电位滴定法综述

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,5.4,电位滴定法,E-V,曲线法,E/V-V,曲线法,2,E/V,2,-V,曲线法,一、电位滴定分析法的装置,1.手动电位滴定装置 2.自动电位滴定装置,二、电位滴定法的步骤,关键,:,确定,滴定反应至化学计量点时,所消耗的,滴定剂（标准溶液）的体积。,步骤：,1,、根据预测定数据,取一定量的待测试液；,2,、用标准溶液进行滴定，并记录相应的电位；,3,、根据所得数据，按以下三种方法来确定终点：,E,-,V,曲线法,E,/,V,-,V,曲线法,2,E,/,V,2,-,V,曲线法,见下例,三、实验数据,（,1,）,加入,AgNO,3,的体积,V（ml）,E/v,E/V,2,E/V,2,V/ml,5.00,0.062,0.002,15.00,0.085,0.004,20.00,0.107,0.008,22.00,0.123,0.015,23.00,0.138,0.016,24.00,0.146,0.050,以,0.1000,mol/LAgNO,3,滴定,25.00,mlNaCl,试样溶液，,E,为电池电动势（,v-,伏或,mv,-,毫伏），,V,为体积,ml,实验数据,（,2,）,加入,AgNO,3,的体积,V（ml）,E/V,E/V,2,E/V,2,V/ml,24.10,0.183,0.11,24.20,0.194,0.39,2.8,24.30,0.233,0.83,4.4,24.40,0.316,0.24,-5.9,24.50,0.340,0.11,-1.3,24.60,0.351,0.07,-0.4,24.70,0.358,0.050,25.00,0.373,0.024,25.50,0.385,1、,E/V,值,一阶微商的计算,E/V,：,为电位（,E）,的变化值与相对应的加入滴定剂体积的增量之比，是,一阶微商,dE,/,dv,的近似值,，例如：,24.1024.20,之间，相应的,E/V,为：,E/V=0.1940.18324.2024.10,=0.11,2、,2,E/V,2,二级微商的计算,例如：,对应于,24.30,ml,;,同理，,对应于,24.40,ml,四、滴定终点确定的方法,h,以加入滴定剂的体积,V（ml）,为横坐标、对应的电动势,E（,mv,）,为纵坐标，绘制,E-V,曲线，曲线上的拐点所对应的体积为滴定终点。,1、,E-V,曲线法,2、,E/V V,曲线,曲线的一部分用外延法绘制，其,最高点,对应,于,滴定终点时所消耗滴定剂的体积。,3、,2,E/V,2,法计算滴定终点时的体积,即：二级微商,2,E/V,2,=0,时的,体积,为,滴定终点体积,，用,内插法,计算：,24.34,ml,即为滴定终点时,AgNO,3,消耗的体积,4、,2,E/V,2,V,曲线,以,二阶微商值,为,纵坐标,，加入,滴定剂的体积,为,横坐标,作图。,2,E/V,2,=0,所对应的体积,即为滴定终点。,测定方法,参比电极,指示电极,酸碱滴定,甘录电极,玻璃电极，锑电极,沉淀滴定,甘汞电极，玻璃电极,银电极，硫化银薄膜电极等离子选择性电极,氧化还原滴定,甘汞电极，钨电极，玻璃电极,铂电极,络合滴定,甘汞电极,铂电极，汞电极，钙离子等离子选择性电极。,四、电位滴定法的应用及电极的选择,五、,电位滴定法的计算示例,吸取,Cl,-,和,I,-,混合液,25.00,mL,稀释到,100.00,mL,以0.1000,mol,L,-1,的,AgNO,3,溶液滴定,.,以银电极作指示电极,饱和甘汞电极作参比电极,加入,AgNO,3,的体积及相应的,E,值如下表所示,.,计算,Cl,-,及,I,-,在化学计量点时,所消耗,AgNO,3,的体积,并计算,原来溶液,中,Cl,-,及,I,-,的浓度,各是多少,?以,g/L,表示之。,(,已知,:,K,sp,AgCl,=1.8,10,-10,K,sp,AgI,=8.3,10,-17,Mr,(,NaCl,)=58.44,Mr,(,NaI,)=149.9),测定数据,V,AgNO,3,E,V,AgNO,3,E,V,AgNO,3,E,mL,mV,mL,mV,mL,mV,0.00,+239,3.20,+87,7.20,-178,0.50,+236,3.30,-93,7.30,-184,1.00,+230,3.40,-102,7.40,-194,1.50,+223,3.50,-107,7.50,-207,2.00,+214,3.70,-114,7.60,-237,2.50,+200,4.00,-118,7.70,-303,2.60,+196,4.50,-123,7.80,-322,2.70,+191,5.00,-129,7.90,-334,2.80,+186,5.50,-135,8.00,-342,2.90,+177,6.00,-142,8.50,-364,3.00,+168,6.50,-152,9.00,-375,3.10,+148,7.00,-167,9.50,-383,7.10,-171,10.00,-389,解题（,1,）,解：首先 计算在两个化学计量点附近的,V,E,E/V,2,E/V,2,的值,如下表所示。,加入,AgNO,3,的体积,/,mL,E/mV,V,E,E/V,2,E/V,2,:,:,:,:,:,:,:,:,:,:,3.00,+168,0.10,-20,-200,3.10,+148,-4100,0.10,-61,-610,3.20,+87,-11900,0.10,-180,-1800,3.30,-93,+17100,0.10,-9,-90,3.40,-102,+400,0.10,-5,-50,3.50,-107,:,:,:,解题（,2,）,加入,AgNO,3,的体积,/,mL,E/mV,V,E,E/V,2,E/V,2,:,:,:,:,:,:,:,:,:,:,7.40,-194,0.10,-13,-130,7.50,-207,0.10,-30,-300,7.60,-237,0.10,-66,-660,7.70,-303,0.10,-19,-190,7.80,-322,0.10,-12,-120,7.90,-334,:,:,:,:,解题（,3,）,从,氯化银,（,K,sp,AgCl,=1.8,10,-10,）,和,碘化银,（,K,sp,AgI,=8.3,10,-17,）,的溶度,积可知,首先沉淀,的是,AgI,然后,才是,AgCl,。,从计算的表中可知：,I,-,离子反应的终点,应,2,E/V,2,等于,-11900+17100,所对应,的滴定剂体积之间，即应在,3.20,3.30,mL,之间.,解题（,4,）,2,E/V,2,=0,所对应,AgNO,3,体积的计算,:,加入,AgNO,3,溶液的体积为,3.20 3.30,mL,时,2,E/V,2,的变化为,:,17100-(-11900)=,29000,设,(3.20+,x)mL,时,2,E/V,2,=0,即为滴定,I,-,离子的终点,则：,0.1:29000=,x:17100,x=0.06mL.,所以滴定至,I,-,离子的化学计量点时,滴定剂体积为,3.26,mL,。,解题（,5,）,Cl,-,离子反应的,终点,应在,2,E/V,2,等于,-3600+4700,所对应的滴定剂体积之间,即应在,7.60 7.70,mL,之间。,加入,AgNO,3,溶液体积为,7.60 7.70,mL,时,2,E/V,2,的,变化,为,:,4700-(-3600)=8300,设,(7.60+,x)mL,时,2,E/V,2,=0,即为,滴定,Cl,-,离子的终点,则,0.1:8300=,x:4700,x=0.06mL,所以滴定,Cl,-,离子至等当点时,滴定剂体积为,7.66-3.26=4.40,mL.,解题（,6,）,答：原来溶液中,C,NaI,为1.95,g/L，,C,NACl,为1.03,g/L。,