北京大學定量分析化學簡明教程習題

1、北京大學定量分析化學簡明教程習題第四章絡(luò)合滴定法1.已知銅氨絡(luò)合物各級不穩(wěn)定常數(shù)為：K不穩(wěn)i=7.8父10-3K不穩(wěn)2=1.4x10-3K不穩(wěn)3=3.3父10-4K不穩(wěn)4=7.4父10-5(1)計算各級穩(wěn)定常數(shù)K1K4和各積累常數(shù)P1-P4；(2)若銅氨絡(luò)合物水溶液中Cu(NH3)42+的濃度為Cu(NH3)32+的10倍，問溶液中NH3是多少？(3)若銅氨絡(luò)合物溶液的Cnh3=1.010-2M,Ccu2+=1.0m10-4M,(忽略Cu2+,NH3的副反應)。計算Cu2+與各級銅氨絡(luò)合物的濃度。此時溶液中以那種形體為最主要？解：(1)穩(wěn)定常數(shù)“114K1=5=1.4父10K不穩(wěn)47.4M10

2、-5.11一一一3K2=-4=3.010K不穩(wěn)33.3M10112K3=3-=7.1x10K不穩(wěn)21.4X10-3一：4=KiK2K3K4=1.43.07.11.31011=3.9101222=Cu(NH3)3、=Cu(NH3)41-"Cu2NH33,-"Cu2NH34工Cu(NH3)42Cu2NH33*一Cu2NH34Cu(NH3)32-2-_Cu(NH3)41-2-Cu(NH3”NH3=103.01010_123.910：，0=0.077(ml/l)1，NH3：2NH32-NH33-4NH34_1-4_2_7_410-612 -8 1.4 104.2 103.0 103

3、.9 10;14.2 103 3.0 104 3.9 104_17.3 104= 1.4 10-5=：iNH3-1-iNH3-2NH32-3NH33 -NH341.41027.3104=1.910-3,NH32q-2=：n：5：71-1NH32NH32-3NH334NH34_4,2103一47.310=0.0583=_Nf31 ：1小也：2NH32：3NH33：4NH33.9 107,3 104= 0.53Cu2 = ：%CCu2：；=1.4 10,10 1.4 10，(mol/l)Cu(NH3)2 =中1CCu2 =1.9 104 10，1.9 101mol/l)Cu(NH3)22* =62

4、CCu2 +=0.058父10” = 5.8父10" (mol/l)Cu(NH3)32=包2+ = 0.41M10" =4.1M10”(mol/l)Cu(NH3)42 - CCu2；i： -0.53 10"=5.3 101mol/l)答：主要形體為 Cu(NH3)32+和 Cu(NH3)42+。4_3,0104-47.310=0.414=小山41YNH32NH323NH334NH342.(1)計算pH5.5時EDTA溶液的lgo(Y(H)值；(2)查出pH1,2,/。時EDTA的IguY(H)值，并在坐標紙上做出lg口Y(H)-pH曲線，由圖查出pH5.5時的Ig

5、otY(H)值，與計算值相比較。解：aY(H)=1+PiH+P2H+2+P3H+3+44H+4+P5H+5+P6H+6_1,Hi.H2H3H4Ka6Ka6Ka5KabKasKa4Ka6Ka5Ka，Ka3H5H6Ka6Ka5Ka4Ka3Ka2尢6&5&4&3(2&1二1.也105.51010皿4_22.0I + -_21.35.也10二1.06.58.獨10二6.519.2810 N.510必5J33.010_23.854.84 , _5.582.78= 1+10 +10 +10 +100.65+10 4.55+10 9.15=104.84+105.58=6.91

6、04+3.81055=4.2101g:y(h)=5.63.計算 lg a Cd(NH3)、lga Cd(OH)和 lg a Cd 值。(Cd2+-OH-絡(luò)合物的lgP1一lg3分別是4.3、7.7、10.3、12.0)。(1)含鎘溶液中NH3=NH4+=0.1;加入少量NaOH于(1)液中至pH為10.0。二Cd(NH3)=1+：1NH3+一：2NH32+-：3NH33+-：4NH34+-：5NH35+-：6NH36=1+102.60-1.0+104.65-2.0+106.04-3.0+106.92-4.0+106.6-5.0+104.9-6.0=102.65+103.04+102.92_3=

7、2.41033.38=10lg二Cd(NH3)=3.38+_當NH3=NH4=0.1pH=pKa=9.25,pOH=4.75acd(0H-)=1+Pi0H-+p2OH-2+p3OH-3+"OH-4=1+1043-475+1077-9+1013-1425+101Z0-19=1+10-0.45+10-1.8=1.4lg«Cd(OH)=0.14口Cd(OH)=CCd(NH3)+久Cd(OH)一1=103.38+100.14-1=103.38lgacd=3.38=3.4(2)pH=10.0+CaH=Ka-Ca=5.610-100.2CbCb_.二0c0.25.610C.二105.6

8、10=0.17(mol/l)=10-0.77二Cd(NH3)=1+：1NH3+:2NH32+：3NH33+4NH34+5NH35+6NH36=1+102.60-0.77+104.65-1.54+106.042.31+106.92-3.08+106.6-3.85+104.9462=1+101.83+103.11+103.73+103.84+102.75+100.284.15=10lg二Cd(NH3)=4.15二cd(0H)=1+1OH-+一：2OH-2+3OH-3+4OH-4=1+104.3-4.0+107.7-8.0+1010.3-12.0+1012.0-16.0=1+100.3+10-0.3

9、+10-1.7=3.520.55=10lg二Cd(OH)=0.55?Cd=Cd(NH3)+=Cd(OH)-1=104.15+100.55-14.15=10lg二Cd=4.154.計算下面兩種情況下的lgK,NiY值(1) pH=9.0,CNH3=0.2M;(2) pH=9.0,Cnh3=0.2M,Ccn-=0.01M。解：(1)lgK'NiY=lgKNiYlgUNilgUY工Ni(NH3)=1+I,1NH3+:2NH32+：3NH33+4NH34+5NH35+6NH36Ka5.610,°又：NH3=CnH3=0.2109KaH5.6101.010-114=0.072=10：-

10、Ni(NH3)=1+102.75-1.14+104.95-2.28+106.64-3.42+107.79-4.56+108.50-5.70+108.49-6.84=1+101.61+102.67+103.22+103.23+102.8+101.653.66=1014.工丫(H)=10lgK卜丫=18.6-3.66-1.4=13.54(2):Ni(CN-)=1+一：4OH-4pH=9.0K4910-10CN*cN-K-Hr0.014.910-0=3.310-3=10-2.5,aNi(CN-)=1+1031.3-10=1021.3CtNi=口Ni(NH3)+aNi(CN-)+aNi(OH-)-2=

11、103.66+1021,100.1-221.3=101.4-Y(H)=10lgKNiY=lgKNiYlg二Ylg二Ni= 18.6-1.4-21.3=8.25.以2M10-2MEDTA滴定用濃度的Pb2+溶液，若滴定開始時溶液的pH=10,酒石酸的分析濃度為0.2M,計算等當點時的lgK,PbY,Pb和酒石酸鉛絡(luò)合物的濃度。（酒石酸鉛絡(luò)合物的lgK為3.8）。解：L=Cl點火匯HKa1HKa1Ka2_4_5CC9.1104.310:0.277«109.1109.14.310=O2(mol/l)到等當點時L=0.1mol/l，Pb(L)=1+：iL3.8-1=1+102.8=102.7-Pb(OH-)=10工Pb=Pb(L)+Pb(OH-)-1=102.8+102.7-13.05=10上y（h）=0.5lgKPbY=lgKPbY-lg二Y-lg二Pb=18.0-0.5-3.05=14.41，八,、pPbeg=-(pCPbeg+