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第六章习题课第六章习题课络合滴定法习题课6-4已知乙酰丙酮(L)与Al3+络合物的累积形成常数lg1~lg3分别为8.6,15.5和21.3,(1)AlL3为主要型体时的pL范围是多少?(2)[AlL]与[AlL2]相等时的pL为多少?(3)pL为10.0时铝的主要型体又是什么?解:(1)AlL3为主要型体时:pLlgK3=5.8(2)[AlL]=[AlL2]时,pL=lgK2=6.9(3)pL=10.08.6∴Al3+为主要型体8.5lg3.21lglglglg9.6lg5.15lglglg6.8lg6.8lglg332133213221221211111KKKKKKKKKKKKKKK因为由相邻两级络合物分布曲线的交点处有pLi=lgKiAl3+—AlL—AlL2—AlL3pL8.66.95.86-13将100mL0.020mol·L-1Cu2+溶液与100mL0.28mol·L-1氨水相混后,溶液中浓度最大的型体是哪一种?其平衡浓度为多少?解:假设溶液中Cu2+与NH3形成的配合物主要是以Cu(NH3)42+形式存在。等体积相混合后,Cu2+和NH3的浓度分别为:1010.02020.02LmolcCu114.0228.03LmolcNH溶液中游离氨的浓度为:110.04010.014.03LmolcNH查表:Cu(NH3)42+lgβ1=104.15;lgβ2=107.63;lgβ3=1010.53;lgβ4=4.7×1012.6744332210][][][][11][LLLLMMMc00.467.1200.353.1000.263.700.115.40101010101010101011Cu70.867.853.763.515.310101010101103.070.800.467.120434])([417.170.800.353.100333])([307.370.800.263.70232])([255.570.800.115.4031])([110101010]NH[10101010]NH[10101010]NH[10101010]NH[24323322323NHCuNHCuNHCuNHCu108.67+107.53+105.63=107.6P349表9指数加法表:8.67-7.53=1.14→查1.14对应值0.030→8.67+0.03=8.708.70-5.63=3.0722243)(243])([CuNHCucNHCu故溶液中主要型体为Cu(NH3)42+:)(11LmolLmol32.032.000.03109.31010106-15在含有Ni2+-NH3络合物的溶液中,若Ni(NH3)42+的浓度10倍于Ni(NH3)32+的浓度,问此体系中游离氨的浓度[NH3]等于多少?解:Ni(NH3)62+的lgβ1-lgβ4分别为:2.80;5.04;6.77;7.963343]][[]][[NHNi])[Ni(NHNHNi])[Ni(NH2233322434334233243NH])[Ni(NH])[Ni(NH][])[Ni(NH])[Ni(NHNH233424333][323233NHNi])[Ni(NH]][[3424243NHNi])[Ni(NH]][[3)(....11909677764336401010101010][NHLmol])[Ni(NH])10[Ni(NHNH233423333][43.5511210107.2HHHKK422-42OCHHOHC20.41210171aHKK6-18计算在pH=1.0时草酸根的值。解法一:查表得)(242lgHOC5221104.6,109.5aaKK20.4421110106.11aHHKK求酸效应系数-42-242OHCHOC2H2H1(H)OC][H][H1242ββα酸效应系数3.63lg(H)OC242α3.633.433.2010101012.005.431.004.20101010101212112][][/1aaaaaKKKKKHH242242OCOC525222106.4105.9106.4105.9105.90.1103.626632104185.2103.8103.8105.9103.62lg10lg3.62(H)OC242根据则解法二:查表得5221104.6,109.5aaKK6-19今有pH=5.5的某溶液,其中Cd2+,Mg2+和EDTA的溶液均为1.0×10-2mol•L-1。对于EDTA与Cd2+的主反应,计算其αY值。解:EDTA与Cd2+的主反应,受到酸效应和共存离子效应的影响。查附录一之表4,pH=5.5时,lgY(H)=5.51由附录一之表3可知,KCdY=1016.46,KMgY=108.7由于络合物的形成常数相差很大,可认为EDTA与Cd2+的反应完成时,溶液中EDTA的浓度非常低,[Mg2+]≈0.010mol•L-16.728.72MgYY(Mg)1010101][Mg1K6.736.75.51Y(Mg)Y(H)Y10110101105.51+106.7=107.6P349表9指数加法表:6.7-5.51=1.19→查1.19对应值0.027→6.7+0.03=6.736-23若将0.020mol•L-1EDTA与0.010mol•L-1Mg(NO3)2(两者体积相等)相混合,问在pH=9.0时溶液中游离Mg2+的浓度是多少?。解:当EDTA溶液与Mg2+溶液等体积相混合之后,EDTA和Mg2+的浓度分别为:1010020200LmolcEDTA..100500201002LmolcMg..查表得,当溶液pH=9.0时,281.lg)(HY7.8lgMgYK0lg)(OHM)('lglglgHYMgYMgYKK故7MgY'1063.2K当EDTA与Mg2+混合后,发生如下配位反应:Mg2++Y4-MgY2-反应前:0.00500.0100mol•L-1反应后:x(0.010-0.0050)0.0050mol•L-1=0.005042.728.17.810050.0]'[][LmolYMgY当反应达平衡时:)(][12Lmol87.42103.810Mg0.0050100.0050YMgYMg7.42MgY'2]'[][][KMgY2YMgMgY']'][[][K∴6-25在一定条件下,用0.01000mol•L-1EDTA滴定20.00mL1.0×10-2mol•L-1金属离子M。已知此时反应是完全的,在加入19.98-20.02mLEDTA时的pM值改变1个单位,计算MY络合物的K’MY。解法一:cY=cM=0.010mol•L-1,,M与Y反应完全,可忽略MY在计量点的解离,VM=20.00mL1spM,0050.0Lmolc2.计量点后0.1%:在加入20.02mLEDTA时,pM改变了1个单位,为5.30+1=6.30。此时:1.计量点前0.1%:VY=19.98mL(计量点时,)8052306305p...spM)(100.5010.098.1900.2098.1900.20]M[16LmolcVVVVMYMYM3ppMspM,0.1%c20.01020.000.020.01019.9820.0019.9820.00M][30.5330.23)105lg(3%1.0pM30.50.1%pM'MYMYM'MY)KV(VV]K[Y'[MY]][M'9.30lg3lg6.30'MY'MYKK3lgp'MY0.1%KM'9.30'MY10K20.0020.0220.00lglgK6'MY30.MYM'MYVVVlglgKpM'解法二:计量点时,8052306305p...spM)'lg(21MYspM,sppMKpc9.30'MY10K30.930.260.11'lgMYKMY'lg30.2280.5K1spM,0050.0Lmolc解法三:由终点误差公式计算(滴定突跃)1|pp|0.1%0.1%M'M'%1.0%100'010.0211010%100'10105.05.0,''MYMYspMpMpMtKKcE21910.'MYK5.02/1|''|'spepMMMppp9.218.300.918.30MY101010108.12'K6MY2.3021010(2.85)K'100.5-10-0.5=2.85P348表8:已知△pM=0.5→查0.5对应值2.855021ppp./|''|'spepMMM6-28在pH=5.0时,用0.002000mol•L-1EDTA滴定0.0020mol•L-1Pb2+溶液时,采用(1)HAc-NaAc缓冲剂,终点时cHAc+cAc-=0.31mol•L-1;(2)总浓度与上相同的六亚甲基四胺-HCl缓冲剂(不与Pb2+络合)分别控制溶液的酸度,选用二甲酚橙为指示剂,计算两种情况时的终点误差,并讨论为什么终点误差不同。已知Pb2+-Ac-络合物的累积形成常数lg1=1.9,lg2=3.3;pH=5.0时,lgY(H)=6.45,lgK’Pb-XO=7.0,lgKPbY=18.04,HAcKa=1.810-5。解:∵26.031.0lgAcAccc17.01020.0LmolcAc31.0HAcAccc00.531.0lg74.4lgppHAcAcHAcAcccccKa8.131.0AcAcccAcAccc8.18.131.056.08.2Acc1.98lg101010110101][Ac][Ac1)Pb(Ac1.981.91.21.43.30.71.9221)Pb(AcαββαpH=5.0时,lgY(H)=6.45,lgKPbY=18.0461.998.145.604.18lglglg'lgPbYPbYPbYKK30.6261.900.3)'lg(21',PbYspPbspKpcpPb(cPb,sp=0.0020/2=10-3.00)因为Pb2+此时有副反应0.7'lgxoPbtKpPb0.598.10.7lg'PbteppPbpPb3.130.60.5'''speppPbpPbpPb%0.1%10010101010%100'101061.900.33.13.1,''PbYspPbpPbpPbtKcE查表,二甲酚橙理论变色点:(2)pH=5.0时,lgY(H)=6.45,lgKPbY=18.0430.030.70.7speppPbpPbpPb59.1145.604.18lglg'lgYPbYPbYKK30.7259.1100.32'lg,PbYspPbspKpcpPb0.7'lgxoP
本文标题:6.络合滴定法习题2011
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