D. Harvey - Modern Analytical Chemistry (794078), страница 44
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Whether an amphiprotic species behaves as an acid or as a base dependson the equilibrium constants for the two competing reactions. For bicarbonate, theacid dissociation constant for reaction 6.8Ka2 = 4.69 × 10–11is smaller than the base dissociation constant for reaction 6.9.Kb2 = 2.25 × 10–8Since bicarbonate is a stronger base than it is an acid (kb2 > ka2), we expect thataqueous solutions of HCO3– will be basic.Dissociation of Water Water is an amphiprotic solvent in that it can serve as anacid or a base. An interesting feature of an amphiprotic solvent is that it is capableof reacting with itself as an acid and a base.H2O(l) + H2O(l)t H3O+(aq) + OH–(aq)The equilibrium constant for this reaction is called water’s dissociation constant, Kw,Kw = [H3O+][OH–]6.10which has a value of 1.0000 ×at a temperature of 24 °C.
The value of Kw variessubstantially with temperature. For example, at 20 °C, Kw is 6.809 × 10–15, but at30 °C K w is 1.469 × 10 –14 . At the standard state temperature of 25 °C, K w is1.008 × 10–14, which is sufficiently close to 1.00 × 10–14 that the latter value can beused with negligible error.10–14The pH Scale An important consequence of equation 6.10 is that the concentrations of H3O+ and OH– are related. If we know [H3O+] for a solution, then [OH–]can be calculated using equation 6.10.EXAMPLE 6.2What is the [OH–] if the [H3O+] is 6.12 × 10–5 M?SOLUTION[OH – ] =pHDefined as pH = –log[H3O+].Kw1.00 × 10 –14= 1.63 × 10 –10=+6.12 × 10 –5[H 3 O ]Equation 6.10 also allows us to develop a pH scale that indicates the acidity of a solution. When the concentrations of H3O+ and OH– are equal, a solution is neitheracidic nor basic; that is, the solution is neutral.
Letting[H3O+] = [OH–]and substituting into equation 6.10 leaves us withKw = [H3O+]2 = 1.00 × 10–141400-CH06 9/9/99 7:40 AM Page 143Chapter 6 Equilibrium ChemistrySolving for [H3O+] gives1[H 3 O + ] =3A neutral solution has a hydronium ion concentration of 1.00 × 10–7 M and a pH of7.00.* For a solution to be acidic, the concentration of H3O+ must be greater thanthat for OH–, or56pHThe pH of an acidic solution, therefore, must be less than 7.00.
A basic solution, onthe other hand, will have a pH greater than 7.00. Figure 6.3 shows the pH scalealong with pH values for some representative solutions.786.116.12t H3O+(aq) + OH–(aq)6.13The equilibrium constant for equation 6.13 is Kw. Since equation 6.13 is obtainedby adding together reactions 6.11 and 6.12, Kw may also be expressed as the productof Ka for CH3COOH and Kb for CH3COO–. Thus, for a weak acid, HA, and its conjugate weak base, A–,Kw = Ka × Kb6.14This relationship between Ka and Kb simplifies the tabulation of acid and base dissociation constants. Acid dissociation constants for a variety of weak acids are listedin Appendix 3B. The corresponding values of Kb for their conjugate weak bases aredetermined using equation 6.14.EXAMPLE 6.3Using Appendix 3B, calculate the following equilibrium constants(a) Kb for pyridine, C5H5N(b) Kb for dihydrogen phosphate, H2PO4–SOLUTION(a) K b, C 5H 5N =(b ) K b , H2PO 4–KwKa, C 5H 5NH +=KwKa, H 3PO 4==1.00 × 10 –14= 1.69 × 10 –95.90 × 10 –61.00 × 10 –14= 1.41 × 10 –127.11 × 10 –3*The use of a p-function to express a concentration is covered in Chapter 2.BloodSeawaterMilk of magnesia111213Household bleach14Figure 6.3pH scale showing values for representativesolutions.Adding together these two reactions gives2H2O(l)“Pure” rainMilkNeutral910Tabulating Values for Ka and Kb A useful observation about acids and bases is thatthe strength of a base is inversely proportional to the strength of its conjugate acid.Consider, for example, the dissociation reactions of acetic acid and acetate.Vinegar4[H3O+] > 1.00 × 10–7 Mt H3O+(aq) + CH3COO–(aq)CH3COO–(aq) + H2O(l) t CH3COOH(aq) + OH–(aq)Gastric juice21.00 × 10 –14 = 1.00 × 10 –7CH3COOH(aq) + H2O(l)1431400-CH06 9/9/99 7:40 AM Page 144144Modern Analytical Chemistry6D.3 Complexation ReactionsligandA Lewis base that binds with a metal ion.A more general definition of acids and bases was proposed by G.
N. Lewis(1875–1946) in 1923. The Brønsted–Lowry definition of acids and bases focuses onan acid’s proton-donating ability and a base’s proton-accepting ability. Lewis theory, on the other hand, uses the breaking and forming of covalent bonds to describeacid–base characteristics. In this treatment, an acid is an electron pair acceptor, anda base is an electron pair donor. Although Lewis theory can be applied to the treatment of acid–base reactions, it is more useful for treating complexation reactionsbetween metal ions and ligands.The following reaction between the metal ion Cd2+ and the ligand NH3 is typical of a complexation reaction.Cd2+(aq) + 4(:NH3)(aq)formation constantThe equilibrium constant for a reactionin which a metal and a ligand bind toform a metal–ligand complex (Kf).stepwise formation constantThe formation constant for ametal–ligand complex in which only oneligand is added to the metal ion or to ametal–ligand complex (Ki).cumulative formation constantThe formation constant for ametal–ligand complex in which two ormore ligands are simultaneously addedto a metal ion or to a metal–ligandcomplex (βi).6.15The product of this reaction is called a metal–ligand complex.
In writing the equation for this reaction, we have shown ammonia as :NH3 to emphasize the pair ofelectrons it donates to Cd2+. In subsequent reactions we will omit this notation.The formation of a metal–ligand complex is described by a formation constant, Kf. The complexation reaction between Cd2+ and NH3, for example, has thefollowing equilibrium constantKf =dissociation constantThe equilibrium constant for a reactionin which a metal–ligand complexdissociates to form uncomplexed metalion and ligand (Kd).t Cd(:NH3)42+(aq)[Cd(NH 3 )42 + ]= 5.5 × 107[Cd2 + ][NH 3 ]46.16The reverse of reaction 6.15 is called a dissociation reaction and is characterized bya dissociation constant, Kd, which is the reciprocal of Kf.Many complexation reactions occur in a stepwise fashion.
For example, the reaction between Cd2+ and NH3 involves four successive reactionst Cd(NH3)2+(aq)Cd(NH3)2+(aq) + NH3(aq) t Cd(NH3)22+(aq)Cd(NH3)22+(aq) + NH3(aq) t Cd(NH3)32+(aq)Cd(NH3)32+(aq) + NH3(aq) t Cd(NH3)42+(aq)Cd2+(aq) + NH3(aq)6.176.186.196.20This creates a problem since it no longer is clear what reaction is described by a formation constant. To avoid ambiguity, formation constants are divided into two categories. Stepwise formation constants, which are designated as Ki for the ith step,describe the successive addition of a ligand to the metal–ligand complex formed inthe previous step.
Thus, the equilibrium constants for reactions 6.17–6.20 are, respectively, K1, K2, K3, and K4. Overall, or cumulative formation constants, whichare designated as βi, describe the addition of i ligands to the free metal ion. Theequilibrium constant expression given in equation 6.16, therefore, is correctly identified as β4, whereβ4 = K1 × K2 × K3 × K4In generalβi = K1 × K2 × . . . × KiStepwise and cumulative formation constants for selected metal–ligand complexesare given in Appendix 3C.1400-CH06 9/9/99 7:40 AM Page 145Chapter 6 Equilibrium ChemistryEquilibrium constants for complexation reactions involving solids are definedby combining appropriate Ksp and Kf expressions. For example, the solubility ofAgCl increases in the presence of excess chloride as the result of the following complexation reactionAgCl(s) + Cl–(aq)t AgCl2–(aq)6.21This reaction can be separated into three reactions for which equilibrium constantsare known—the solubility of AgCl, described by its KspAgCl(s)t Ag+(aq) + Cl–(aq)and the stepwise formation of AgCl2–, described by K1 and K2t AgCl(aq)AgCl(aq) + Cl–(aq) t AgCl2–(aq)Ag+(aq) + Cl–(aq)The equilibrium constant for reaction 6.21, therefore, is equal to Ksp × K1 × K2.EXAMPLE 6.4Determine the value of the equilibrium constant for the reactionPbCl2(s)t PbCl2(aq)SOLUTIONThis reaction can be broken down into three reactions.
The first of thesereactions is the solubility of PbCl2, described by its KspPbCl2(s)t Pb2+(aq) + 2Cl–(aq)and the second and third are the stepwise formation of PbCl2 (aq), described byK1 and K2t PbCl+(aq)PbCl+(aq) + Cl–(aq) t PbCl2(aq)Pb2+(aq) + Cl–(aq)Using values for Ksp, K1, and K2 from Appendices 3A and 3C, we find theequilibrium constant to beK = Ksp × K1 × K2 = (1.7 × 10–5)(38.9)(1.62) = 1.1 × 10–36D.4 Oxidation–Reduction ReactionsIn a complexation reaction, a Lewis base donates a pair of electrons to a Lewis acid.In an oxidation–reduction reaction, also known as a redox reaction, electrons arenot shared, but are transferred from one reactant to another.
As a result of this electron transfer, some of the elements involved in the reaction undergo a change in oxidation state. Those species experiencing an increase in their oxidation state are oxidized, while those experiencing a decrease in their oxidation state are reduced. Forexample, in the following redox reaction between Fe3+ and oxalic acid, H2C2O4,iron is reduced since its oxidation state changes from +3 to +2.2Fe3+(aq) + H2C2O4(aq) + 2H2O(l)t 2Fe2+(aq) + 2CO2(g) + 2H3O+(aq)6.22redox reactionAn electron-transfer reaction.1451400-CH06 9/9/99 7:40 AM Page 146146Modern Analytical ChemistryoxidationA loss of electrons.reductionA gain of electrons.reducing agentA species that donates electrons toanother species.oxidizing agentA species that accepts electrons fromanother species.Oxalic acid, on the other hand, is oxidized since the oxidation state for carbon increases from +3 in H2C2O4 to +4 in CO2.Redox reactions, such as that shown in equation 6.22, can be divided into separate half-reactions that individually describe the oxidation and the reductionprocesses.H2C2O4(aq) + 2H2O(l) → 2CO2(g) + 2H3O+(aq) + 2e–Fe3+(aq) + e– → Fe2+(aq)It is important to remember, however, that oxidation and reduction reactions always occur in pairs.* This relationship is formalized by the convention of calling thespecies being oxidized a reducing agent, because it provides the electrons for the reduction half-reaction.
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