定量分析化学05.ppt

第,5,章 酸碱平衡及酸碱滴定法,5.1,滴定分析中化学平衡,5.2,平衡浓度及分布分数,5.3,酸碱溶液的,H,+,浓度计算,5.4,对数图解法,5.5,缓冲溶液,5.6,酸碱指示剂,5.7,酸碱滴定原理,5.8,终点误差,5.9,酸碱滴定法的应用,5.10,非水溶液酸碱滴定简介,5.1,滴定分析中化学平衡,四大平衡体系：,酸碱平衡,配位平衡,氧化还原平衡,沉淀平衡,四种滴定分析法：,酸碱,滴定法,配位,滴定法,氧化还原,滴定法,沉淀,滴定法,酸 共轭碱,+,质子,1,酸碱平衡,HF,F,-,+H,+,H,2,PO,4,-,HPO,4,2-,+H,+,H,6,Y,2+,H,5,Y,+,+H,+,NH,4,+,NH,3,+H,+,通式,:,HA A +H,+,酸碱半反应,例,:HF,在水中的离解反应,半反应,:,HF F,-,+H,+,半反应,:H,+,+H,2,O H,3,O,+,总反应,:HF+H,2,O F,-,+H,3,O,+,简写,:HF F,-,+H,+,酸碱反应的实质是质子转移,2,酸碱反应类型及平衡常数,HA +H,2,O,A,-,+H,3,O,+,A +H,2,O,HA,+OH,-,一元弱酸,(,碱,),的解离反应,a,H,+,a,A,-,K,a,=,a,HA,a,HA,a,OH,-,K,b,=,a,A,-,多元酸碱的解离反应,p,K,b1,+,p,K,a3,=14.00,p,K,b2,+p,K,a2,=14.00,p,K,b3,+p,K,a1,=14.00,H,3,PO,4,H,2,PO,4,-,HPO,4,2-,PO,4,3-,K,b2,K,b1,K,b3,K,a1,K,a2,K,a3,K,bi,=,K,w,K,a(n-i+1),H,2,O+H,2,O,H,3,O,+,+OH,-,(25,C),共轭酸碱对,(HA-A),的,K,a,与,K,b,的关系为,溶剂分子的质子自递反应,a,H,+,a,A,-,K,a,K,b,=,K,w,a,HA,a,HA,a,OH,-,a,A,-,K,w,=,a,H,+,a,OH,-,=,1.010,-14,p,K,a,+,p,K,b,=,p,K,w,=14.00,H,+,+OH,-,H,2,O,H,+,+Ac,-,HAc,OH,-,+,HAc,H,2,O,+Ac,-,酸碱中和反应,(,滴定反应,),K,t,滴定反应常数,1,K,t,=,10,14.00,K,w,1,K,t,=,K,a,K,b,K,w,1,K,t,=,K,b,K,a,K,w,3,活度与浓度,a,i,=,g,i,c,i,活度：在化学反应中表现出来的有效浓度，,通常用,a,表示,溶液无限稀时,:,g,=1,中性分子,:,g,=1,溶剂活度,:,a,=1,Debye-H,ckel,公式：,（稀溶液,I0.1 mol/L,）,I,：离子强度,I=1/2c,i,Z,i,2,z,i,：离子电荷,B:,常数,(=0.00328 25,),与温度、介电常数有关,：离子体积参数,(pm),-,lg,g,i,=0.512z,i,2,I,1+B,I,-,lg,g,i,=0.512z,i,2,I,活度常数,K,与温度有关,反应：,HA,B HB,+,+A,-,平衡常数,a,HB,+,a,A,-,K,=,a,B,a,HA,浓度常数,K,c,与温度和离子强度有关,HB,+,A,-,K,c,=,=,BHA,a,HB,+,a,A,-,=,a,B,a,HA,g,B,g,HA,-,g,H,B,+,g,A,-,K,g,H,B,+,g,A,-,物料平衡,(Material(Mass)Balance):,各物种的平衡浓度之和等于其分析浓度。,电荷平衡,(Charge Balance):,溶液中正离子所带正电荷的总数等于负离子所带负电荷的总数,(,电中性原则,),。,质子平衡,(Proton Balance):,溶液中酸失去质子数目等于碱得到质子数目。,4,质子条件式,质量平衡方程,（,MBE,）,2,10,-3,mol/L ZnCl,2,和,0.2 mol/L NH,3,Cl,-,=,4,10,-3,mol/L,Zn,2+,+Zn(NH,3,),2+,+Zn(NH,3,),2,2+,+Zn(NH,3,),3,2+,+Zn(NH,3,),4,2+,=2,10,-3,mol/L,NH,3,+Zn(NH,3,),2+,+,2,Zn(NH,3,),2,2+,+,3,Zn(NH,3,),3,2+,+,4,Zn(NH,3,),4,2+,=,0.2 mol/L,物料平衡,各物种的平衡浓度之和等于其分析浓度。,电荷平衡方程,（,CBE),Na,2,C,2,O,4,水溶液,Na,+,+H,+,=OH,-,+HC,2,O,4,-,+,2,C,2,O,4,2-,电荷平衡,溶液中正离子所带正电荷的总数等于负离子所带负电荷的总数,(,电中性原则,),。,质子平衡,溶液中酸失去质子数目等于碱得到质子数目。,质子条件式,（,PBE,）,(1),先选零水准,(,大量存在,参与质子转移的物质,),，一般选取,投料组分,及,H,2,O,(2),将零水准得质子产物写在等式一边,失质子产物写在等式另一边,(3),浓度项前,乘上得失质子数,例：,Na,2,HPO,4,水溶液,H,+,+H,2,PO,4,-,+,2,H,3,PO,4,=OH,-,+PO,4,3-,零,水准：,H,2,O,、,HPO,4,2-,Na(NH,4,)HPO,4,H,+,+H,2,PO,4,-,+2H,3,PO,4,=OH,-,+NH,3,+PO,4,3-,Na,2,CO,3,H,+,+HCO,3,-,+2H,2,CO,3,=OH,-,5.2,平衡浓度及分布分数,酸度对弱酸,(,碱,),形体分布的影响,1,酸度和酸的浓度,酸度：溶液中,H,的平衡浓度或活度，通常用,pH,表示,pH=-,lg,H,+,酸的浓度：酸的分析浓度，包含未解离的和已解离的 酸的浓度,对一元弱酸：,c,HA,HA+A,-,2,分布分数,分布分数：溶液中某酸碱组分的平衡浓度占其分析浓 度的分数，用,表示,“,”,将平衡浓度与分析浓度联系起来,HA,HA,c,HA,A,-,=,A-,c,HA,一元弱酸溶液,多元弱酸溶液,HAc,HAc,K,a,HAc,+,H,+,=,分布分数一元弱酸,HAc,Ac,-,H,+,+,c,HAc,=,HAc+Ac,-,HAc,HAc,=,c,HAc,def,H,+,=,H,+,+,K,a,HAc,HAc+Ac,-,HAc,=,Ac,-,Ac,-,Ac,-,=,c,HAc,HAc+Ac,-,def,H,+,+,K,a,K,a,=,Ac,-,HA,A,-,1,分布分数的一些特征,仅是,pH,和,p,K,a,的函数，与酸的分析浓度,c,无关,对于给定弱酸，,仅与,pH,有关,H,+,=,H,+,+,K,a,HA,H,+,+,K,a,K,a,=,A,-,例,计算,pH4.00,和,8.00,时,HAc,的,HAc,、,Ac,-,解,:,已知,HAc,的,K,a,=1.7510,-5,pH=4.00,时,pH=8.00,时,HAc,=5.710,-4,Ac,-,1.0,H,+,HAc,=,=0.85,H,+,+,K,a,K,a,Ac,-,=,=0.15,H,+,+,K,a,对于给定弱酸，,对,pH,作图,分布分数图,pH,HA,A,-,p,K,a,-2.0,0.99,0.01,*,p,K,a,-1.3,0.95,0.05,p,K,a,-1.0,0.91,0.09,*,p,K,a,0.50,0.50,p,K,a,+1.0,0.09,0.91,*,p,K,a,+1.3,0.05,0.95,p,K,a,+2.0,0.01,0.99,不同,pH,下的,HA,与,A,-,*,p,K,a,0.50,0.50,*,p,K,a,+1.3,0.05,0.95,*,p,K,a,-1.3,0.95,0.05,HAc,的,分布分数,图（,p,K,a,=4.76,）,3.46 6.06,p,K,a,1.3,pH,HAc,Ac,-,4.76,分布分数图,优势区域图,HF,的分布分数图（,p,K,a,=3.17,）,HF,F,-,p,K,a,3.17,pH,优势区域图,1.0,0.5,0.0,0 2 4 6 8 10 12 pH,3.17,HF,F,-,HCN,的分布分数图（,p,K,a,=9.31,）,p,K,a,9.31,优势区域图,HCN,CN,-,pH,0 2 4 6 8 10 12 pH,1.0,0.5,0.0,9.31,HCN,CN,-,HA,的,分布分数,图（,p,K,a,）,分布分数图的特征,两条分布分数曲线相交于,（,p,k,a,，,0.5,）,pH,p,K,a,时，溶液中以,A,-,为主,阿司匹林,pH=1.0,pH=8.4,假设,H,+,A,-,等离子不能穿透隔膜,HA,分子可自由通过隔膜,.,达平衡时,隔膜两边的,HA,浓度相等，,即,HA,血,=HA,胃,阿司匹林是一种弱酸,(,即乙酰水杨酸,),p,K,a,=3.5.,计算阿司匹林在胃中的吸收比率。,pH=7.4,pH=1.0,血浆,隔膜,胃,药物的吸收方式可表示为,H,+,+A,-,HA,HA H,+,+A,-,HA=,血,HA,c,血,=,胃,HA,c,胃,=,c,血,c,总药,c,血,c,血,+,c,胃,c,血,c,胃,=1+,血,HA,胃,HA,=1+,H,胃,(,H,血,+,K,a,),(,H,胃,+,K,a,),H,血,=1+,=,c,血,c,总药,99.99%,分布分数多元弱酸,二元弱酸,H,2,A,H,2,A,H,+,+,HA,-,H,+,+,A,2-,c,H,2,CO,3,=H,2,CO,3,+HCO,3,-,+CO,3,2-,H,2,A,=,H,2,A,c,H,2,A,A,2-,HA,-,def,HA,-,=,c,H,2,A,def,A,2-,=,c,H,2,A,def,物料平衡,酸碱解离平衡,二元弱酸,H,2,A,H,2,A,H,+,+,HA,-,H,+,+,A,2-,c,H,2,CO,3,=H,2,CO,3,+HCO,3,-,+CO,3,2-,H,2,A,=,H,2,A,c,H,2,A,A,2-,HA,-,def,HA,-,=,c,H,2,A,def,A,2-,=,c,H,2,A,def,H,+,2,H,+,2,+H,+,K,a1,+,K,a1,K,a2,=,=,=,H,+,2,+H,+,K,a1,+,K,a1,K,a2,H,+,2,+H,+,K,a1,+,K,a1,K,a2,H,+,K,a1,K,a1,K,a2,n,元弱酸,H,n,A,H,n,A,H,+,+,H,n-1,A,-,H,+,+,HA,(n+1)-,H,+,+,A,n,-,H,+,n,=,0,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,H,+,n-1,K,a1,=,1,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,=,n,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,K,a1,K,a2,.,K,an,分布分数定义,物料平衡,酸碱解离平衡,H,2,CO,3,的,分布分数,图,1.0,0.0,0 2 4 6 8 10 12 pH,H,2,CO,3,HCO,3,-,CO,3,2,-,H,2,CO,3,HCO,3,-,CO,3,2,-,6.38,p,K,a1,10.25,p,K,a2,p,K,a,=3.87,pH,优势区域图,酒石酸,(H,2,A),的,-pH,图,H,2,A,HA,-,A,2,-,p,K,a1,p,K,a2,p,K,a,=1.33,pH,1.0,0.5,0.0,0 2 4 6 8 10 12 pH,H,2,A,HA,-,A,2,-,3.04 4.37,磷酸,(H,3,A),的分布系数图,H,3,PO,4,H,2,PO,4,-,HPO,4,2,-,PO,4,3,-,2.16,p,K,a,5.05,7.21,p,K,a,5.11,12.32,p,K,a1,p,K,a2,p,K,a3,1.0,0.0,0 2 4 6 8 10 12 pH,H,3,PO,4,H,2,PO,4,-,HPO,4,2,-,PO,4,3,-,优势区域图,分布分数的总结,仅是,pH,和,p,K,a,的函数，与酸的分析浓度,c,无关,对于给定弱酸，,仅与,pH,有关,H,+,n,=,0,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,H,+,n-1,K,a1,=,1,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,=,n,H,+,n,+H,+,n-1,K,a1,+,K,a1,K,a2,.,K,an,K,a1,K,a2,.,K,an,5.3,酸碱溶液,H,+,的计算,酸碱溶液的几种类型,一,.,强酸碱,二,.,一元弱酸碱,HA,多元弱酸碱,H,2,A,H,3,A,三,.,两性物质,HA,-,四,.,共轭酸碱,HA+A,-,五,.,混合酸碱,强,+,弱,.,弱,+,弱,1,强酸碱,溶液,强酸,(,HCl,):,强碱,(,NaOH,):,c,HCl,=10,-5.0,and,10,-8.0,molL,-1,pH=?,质子条件,:H,+,+,c,NaOH,=OH,-,最简式,:OH,-,=,c,NaOH,质子条件,:H,+,=,c,HCl,+OH,-,最简式,:H,+,=,c,HCl,2,弱酸,(,碱,),溶液,展开则得一元三次方程,数学处理麻烦,!,一元弱酸,(HA),质子条件式,:,H,+,=A,-,+OH,-,平衡关系式,精确表达式,:,H,+,K,a,HA,+,K,w,H,+,=+,H,+,K,a,HA,H,+,K,w,H,+,+,K,a,c,a,H,+,HA=,若,:,K,a,c,a,10,K,w,忽略,K,w,(,即忽略水的酸性,),HA=,c,a,-A,-,=,c,a,-(H,+,-OH,-,),c,a,-H,+,近似计算式,:,展开得一元二次方程,H,+,2,+,K,a,H,+,-,c,a,K,a,=0,，求解即可,最简式,:,若,:,c,a,/,K,a,100,则,c,a,-,H,+,c,a,H,+,K,a,HA,+,K,w,精确表达式：,H,+,K,a,(c,a,-H,+,),H,+,K,a,c,a,若,:,K,a,c,a,10,K,w,但,c,a,/,K,a,100,酸的解离可以忽略,HA,c,a,得近似式,:,H,+,K,a,HA,+,K,w,精确式：,H,+,K,a,c,a,+,K,w,(1),K,a,c,a,10,K,w:,(2),c,a,/,K,a,100,:,(3),K,a,c,a,10,K,w,c,a,/K,a,100,:,H,+,K,a,HA,+,K,w,精确表达式：,H,+,K,a,(c,a,-H,+,),H,+,K,a,c,a,+,K,w,(,最简式,),H,+,K,a,c,a,例,计算,0.20molL,-1,Cl,2,CHCOOH,的,pH.(p,K,a,=1.26),如不考虑酸的离解,(,用最简式,:pH=0.98),则,E,r,=29%,解,:,K,a,c,=10,-1.26,0.20=10,-1.96,10,K,w,c,/,K,a,=0.20/10,-1.26,=10,0.56,100,故近似式,:,解一元二次方程,:H,+,=10,-1.09,则,pH=1.09,H,+,K,a,(c,a,-H,+,),一元弱碱,(B,-,),质子条件,:,得精确表达式,:,H,+,+HB=OH,-,H,+,+=,K,a,H,+,B,-,H,+,K,w,H,+,K,w,1+B,-,/,K,a,酸碱平衡关系,若,:,c,/,K,a,20,“1”,可以忽略,(,水的碱性,),并且,:B,-,=,c,b,-HB=,c,b,(OH,-,-H,+,),c,b,OH,-,近似计算式,:,最简式,:,若,K,b,/,c,b,10,K,w:,(2),c,/,K,b,100,:,(3),K,b,c,10,K,w,c,/,K,b,100,:,OH,-,=,K,b,(,c,b,-OH,-,),OH,-,=,K,b,c,b,+,K,w,H,+,=,K,a,K,w,c,b,OH,-,=,K,b,c,b,最简式,:,多元弱酸溶液,二元弱酸,(H,2,A),质子条件,:,H,+,=HA,-,+2A,2-,+OH,-,2,K,a2,H,+,=,K,a1,H,2,A (1+)+,K,w,H,+,酸碱平衡关系,K,w,K,a1,H,2,A,H,+,=+,2,K,a1,K,a2,H,2,A,H,+,H,+,2,H,+,0.05,可略,近似式,:,以下与一元酸的计算方法相同,K,a1,c,a,10,K,w,2,K,a2,H,+,=,K,a1,H,2,A (1+)+,K,w,H,+,2,K,a2,H,+,=,K,a1,H,2,A (1+)+,K,w,H,+,2K,a2,H,+,2K,a2,H,+,H,+,=,K,a1,H,2,A,(,忽略二级及以后各步离解,),K,a1,c,a,10,K,w,，,则：,0.05,则：,c,a,/,K,a1,100,计算饱和,H,2,CO,3,溶液的,pH,值,(0.040 mol/L),H,+,K,a1,c,a,2,K,a2,H,+,=,K,a1,H,2,A (1+),H,+,H,+,=,K,a1,H,2,A,2K,a2,H,+,2K,a2,K,a1,c,a,3,两性物质溶液,两性物质：,在溶液中既起酸,(,给质子,),、又起碱（得质子）的作用,多元酸的酸式盐,Na,2,HPO,4,NaH,2,PO,4,弱酸弱碱盐,NH,4,Ac,氨基酸,质子条件,:H,+,+H,2,A=A,2,-,+OH,-,精确表达式,:,酸碱平衡关系式,酸式盐,NaHA,K,w,H,+,HA,-,H,+,+=+,K,a2,HA,-,K,a1,H,+,H,+,H,+,K,a1,(,K,a2,HA,-,+,K,w,),K,a1,+HA,-,若,:,K,a1,K,a2,HA,-,c,(,p,K,a,3.2),近似计算式,:,如果,c,10,K,a1,则“,K,a1,”,可略,得,最简式,:,H,+,=,K,a1,K,a2,c,K,a1,+c,H,+,=,K,a1,(,K,a2,c,K,w,),K,a1,+,c,若,K,a2,c,10,K,w,则,K,w,可忽略,H,+,K,a1,K,a2,精确式：,K,a1,K,a2,HA,-,c,K,a2,c,1,0,K,w,c,10,K,a1,pH=1/2(p,K,a1,+p,K,a2,),H,+,K,a1,(,K,a2,c+,K,w,),K,a1,+c,H,+,K,a1,K,a2,c,K,a1,+c,H,+,K,a1,K,a2,H,+,K,a1,(,K,a2,HA,-,+,K,w,),K,a1,+HA,-,弱酸弱碱盐,NH,4,Ac,质子条件式,:,H,+,+,HAc,=NH,3,+OH,-,K,a,c,10,K,w,c,10,K,a,H,+,K,a,(,K,a,c+,K,w,),K,a1,+c,酸碱平衡关系,NH,4,+,Ac,-,c,H,+,=,K,a,K,a,c,K,a,+c,H,+,K,a,K,a,K,a,NH,4,+,K,a,HAc,例 计算,0.0010 mol/L CH,2,ClCOONH,4,溶液的,pH,CH,2,ClCOOH:,K,a,=1.410,-3,NH,3,:,K,b,=1.810,-4,K,a,=5.610,-10,K,a,c,10,K,w,c,10,K,w,c,/,K,a1,10,H,+,K,a1,K,a2,c,K,a1,+c,H,+,K,a1,(,K,a2,c+,K,w,),K,a1,+c,酸碱平衡关系,H,+,K,a1,K,a2,4.,共轭酸碱体系,H,+,=,K,a,=,K,a,HA,A,-,c,a,-H,+,+OH,-,c,b,+H,+,-OH,-,H,+,=,K,a,HA,A,-,c,a,mol,/L HA+,c,b,mol,/L,NaA,PBE,：,HA=,c,a,+OH,-,-H,+,A,-,=,c,b,+H,+,-OH,-,物料平衡,:HA+A,-,=,c,a,+,c,b,电荷平衡,:H,+,+Na,+,=OH,-,+A,-,HA=,c,a,-H,+,+OH,-,A,-,=,c,a,+,c,b,-HA,=,c,b,+H,+,-OH,-,+),pH 8(,碱性,),略去,H,+,若,c,a,20H,+,;,c,b,20H,+,或,c,a,20OH,-,;,c,b,20OH,-,最简式,H,+,=,K,a,c,a,c,b,计算方法：,(1),先按最简式计算,OH,-,或,H,+,。,(2),再计算,HA,或,A,-,看其是否可以忽略,.,如果不能忽略,再按近似式计算。,通常情况下，由共轭酸碱对组成的缓冲溶液可以用最简式直接计算,pH,例,(1)0.10 mol/L NH,4,Cl 0.20 mol/L NH,3,先按最简式,:,(2),0.080 mol/L,二氯乙酸,0.12mol/L,二氯乙酸钠,先用最简式求得,H,+,0.037 mol/L,c,a,OH,+,c,b,OH,-,结果合理,pH=9.56,应用近似式,:,解一元二次方程，,H,+,=10,-1.65,molL,-1,pH=1.65,pH=,p,K,a,+,lg,=9.56,c,a,c,b,H,+,=,K,a,c,a,-H,+,c,b,+H,+,强酸,(,HCl,)+,弱酸,(HA),质子条件,:H,+,=,c,HCl,+A,-,+OH,-,(,近似式,),忽略弱酸的离解,:H,+,c,HCl,(,最简式,),5.,混合酸碱体系,K,w,H,+,=,c,HCl,+,K,a,c,a,K,a,+H,+,H,+,酸碱平衡关系,强碱,(,NaOH,)+,弱碱,(B,-,),质子条件,:H,+,+HB+,c,NaOH,=OH,-,忽略弱碱的离解,:,OH,-,c,(NaOH,),(,最简式,),K,w,OH,-,=,c,HCl,+,K,b,c,b,K,b,+OH,-,OH,-,两弱酸,(HA+HB),溶液,质子条件,:,H,+,=A,-,+B,-,+OH,-,HA,c,HA,HBc,HB,酸碱平衡关系,K,w,K,HA,HA,H,+,=+,K,HB,HB,H,+,H,+,H,+,H,+,K,HA,c,HA,H,+,K,HA,c,HA,+,K,HB,c,HB,K,HA,c,HA,K,HB,c,HB,弱酸,+,弱碱,(HA+B,-,),溶液,质子条件,:,H,+,+HB=A,-,+OH,-,HA,c,HA,HBc,HB,酸碱平衡关系,K,w,H,+,HB,H,+,+=+,K,HA,HA,K,HB,H,+,H,+,H,+,K,HA,K,HB,c,HA,/c,B,综合考虑、分清主次、合理取舍、近似计算,酸碱溶液,H,+,的计算总结,质子条件,物料平衡,电荷平衡,酸碱平衡关系,H,+,的精确表达式,近似处理,H,+,的近似计算式和最简式,5.4,对数图解法,图解法,1,强酸强碱的浓度对数图,0.1mol/L,HCl,Cl,-,H,+,OH,-,lg,Cl,-,=-1,lg,H,+,=-pH,lg,OH,-,=pH-14,2,一元弱酸（碱）的浓度对数图,0.01mol/L,HAc,HAc,Ac,-,H,+,OH,-,+,a,+,a,H ,HAc,H ,c,K,=,+,a,+,a,K,a,Ac,-,H ,c,K,=,+,一元弱酸（碱）的浓度对数图绘制,1,确定体系点,S,（,p,Ka,，,lgc,a,）,2,过,S,，画,斜率为,0,，,1,的三条直线,3 S,附近,lgc,与,pH,的曲线关系,准确：逐个计算,S,点附近的点（繁琐）,近似：确定点,O,（,p,k,a,，,lgc,a,-0.3,），,及与直线的切点,pH,pKa,1.3,3,多元弱酸（碱）的浓度对数图,0.01mol/L H,2,A (p,K,a1,=4,p,K,a2,=8),H,2,A HA,-,A,2-,H,+,OH,-,1,计算,pH,值,2,计算各种分布形式的平衡浓度及分布系数,4,对数图解法的应用,1 pH,值计算,0.01mol/L,HAc,H,+,=Ac,-,+OH,-,0.01mol/L,NaAc,H,+,+,HAc,=OH,-,P,C,H,2,A,0.01mol/L,pH,9.0,logH,2,A=-8.2,logHA,-,=-3.2,A,2-,=C,H,2,A,-H,2,A-HA,-,2,平衡浓度及分布系数的计算,缓冲溶液：,能减缓强酸强碱的加入或稀释而引起的,pH,变化,5.5,酸碱,缓冲溶液,高浓度强酸强碱,共轭弱酸碱,1,缓冲溶液,pH,计算,C,a,mol/L,的,HA,与,C,b,mol/L,的,A,-,将,0.30mol/L,的,吡啶,与,0.10mol/L,的,HCl,等体积混合配置成缓冲溶液，求其,pH,值。吡啶的,p,K,b,8.77,。,pH=,p,K,a,+,lg,C,a,C,b,2,缓冲容量,缓冲容量：,衡量缓冲溶液缓冲能力大小，用,表示,dc/,dpH,加合性：,=,H,+,OH,-,HA,=2.3H,+,+2.3OH,-,+2.3,HA,A,c,HA,对于,pH,在,p,K,a,1,范围内的,HA,=2.3,HA,A,c,HA,HA,A,的缓冲体系,有极大值,pH,p,Ka,时，即,HA=A,极大,0.58cHA,标准缓冲溶液,校准酸度计,3,常用缓冲溶液,pH,标 准 溶 液,pH,（,25,）,饱和酒石酸氢钾,(0.034molL,-1,),3.56,0.050 molL,-1,邻苯二甲酸氢钾,4.01,0.025 molL,-1,KH,2,PO,4,0.025 molL,-1,Na,2,HPO,4,6.86,0.010,molL,-1,硼砂,9.18,常用缓冲溶液,8.5,10,9.25,NH,4,+,+NH,3,8.5,10,9.24,硼砂,(H,3,BO,3,+H,2,BO,3,-,),7.5,9,8.21,三羟甲基甲胺,+,HCl,6.5,8,7.21,H,2,PO,4,-,+HPO,4,2-,4.5,6.0,5.13,六次甲基四胺,+,HCl,4,5.5,4.76,HAc+NaAc,3,4.5,3.77,甲酸,+,NaOH,2,3.5,2.86,氯乙酸,+,NaOH,1.5,3.0,2.35,氨基乙酸,+,HCl,缓冲范围,p,K,a,缓冲溶液,4,缓冲溶液的选择原则,不干扰测定,例如：,EDTA,滴定,Pb,2+,不用,HAc,-Ac,-,有较大的缓冲能力，足够的缓冲容量,较大浓度,(0.01,1molL,-1,);,pHp,K,a,即,c,a,c,b,11,HAc,NaAc,:,p,K,a,=4.76,(45.5),NH,4,OHNH,3,:,p,K,b,=4.75,(810),(CH,2,),6,N,4,(CH,2,),6,N,4,H,+,:,p,K,b,=8.87,（,4.56,）,甲基橙,(,MO,),1,作用原理,:,酸式和其共轭碱式具有明显不同的颜色,5.6,酸碱,指示剂,H,+,OH,-,红,3.1,橙,4.0,4.4,黄,p,K,a,=3.4,甲基橙的,-pH,图,0 2 4 6 8 pH,p,K,a,=3.4,红,橙,黄,3.1 4.4,0.66,0.34,0.91,0.09,0.8,0.2,1.0,0.5,0.0,2,指示剂变色范围,HIn,H,+,+In,-,K,HIn,=,In,-,/,HIn,10,显示,In,-,色,In,-,/,HIn,0.1,显示,HIn,色,理论变色范围：,pH=,p,K,HIn,1,HIn,H,+,In,-,=,H,+,K,HIn,HIn,In,-,甲基橙,MO,甲基红,MR,酚酞,PP,3.1,4.4,4.4,6.2,8.0,9.6,4.0,5.0,9.0,常用单一酸碱指示剂,百里酚酞,:,无色,9.4-,10.0(,浅蓝,),-,10.6,蓝,3,影响指示剂变色范围的因素,指示剂用量,:,宜少不宜多，,对单色指示剂影响较大,例：,50,100mL,溶液中,23,滴,PP,，,pH9,变色，,而,1015,滴,PP,，,pH8,变色,离子强度,：影响,p,K,HIn,温度,其他,4,混合指示剂,溴甲酚绿,甲基红,5.0,-,-5.1,-,5.2,橙红,灰,绿,（黄红）（绿,+,橙红）（蓝黄）,用于,Na,2,CO,3,标定,HCl,时,指示终点,通过颜色互补，使变色范围变窄，,变色更敏锐,指示剂选择,:,pH,ep,与,pH,sp,尽可能接近，以减小滴定误差,滴定曲线：,溶液,pH,随滴定分数,(,a,),变化的曲线,5.7,酸碱,滴定原理,化学计量点,(sp),滴定突跃,滴定突跃,SP,0.1000 molL,-1,NaOH,滴定,20.00,mL,0.1000molL,-1,HCl,H,+,+OH,-,=H,2,O,K,t,=1/,K,w,=10,14.00,PBE:H,+,=OH,-,+,c,HCl,-c,NaOH,滴定分数：,a,=,c,T,/c,A,=,c,NaOH,/c,HCl,滴定曲线方程：,K,t,H,+,2,+,K,t,c,HCl,（,a,-1,）,H,+,-1=0,此例：,a,=,c,NaOH,/c,HCl,=V,NaOH,/20,1,强酸碱滴定,(3)sp,时,:,a,=1 H,+,=OH,-,=,K,t,-0.5,pH=7.00,(4)sp,后,:OH,-,=,c,NaOH,(,过量,),(1),滴定前,:,a,=0 H,+,=,c,HCl,=0.1000molL,-1,pH=1.00,(2),滴定开始到,sp,前,:H,+,=,c,HCl,(,剩余,),-0.1%,时,:,a,=0.999 H,+,=5.010,-5,mol/L pH=4.30,+0.1%,时,:,a,=1.001 OH,-,=5.010,-5,mol/L pH=9.70,0.1000molL,-1,NaOH,滴定,20.00mL 0.1000molL,-1,HCl,12.52,20.00,2.000,40.00,11.68,2.00,1.100,22.00,sp,后,:OH,-,=,c,NaOH,(,过量,),10.70,0.20,1.010,20.20,9.70,0.02,1.001,20.02,7.00,0.00,0.00,1.000,20.00,sp:H,+,=OH,-,=10,-7.00,4.30,0.02,0.999,19.98,3.00,0.20,0.99,19.80,sp,前,:H,+,=,c,HCl,（剩余）,2.28,2.00,0.90,18.00,滴定前,:H,+,=,c,HCl,1.00,20.0,0.00,0.00,H,+,计算,pH,过量,NaOHmL,剩余,HCl,mL,a,NaOH,mL,突跃,强酸碱滴定曲线,0.1000molL,-1,HCl,0.1000molL,-1,NaOH,突跃,:9.74.3,pH,12,10,8,6,4,2,0,0 1,2,滴定分数,a,9.7 sp+0.1%,4.3 sp-0.1%,sp,7.0,突跃,PP 9.0,MR 6.2,MO 4.4,PP 8.0,MR 5.0,指示剂选择？,浓度对滴定突跃的影响,0 1 2,pH,12,10,8,6,4,2,10.7,9.7,8.7,7.0,5.3,4.3,3.3,0.01molL,-1,0.1molL,-1,1molL,-1,浓度增大,10,倍，,突跃增加,2,个,pH,单位。,5.38.7,4.39.7,3.310.7,0.1000 mol/,LNaOH,滴定,0.1000mol/LHA(,K,a,),PBE:H,+,=OH,-,+A,-,-,c,NaOH,滴定分数：,a,=,c,T,/c,A,=,c,NaOH,/c,HA,滴定曲线方程：,H,+,3,+(,K,a,+,a,c,HA,)H,+,2,+,K,a,c,HCl,(1-,a,)+,K,w,H,+,-,K,a,K,w,=0,2,一元弱酸碱的滴定,0.1000 mol/,LNaOH,滴定,0.1000 mol/LHA(,p,K,a,=4.76),11.68,A,-,+OH,-,1.100,22.00,10.70,A,-,+OH,-,1.010,20.20,10.00,A,-,+OH,-,1.002,20.04,9.70,A,-,+OH,-,1.001,20.02,8.72,A,-,1.000,20.00,7.76,HA+A,-,0.999,19.98,7.46,HA+A,-,0.998,19.96,6.67,HA+A,-,0.99,19.80,5.71,HA+A,-,0.90,18.00,4.76,HA+A,-,0.50,10.00,2.88,HA,0.00,0.00,H,+,计算式,pH,组成,a,NaOH,mL,-0.1%:pH=p,K,a,+3,H,+,K,a,c,a,H,+,=,K,a,HA,A,-,sp,后,:OH,-,=,c,NaOH,(,过量,),OH,-,=,K,b,c,b,sp,前,滴定前,sp,强碱滴定弱酸滴定曲线,0 1 2,pH,12,10,8,6,4,2,0,HA A,-,A,-,+OH,-,突跃,9.7,8.7,7.7,4.3,HA,HCl,MO,MR,PP,浓度,:,增大,10,倍，突跃增加,1,个,pH,单位（上限）,K,a,：,增大,10,倍，突跃增加,1,个,pH,单位（下限）,影响滴定突跃的因素,滴定突跃：,p,K,a,+3,-,lg,K,w,/c,NaOH,(,剩余,),弱酸准确滴定条件：,cK,a,10,-8,对于,0.1000molL,-1,的,HA,K,a,10,-7,才能准确滴定,0.1molL,-1,HCl,NH,3,0.1molL,-1,p,K,b,=4.75,0 50,100 150 200%,6.25,4.30,5.28,6.2,4.4,3.1,pH,NaOH,NH,3,突跃处于弱酸性,选,甲基红,或,甲基橙,作指示剂,.,强酸滴定弱碱,8.0,3,多元酸,(,碱,),的滴定,多元酸能分步滴定的条件,滴定计量点的,pH,值的计算,指示剂的选择,多元酸能分步滴定的条件,:,被滴定的酸足够强,cK,an,10,-8,相邻两步解离相互不影响，,lg,K,a,足够大,若,pH=0.2,允许,E,t,=0.3%,则需,lg,K,a,5,滴定化学计量点的,pH,值的计算,以,H,3,A,为例：,前提：可以分步滴定,sp1:H,2,A,-,按,酸式盐,计算,pH=1/2(p,K,a1,+p,K,a2,),sp2:HA,2-,按,酸式盐,计算,pH=1/2(p,K,a2,+p,K,a3,),sp3:A,3-,按,多元碱,计算,OH,-,=,K,b1,c,K,w,/K,a3,c,pH,0 100 200 300 400 (T%),10.0,9.4,5.0,4.4,p,K,a,lg,K,a,2.12,7.20,12.36,5.08,5.16,H,3,A+H,2,A,-,H,2,A,-,+HA,2-,HA,2-,+A,3-,A,3-,+NaOH,sp2,sp1,NaOH,滴定,0.1molL,-1,H,3,PO,4,c,K,a3,10,-8,c,K,a1,10,-8,pH,sp2,=9.66,百里酚酞,至,浅蓝,(10.0),pH,sp1,=4.70,MO,至,黄,(4.4),混合酸分步滴定,:,两弱酸混合（,HA,HB,）,被滴定的酸足够强,cK,a,10,-8,c,1,K,a,/c,2,K,a,10,5,强酸弱酸（,H,+,HA,）,K,a,10,-7,测总量,K,a,10,-7,测强酸量,5.8,终点误差,终点误差：,指示剂确定的滴定终点,(EP),与化学计量点,(SP),之