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1、溶液中的化学平衡Chemical Equilibrium in Solution1Protonic Concept of Acids and Bases 酸碱质子理论Acids have sour taste,change color of dyes.Acids have sour taste,change color of dyes.Bases have bitter taste,restore color of dye Bases have bitter taste,restore color of dye that acids change.that acids change.In 19
2、23,Bronsted and Lowry put forth:In 1923,Bronsted and Lowry put forth:Acid base+proton Acid base+proton2Example:Example:HCl Cl HCl Cl+H+H+hydrochloric acid hydrochloric acid HPO HPO4422 PO PO4433+H+H+hydrogen phosphate hydrogen phosphate ionion NH NH44+NH NH33+H+H+ammonium ion ammonium ion H H22O OHO
3、 OH+H+H+waterwaterAn acid will have no meanings if the conjugate base An acid will have no meanings if the conjugate base isnt.isnt.oror3*amphoteric amphoteric 两性物质两性物质:Species like HSOSpecies like HSO4422 either either donates or accepts a proton,called amphoteric substance.donates or accepts a pro
4、ton,called amphoteric substance.HPO HPO4422 PO PO4433+H+H+;HPO;HPO4422+H+H+HH22POPO44*the nature of acid-base reactions*the nature of acid-base reactions 酸碱反应的本质酸碱反应的本质:Acid-base reactions originate in the transfer of protons.Acid-base reactions originate in the transfer of protons.例如例如，HCl+HHCl+H22
5、O O H H33OO+Cl+Cl H H22O acts as a baseO acts as a base又又如如:H H22O+NHO+NH33 NH NH44+OH+OH H H22O acts as an acidO acts as an acid A competition for proton exists both for acid A competition for proton exists both for acid11 and acid and acid22 and for baseand for base11 and base and base2 2 由此，由此，酸酸
6、11和酸和酸22、碱、碱11和碱和碱22均参与均参与质子的争夺质子的争夺.4*acidity strength of acids and bases*acidity strength of acids and bases 酸碱性的强弱酸碱性的强弱:Acid Acid11+Base+Base22 Acid Acid22+Base+Base11 K K 1 1Conclusion:Conclusion:ororFor example:HCl+HFor example:HCl+H22O O H H33OO+Cl+Cl K K 1 1 H H22O+NHO+NH33 NH NH44+OH+OH KK
7、H H22O;O;由由(2)(2)式式，NHNH44+H H22OO。若对三者排序，则需若对三者排序，则需:HCl+NH HCl+NH33=NH=NH44+Cl+Cl 5Acid-Base Equilibrium 酸碱平衡The dissociation of monoprotonic acidsThe dissociation of monoprotonic acids一一元酸的解离元酸的解离:The common monoprotonic acids are:acetic acid,boric The common monoprotonic acids are:acetic acid,bo
8、ric acid,nitrous acid,hypochlorous acid,formic acid,lactic acid,nitrous acid,hypochlorous acid,formic acid,lactic acid,benzoic acid.acid,benzoic acid.醋酸、硼酸、亚硝酸、次氯酸、蚁酸、醋酸、硼酸、亚硝酸、次氯酸、蚁酸、乳酸、苯甲酸乳酸、苯甲酸The common biprotonic acids are:hydrogen sulfide,The common biprotonic acids are:hydrogen sulfide,carbon
9、ic acid,sulfurous acid,oxalic acid.carbonic acid,sulfurous acid,oxalic acid.硫化氢、碳酸、硫化氢、碳酸、亚硫酸、草酸等亚硫酸、草酸等6ExampleExample:Calculate the pH value of 0.1 M solution of:Calculate the pH value of 0.1 M solution of acetic acid.acetic acid.计算计算0.1 0.1 molmolLL11 HAc HAc的的pHpH值值。KKaa=1.81.8101055 SolutionSol
10、ution:In the solution of acetic acid,there exist,:In the solution of acetic acid,there exist,Initial concentration 0.1 0 0Initial concentration 0.1 0 0concentration at equilibrium 0.1concentration at equilibrium 0.1 x x x x x x采用牛顿逼近法。先假定采用牛顿逼近法。先假定0.1 0.1 x=0.1,x=0.1,得得7所以，所以，HH+=1.32=1.32101033 mo
11、lmolLL11.验证假设。将验证假设。将HAc=0.1 HAc=0.1 1.321.32101033=0.99=0.99 molmolLL11代代入上式，得入上式，得 HH+=1.32=1.32101033 molmolLL11.根据根据pHpH值的定义值的定义:Rewrite the above equation,we haveRewrite the above equation,we havepH=pH=loglogHH+=loglog(1.321.32101033)=2.88)=2.88QuestionQuestion:Why the self-dissociation of wate
12、r is not:Why the self-dissociation of water is not taken into consideration?taken into consideration?8The dissociation of weak baseammoniaThe dissociation of weak baseammonia 弱碱弱碱的解离的解离:Calculation equation of the pH value of weak bases:Calculation equation of the pH value of weak bases:ExampleExamp
13、le:Calculate the pH value of 0.1:Calculate the pH value of 0.1 molmolLL11 NaAc NaAc.SolutionSolution:the acetic acid is a weak acid,so its:the acetic acid is a weak acid,so its conjugated base is a rather stronger baseconjugated base is a rather stronger base 醋酸是弱酸，醋酸是弱酸，其共轭碱其共轭碱AcAc的碱性要强些的碱性要强些.9因因
14、酸及其它的共轭碱的解离常数的关系是酸及其它的共轭碱的解离常数的关系是:KKww=KKaa KKbb(自行证明自行证明)。利用公式利用公式:所以，所以，OHOH=7.45=7.45101066 molmolLL11.pOH=pOH=loglogOHOH=loglog(7.45(7.45101066)=5.12)=5.12 pH=14 pH=14 pH=8.88 pH=8.8810pH in Our Daily LifeBasic Basic 碱性碱性AcidicAcidic 酸性酸性1 2 3 4 5 6 7 8 9 10 11 12 13 141 2 3 4 5 6 7 8 9 10 11 1
15、2 13 14Lye Lye碱液 碱液Ammonia AmmoniaMilk of magnesia Milk of magnesia氧化镁乳剂 氧化镁乳剂Sea water Sea waterHuman blood Human bloodmilk milk“pure”“pure”acid rain acid rain pH=5.6 pH=5.6Fish die Fish dievinegar vinegarMost Most acid acid rains rainsBattery acid Battery acidBaking soda Baking sodaSalt lake Salt
16、lakeHuman urine Human urine Saliva Saliva 唾液 唾液Tomato Tomato juice juiceAverage Average acid rain acid rain Apples ApplesLemon juice Lemon juice1 1Buffer Solutions缓冲溶液 Calculate the pH of mixture of 0.1 molmolLL11 HAc and 0.1 molmolLL11 NaAc.Initial concentration 0.1 0 0.1Initial concentration 0.1 0
17、 0.1concentration at equilibrium 0.1concentration at equilibrium 0.1x x 0.1+xx x 0.1+x采用牛顿逼近法。先假定采用牛顿逼近法。先假定0.10.1x=0.1,0.1+x=0.1,x=0.1,0.1+x=0.1,得得12Rewrite the above equation,we haveRewrite the above equation,we have所以，所以，HH+=1.8=1.8101055 molmolLL11.验证假设。将验证假设。将HAc=0.1 HAc=0.1 1.81.8101055=0.0999
18、8=0.09998 molmolLL11、NaAc=0.1000 NaAc=0.1000 molmolLL11代入上式，得代入上式，得 HH+=1.8=1.8101055 molmolLL11.根据根据pHpH值的定义值的定义:pH=pH=loglogHH+=loglog(1.81.8101055)=4.74)=4.7413The calculation equation of the pH of buffers is:The calculation equation of the pH of buffers is:两边同乘两边同乘 OHOH，因因OHOHHH+=KKww，而酸及其它的共轭碱的
19、解离常数而酸及其它的共轭碱的解离常数的关系是的关系是:KKww=KKaa KKbb(自行证明自行证明)。上式可写为。上式可写为:14Factors Affecting on Hydrolysis of Salts 影响盐类水解的因素The nature of saltsThe nature of saltsdilution with waterdilution with watertemperaturetemperaturethe pH value of the solutionthe pH value of the solution15Nature of SaltsHydrolysis is
20、 the reversed reaction of neutralization.Only very small amount of salts hydrolyzed.The weaker the resulting acid or base is,the greater the degree of hydrolysis,h,the salt solution has.h=salthydrolyzed/saltinit.100%E.g.for the solution of NaAc,h=0.0087%;for the solution of KCN,h=1.2%16Dilution with
21、 WaterIf the solution of NaAC is diluted to the volume of beingtwo times to the original,Ac+H2O HAc+OHKhoNumeratordenominatorThe numerator now is 1/4 of the original,whereas thedenominator is half of the original.The reaction quotientbecomes smaller.Finally,the equilibrium will shift to the right.17
22、temperaturethe hydrolytic reaction is endothermic,it is favored as temperature rises.The pH value of solutionsif the solution is acidic,hydrolysis is favored at high pH;if the solution is basic,hydrolysis is favored at low pH.18The Precipitation and Dissolution Reactions of Insoluble Electrolytes难溶电
23、解质的沉淀-溶解反应 A slightly soluble electrolyte dissolves until aA slightly soluble electrolyte dissolves until asaturated solution is formed.An multi-phase saturated solution is formed.An multi-phase equilibrium with the undissolved solute establishes.equilibrium with the undissolved solute establishes.m
24、mAAn+n+(aqaq)+)+nnBBmm-(aqaq)dissolution dissolutionprecipitation precipitationAmBn(AmBn(ss)19The Solubility and Solubility Product Constant 溶解度与溶度积Solubility(Solubility(SS):the maximum amount of):the maximum amount of dissolved materials,dissolved materials,usually in mol L 1.For different types of
25、 slight soluble electrolytes,For different types of slight soluble electrolytes,AB type:AB type:Kspo=S 2ABAB22、AA22B type:B type:Kspo=4S 3ABAB33、AA33B type:B type:Kspo=27S420The Rule of Solubility Product溶度积规则For a slightly soluble electrolyte having the For a slightly soluble electrolyte having the
26、 precipitation,precipitation,A AmmBBnn(s)(s)mmAAn+n+(aq)+(aq)+nnBBm-m-(aq)(aq)the reaction quotient the reaction quotient QQ=A=Ann+mmBBm-m-n n could becould be21Solubility Product溶度积If Q=Ksp,the dissolution is at equilibrium.If Q Ksp,the dissolution isnt at equilibrium.The solution is supersaturated
27、.The solid precipitates.Note:The supersaturated solution is in a metastable state.22ExampleExample:5 mL of 0.2 5 mL of 0.2 molmolLL11 solution of MgCl solution of MgCl22 and and equal volume of 0.1 equal volume of 0.1 molmolLL11 solution of ammonia is solution of ammonia is mixed.Does Mg(OH)mixed.Do
28、es Mg(OH)22 precipitates in the mixture?precipitates in the mixture?Calculate the amount of solid NHCalculate the amount of solid NH44Cl that is required to Cl that is required to prevent the precipitation of Mg(OH)prevent the precipitation of Mg(OH)22.(K.(Kspsp=5.615.6110101212,K,Kbb=1=1.8.8101055.
29、SolutionSolution:Just after mixing of the two solutions,Just after mixing of the two solutions,Mg Mg2+2+=0.2=0.2 2=0.1 2=0.1 molmolLL11;NH NH33=0.1=0.1 2=0.05 2=0.05 molmolLL11 the hydroxide ions supplied by ammonia.Use simplified the hydroxide ions supplied by ammonia.Use simplifiedcalculation.calc
30、ulation.23Calculate the Calculate the QQ just after mixing,just after mixing,QQ=Mg=Mg2+2+OH OH22=0.1=0.1(9.4(9.4101044)22=8.8 8.8 101088 KKspspSo,magnesium hydroxide will precipitate.So,magnesium hydroxide will precipitate.Mg(OH)Mg(OH)2 2(s)+2NH(s)+2NH4 4+(aq)Mg(aq)Mg2+2+(aq)+2NH(aq)+2NH3 3H H2 2O(a
31、q)O(aq)is composed of the two equilibria:is composed of the two equilibria:NHNH44+=0.54 molL=0.54 molL11,weight=0.54,weight=0.540.010.0153.5=0.2953.5=0.29 克克 The dissolution equilibrium of Mg(OH)The dissolution equilibrium of Mg(OH)22 as following as following Mg(OH)Mg(OH)22(s)Mg(s)Mg2+2+(aq)+2OH(aq
32、)+2OH(aq)(aq)-2-2 NH NH33HH22O(aq)NHO(aq)NH44+(aq)+OH(aq)+OH(aq)(aq)using the rule of the Multi Equilibria,we have using the rule of the Multi Equilibria,we have 24Dissolution of the Precipitates沉淀的溶解Dissolution by the formation of a very weak electrolyteDissolution by redox reactionsDissolution by
33、the formation of complexes25Dissolution by the Formation of Very Weak ElectrolytesExample:Magnesium hydroxide Mg(OH)2,dissolves in the solutionof ammonium chloride(NH4)Cl.Example:Mg(OH)2(s)Mg 2+2OH-2NH4Cl+2NH3.H2O2Cl-+2NH4+It is because Q Kspo26Dissolution by Redox Reactions Some metal sulfides 硫化物
34、dissolves in hydrochloric acidHCl,E.g.FeS+2H+(aq)Fe2+(aq)+H2S(aq)ferrous sulfide ferrous ion hydrosulfric acidcopper sulfide has a extremely small Kspo,it only dissolves in concentrated nitric acid,HNO3,3CuS(s)+8H+(aq)3Cu2+(aq)+3S(s)+2NO(g)+4H2O+6NO3-(aq)27Dissolution by Formation of ComplexesA slig
35、htly soluble electrolyte,AgCl,dissolves in 6 Mammonia.AgCl(s)+2NH3(aq)Ag(NH3)2+(aq)+Cl-(aq)28Fractional Precipitation分步沉淀In order to use the natural resources,the In order to use the natural resources,the impurities in mineral ores such as calciteimpurities in mineral ores such as calcite(石灰石石灰石),zi
36、nc blende(),zinc blende(闪锌矿闪锌矿),pyrite(),pyrite(黄铁黄铁矿矿),apatite(),apatite(磷灰石磷灰石),gypsum(),gypsum(石膏石膏)should be eliminated.should be eliminated.If a precipitant is added to the solution If a precipitant is added to the solution containing more than one precipitate ions,containing more than one prec
37、ipitate ions,the precipitates are formed at different the precipitates are formed at different stages.This effect is called fractional stages.This effect is called fractional precipitation,or ion separation.precipitation,or ion separation.29Example:When sodium iodide(NaI)is added dropping Example:Wh
38、en sodium iodide(NaI)is added dropping to a solution containing the same concentrations of to a solution containing the same concentrations of silver(Agsilver(Ag+)and lead(Pb)and lead(Pb2+2+)ions(e.g.0.01)ions(e.g.0.01 mol L1).AgI(s)Ag AgI(s)Ag+(aq)+I(aq)+I(aq)(aq)PbI PbI22(s)Pb(s)Pb2+2+(aq)+2I(aq)+
39、2I(aq)(aq)the I the I required to precipitate the Ag required to precipitate the Ag+ion and Pb ion and Pb2+2+are,are,Therefore,PbI2 begin to precipitate until all the Ag+ions are already precipitated(now Ag+105 mol L1).30Coordination complexes and equilibrium between complexes in aqueous solution配位化
40、合物及水溶液中的配位平衡31Definition 定义A coordination compound 配位化合物 is one in which the metal ion or atom is bonded to one or more neutral molecules or anions so as to be a defined and integral structural unit.Often,we call it coordination complex.配位化合物是中心原子与一定数量的分子或离子形成的具有一定的空间结构的配离子或分子，有时称络合物。32Central Atom,
41、Ligand,and Coordination Bonding中心原子、配体和配位键The metal ion or atom is called central ion or atom.金属原子或离子为中心离子或原子The molecules or ions bonded to the central metal ion are called ligands 配体.For example,NH3,Cl etc.与中心原子键合的分子或离子称为配体。The ligands donate the lone pair electrons to the central metal to form th
42、e covalent bondings,called coordination bondings.配体提供原子对与金属原子形成的共价键称为配位键。33Coordination Ion,Coordination Atom 配离子，配位原子Mn+NH3 M NH3n+,coordination ion 配离子In ligands,the atom which donates its lone pair electrons to the central ion is called coordination atom 配体中，提供电子对与金属原子形成配位键的原子，称 为配位原子.If a ligand
43、 has more than one coordination atom is called polydentate or“many-toothed”配体中，若有多个配位原子，称为多啮配体.34Chelate螯合物the complex formed between a central and one or several polydentate ligand(s)多啮配体 is called a chelate.与多啮配体形成的配合物，称为螯合物。For example,Cu(en)22+,二乙二胺合铜(II)Here,en:NH2CH2CH2NH2,ethylenediamine 乙二胺.
44、35Some Common Ligands常见配体 monodentate 单啮配体:H2O aqua,NH3 ammine,Cl chloro,Br bromo,I iodo,CN cyano,SCN thiocyanato,CO carbonyl.bidentate 双啮配体:CO32 carbonato,C2O42 oxalato,en:ethylenediamine.polydentate 多啮配体:EDTA ethylenediamine tetraacetic acid.Coordination number is the number of nearest coordinatio
45、n atoms surrounding the central ion.E.g.the coordination number is 4 in Cu(en)22+.离 中心原子距离最近的配位原子数配位数。如 Cu(en)22+的配位数为4。36Bidentate and Polydentate LigandsOxalato 草酸根 2,2-bipyridine 联吡啶EDTA,乙二胺四乙酸37Nomenclature of Coordination Compounds 配合物命名if a coordiantion compound is ionic,name the cation first
46、and the anion second.先命名阳离子，再阴离子name the ligands first,followed by the central metal.在配离子中，先命名配体，再命名金属。name the ligands alphabetically按字母顺序命名配体.Negative ligands(anions)have names formed by adding“o”to the stem name of the group.For example,O2 oxo,NO3 nitrato.use the numeric prefix to indicate the nu
47、mber of ligands present.用数字前缀表示配体数量。2:di 或 bis；3:tri 或 tris；4:tetra 或 tetrakis etc.38Examples of the Names of Complexes配合物命名范例K3Ag(S2O3)potassium bis(thiosulfato)argentate(I).二硫代硫酸根合银(I)酸钾。Co(en)(NO2)Cl2 dichloroethylenediaminenitrocobaltate(III).二氯二硝基乙二胺合钴(III)配离子。Cu(NH3)4SO4 tetraamminecopper(II)s
48、ulfate 硫酸四氨合铜(II)Co2(CO)8 octacarbonyl dicobalt(0)八羰基合二钴。39Coordination Equilibrium配离子的配合平衡Cu(NH3)42+Cu2+4NH3 dissociationcomplexationIf we define the overall formation constant K稳,Cu2+4NH3 Cu(NH3)42+dissociationcomplexation40Formation constant and disassociation constant稳定常数和不稳定常数 The formation of
49、a complex is in step,each step has its own stepwise formation constant n 累积稳定常数,K稳=1 2 3 n.Cu(NH3)42+Cu2+4NH3 dissociationcomplexation41CalculationsExample 1:0.1 mole of solid copper sulfate is added to the solution of 1 L of 6.0 mol L1 solution of ammonia.Which forms,solid Cu(IO3)2 or CuS,when the
50、solid of iodates(IO32)and sulfides(S2)are added to the resulting solution respectively.(see page 124 on Textbook)Cu2+4NH3 Cu(NH3)42+initial 0.1 6.0 0at equilibrium x 5.6 0.1Cu2+free=4.87 1018 mol L1 42If solid iodate is added to the solution,Cu2+4NH3 Cu(NH3)42+(1)Cu(IO3)2 Cu2+IO32(2)the precipitatio