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1、Buffer solution and some related calculations1.The common-ion effect (1)The homogenous common-ion effect A strong electrolyte with the same common ions as the weak electrolyte is added to the weak electrolyte solution,the ionization equilibrium of weak electrolyte will move to the left,resulting in
2、its decreased ionization degree.Examples:NH3H2O-NH4Ac NH3H2O =NH4+OH-NH4Ac =NH4+Ac-Known that:Ka(HAc)=1.8 10-5 Solve:HAc(aq)+H2O(l)=Ac-(aq)+H3O+(aq)Initial con.:0.10Eq.con.:0.10-x xxEq.con.after added NH4Ac:0.10-x0.10+xxKa(HAc)=c(Ac-)c(H3O+)c(HAc)=x(0.10+x)0.10-xx=1.8 105 c(H3O+)=1.8 105 mol L-1-1 p
3、H=4.74=x0.1-x=0.018%HAc(aq)+H2O(l)=Ac-(aq)+H3O+(aq)Eq.con.:0.10-xxxKa(HAc)=(x)20.10-xpH=2.89=1.3%The decreased solubility of AgCl(2).The multi identical-ion effects A strong electrolyte with the same ions as the insoluble electrolyte is added to the solution of the insoluble electrolyte,resulting in
4、 the decreased solubility of the insoluble electrolyte.Example:AgCl-NaCl AgCl(s)=Ag+Cl-NaCl =Na+Cl-50 mL HAc-NaAc c(HAc)=c(NaAc)=0.10molL-1,pH=4.74The Buffer solution:which can keep the relative unchanged of its pH value.In other words,the pH value of the solution system will not be significantly ch
5、anged by adding a small amount of strong acid or base.Added 1 drop(0.05 mL)1molL-1 HClAdded 1 drop(0.05mL)1molL-1 NaOHExperiment:pH=7 pH=3 pH=11 pH=4.73 pH=4.752.Buffer solution50 mL pure H2ODefinition:can stabilize the pH value of the system.HAc-NaAc,NH3H2O-NH4Cl,NaH2PO4-Na2HPO4Buffer pairs:a pair
6、of conjugated acids and bases in the buffer solution.How to separate Al3+and Mg2+?Al3+Mg2+NH3H2O-NH4Cl Al(OH)3Mg2+2.Buffer solution3.Calculation of pH value of buffer solutionBuffer solution composed of weak acid and its conjugated baseHAc(aq)+H2O(l)=H3O+(aq)+Ac-(aq)pH=pKa-lgCa和和Cs为平衡浓度，约等于酸和共轭碱的开始浓度为平衡浓度，约等于酸和共轭碱的开始浓度Strong acid and weak basic solution NH3-NH4Cl pOH=pKb-lgpH=14-pOH=14-pKb+lg
