0.01 解:Cdl-= 10 MolL 10 5.1 axca=1+101+101059ax(=1046 Ig ax.9 lgK(ZnX)=145-59-86>8,能准确滴定, 指示剂:XO pZn=1/2(lgK'+pc(Zn)=1/2(86+2.0)=53 Zn2}=1053mo/L, acdo= Cd+2lCdHiy/Cd ∑[CdIL=105171-101=1020mo/L解: [Cd]=  X(Cd)=1+10-7.1+13.0=105.9>  X(H)=104.6 lg  X=5.9 lgK(ZnX)=14.5-5.9=8.6>8, 能准确滴定, 指示剂:XO pZn=1/2(lgK +pc(Zn))=1/2(8.6+2.0)=5.3 [Zn2+]=10-5.3 mol/L, -7.1 5.1 0.01 10 mol/L 10 =  Cd(I)={[Cd]+∑[CdIi]}/[Cd] ∑[CdIi ] =105.1-7.1 -10-7.1=10-2.0mol/L