
Equilibrium is achieved when the concentrations of Pb2+and IO3– are sufficient to satisfy the solubility product for Pb(IO3)2. At equilibriumthe solution is saturated with Pb(IO3)2. How can we determine the concentrationsof Pb2+ and IO3–, and the solubility of Pb(IO3)2 in a saturated solution prepared byadding Pb(IO3)2 to distilled water?We begin by writing the equilibrium reactionPb(IO3)2(s)t Pb2+(aq) + 2IO3–(aq)and its equilibrium constantKsp = [Pb2+][IO3–]2 = 2.5 × 10–136.33As equilibrium is established, two IO3– ions are produced for each ion of Pb2+. If weassume that the molar concentration of Pb2+ at equilibrium is x then the molar concentration of IO3– is 2x. To help keep track of these relationships, we can use thefollowing table.PbI2(s)tPb2+(aq)+2IO3–(aq)Initial concentrationChange in concentrationsolidsolid0+x0+2xEquilibrium concentrationsolid0+x=x0 + 2x = 2xSubstituting the equilibrium concentrations into equation 6.33(x)(2x)2 = 2.5 × 10–13and solving gives4x3 = 2.5 × 10–13x = 3.97 × 10–5The equilibrium concentrations of Pb2+ and IO3–, therefore, are[Pb2+] = x = 4.0 × 10–5 M[I–] = 2x = 7.9 × 10–5 MSince one mole of Pb(IO3)2 contains one mole of Pb2+, the solubility of Pb(IO3)2is the same as the concentration of Pb2+ ; thus, the solubility of Pb(IO3 ) 2 is4.0 × 10–5 M.1400-CH06 9/9/99 7:40 AM Page 157Chapter 6 Equilibrium Chemistry6G.2 A More Complex Problem: The Common Ion EffectCalculating the solubility of Pb(IO3)2 in distilled water is a straightforward problem since the dissolution of the solid is the only source of Pb2+ or IO3–.
How is thesolubility of Pb(IO 3 ) 2 affected if we add Pb(IO 3 ) 2 to a solution of 0.10 MPb(NO3)2? Before we set up and solve the problem algebraically, think about thechemistry occurring in this system, and decide whether the solubility of Pb(IO3)2will increase, decrease, or remain the same. This is a good habit to develop.Knowing what answers are reasonable will help you spot errors in your calculations and give you more confidence that your solution to a problem is correct.We begin by setting up a table to help us keep track of the concentrations ofPb2+ and IO3– in this system.PbI2(s)tPb2+(aq)+2IO3–(aq)Initial concentrationChange in concentrationsolidsolid0.10+x0+2xEquilibrium concentrationsolid0.10 + x = x0 + 2x = 2xSubstituting the equilibrium concentrations into the solubility product expression(equation 6.33)(0.10 + x)(2x)2 = 2.5 × 10–13and multiplying out the terms on the left leaves us with4x3 + 0.40x2 = 2.5 × 10–136.34This is a more difficult equation to solve than that for the solubility of Pb(IO3)2 in distilled water, and its solution is not immediately obvious.
A rigorous solution to equation 6.34 can be found using available computer software packages and spreadsheets.How might we solve equation 6.34 if we do not have access to a computer? Onepossibility is that we can apply our understanding of chemistry to simplify the algebra.From Le Châtelier’s principle, we expect that the large initial concentration of Pb2+ willsignificantly decrease the solubility of Pb(IO3)2.
In this case we can reasonably expect theequilibrium concentration of Pb2+ to be very close to its initial concentration; thus, thefollowing approximation for the equilibrium concentration of Pb2+ seems reasonable[Pb2+] = 0.10 + x ≈ 0.10 MSubstituting into equation 6.34(0.10)(2x)2 = 2.5 × 10–13and solving for x gives0.40x2 = 2.5 × 10–13x = 7.91 × 10–7Before accepting this answer, we check to see if our approximation was reasonable.In this case the approximation 0.10 + x ≈ 0.10 seems reasonable since the differencebetween the two values is negligible.
The equilibrium concentrations of Pb2+ andIO3–, therefore, are[Pb2+] = 0.10 + x ≈ 0.10 M[I–] = 2x = 1.6 × 10–6 M1571400-CH06 9/9/99 7:40 AM Page 158158Modern Analytical Chemistrycommon ion effectThe solubility of an insoluble saltdecreases when it is placed in a solutionalready containing one of the salt’s ions.The solubility of Pb(IO3)2 is equal to the additional concentration of Pb2+ in solution, or 7.9 × 10–7 mol/L. As expected, the solubility of Pb(IO3)2 decreases in thepresence of a solution that already contains one of its ions. This is known as thecommon ion effect.As outlined in the following example, the process of making and evaluating approximations can be extended if the first approximation leads to an unacceptablylarge error.EXAMPLE 6.9Calculate the solubility of Pb(IO3)2 in 1.0 × 10–4 M Pb(NO3)2.SOLUTIONLetting x equal the change in the concentration of Pb 2+ , the equilibriumconcentrations are[Pb2+] = 1.0 × 10–4 + x[IO3–] = 2xand(1.0 × 10–4 + x)(2x)2 = 2.5 × 10–13We start by assuming that[Pb2+] = 1.0 × 10–4 + x ≈ 1.0 × 10–4 Mand solve for x, obtaining a value of 2.50 × 10–5.
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