第八章沉淀滴定法讲解

K[M2+]1.1.已知=4水=200，忽略离子强度的影响，计算CaSO4的固有溶[Ca2+][SO2]4[CaSO](1)[CaSO]＝s4水0ss=0=0=200ss[Ca2+][SO2]K4sps=K=2009.1106=1.82103mol.L1sp(2)未解离形式%=s0100%=s0100%s+[Ca2+]s+K0spNaOH，忽略体积变化和各种氢氧基络合物，计算下列不同情况生成沉淀时KK[M2+]KK[M2+]KK[M2+]2Ba1+0.00328根500根44444sp亭s=K0K444444spy.yspy.y41.14.考虑酸效应，计算下列微溶化合物的溶解度：aKKspKaKK+[H+]asp42002121p81.11010ss1KKaaa1.31077.11015+1.3107100.5+100.52K=[Cu2+][S2]=9.21022s=610362BaSO2487.11015=s.=10-7.4s1072+10-71.3107+7.110151.3107K=[Cu2+][S2]=10-7.4.s2=61036HH[H+]21041042(2)考虑到MnS的水解反应21=K.K.sp,MnSwKa[Ag+][Br]=s(s+s)=5.01013112AgClsss=1.81010212LCO2-KKK6.41056.41055.910224aaa221s=Ka=2.0109100.41=7.2105mol.L1spCO2-24CO1056.41055.910224K[CO2-]2433CaCO3233=[Ca2+][CO2-].[OH-][H+].[HCO-]33[CO2-][H+]3KKpCaCOwK淀，应加入氨的总浓度是多少(忽略溶液体积变化)？K=[Ag+][Cl]=[Cl]spa3Ag(NH)3a==105.74a+103.24[NH]+107.05[NH]2=105.74Ag(NH)33NH3[Ag+]=s的溶液。问：(1)EDTA对沉淀的络合效应是否可以忽略？(2)能否生成33Y(H)aKY1aYEDTA的络合效应可以忽略损失24HSOaaHSO2HSOaaHSO2422s=[Ba2+]=Kspa+103.242.0+107.052.02=107.65Ag(NH)s=Ka=5.010一13107.65＝4.7310-3M3spAg(NH)32)pH=8.0a=102.27[EDTA]=0.010=10-4.27Ba(EDTA)s=Ka=1.110一10103.59=6.510-4MspBa(EDTA)33pH9.710.0lgaY(H)1.381.21.00.800.700.45[BaY][BaY]K==KBaY[Ba][Y][Ba]c/aYY(H)[Ba][BaY]0.10cY(H)BaYYY(H)42424a,HF0.005-[H+]+[OH-]0.005-[H+]3Ag(NH)MNHM34pH=pK+lg[NH3]=9.26+lg0.05=8.26a[NH+]0.54因此没有沉淀生成pHpKlgNHlg.2=9.56a[NH+]0.143Ag(NH)3sKKHaaa1[NH]=Kac=10一9.26人0.3=10一0.97M3K+[H+]NH310一9+10一9.26a3CdNH)3aS2一103.6533Zn(OH)1234Zn(OH)a=1+100.4+102.1+102.2+100.5Zn(OH)Zn(OH)K=[Zn2+][OH]2=s[OH]2aaZn(OH)主要存在形式为Zn(OH)与Zn(OH)32(1)K=[Ag+][Cl]=[Cl]ssp1+[Cl]+[Cl]22s=K(1++[Cl])sp[Cl]120.1(2)s=K(1++[Cl])sp[Cl]12其中当1＝[Cl]时，才能使s最小，此时[Cl]221KIO1I1Ag+x可以得到分子式为KIO3 (4)根据(C9H6NO)3Al测定Al2O3。234M151.99F=Cr2O342227227Mg22742434343434343254343434343434)1AlO--2(CHNO)Al23963F=AlO23963(CH96323.推导一元弱酸盐的微溶化合物MA在下列溶液中溶解度的计算公式：1)强酸溶液中；2)酸性溶液中和过量沉淀剂A-存在下；M；4)在过量络合剂L存在下(只形成ML络合物)的酸性溶液中KKKK+[H+]c2KM2+a(4)K=[M2+][A-]2=s〉(2s〉Ka)2gAsO淀为Ag3AsO4。将沉淀过滤、洗涤，最后将沉淀溶于酸中。以0.1000mol/L434As〉100%=22.70%0.5000CaCOMgCO2424mCaC2O42424CaCO40.6240MgCO240.62403eOOAlO231AlO2Al3 NaClNaBrNaClNaBrNaBr0.6280lm＝m＝AgClM