第五章+水溶液中的离子平衡2.ppt

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,*,两个一元弱酸的混合体系：,HA,1,+HA,2,HA,1,(aq)+H,2,O(l)H,3,O,+,(aq)+A,1,-,(aq),x+y x,HA,2,(aq)+H,2,O(l)H,3,O,+,(aq)+A,2,-,(aq),x +y y,如果,K,a1,C,HA1,K,a2,C,HA2,，则,(HA,C,HA,),HAc,K,a1,=1.8,10,-5,;HCN,K,a2,=4.93,10,-10,两个一元弱碱的混合体系：,B,1,+B,2,如果,K,b1,C,B1,K,b2,C,B2,，则,P332/例,12-6,HCOOH(aq,)+H,2,O(l)H,3,O,+,(aq)+HCOO,-,(,aq,),平衡,0.100,x,4.4110,-3,x,x,=HCOO,-,=3.86,10,-3,(mol/L),HCOOH=0.100,x,=,9.61,10,-2,(mol/L),CH,3,COOH(aq)+H,2,O(l)H,3,O,+,(aq)+CH,3,COO,-,(aq),平衡,0.100,y,4.4110,-3,y,y,=CH,3,COO,-,=3.98,10,-4,(mol/L),CH,3,COOH=0.100,y,=,9.96,10,-2,(mol/L),弱酸（,HA,）和强酸HX,）的混合溶液,如果,K,a,比较小，,HA,电离产生的,H,3,O,+,也比较小，可忽,略,(,强酸,浓度不是太小,),，因此,H,3,O,+,=C,HX,（酸性溶液,中水自电离产生的,H,3,O,+,忽略）,如果,K,a,比较大，,HA,电离产生的,H,3,O,+,也比较大，不能忽略。酸性溶液中水自电离产生的,H,3,O,+,忽略,例,3:,由,HAc,和,HCl,组成的溶液，其中,HAc,的浓度为,0.1 mol/L,，,HCl,的浓度为,0.1 mol/L,，,计算溶液中各离子的平衡浓度？（,HAc,的,K,a,=1.8,10,-5,）,解,:,HAc,+H,2,O,H,3,OAc,-,平衡：,0.1,x,C,HCl,x,K,a,=C,HX,x,/(0.1,x,)=1.8,10,-5,x,=1.8,10,-5,mol/L,H,3,O,+,=,C,HCl,=0.1 mol/L,Ac,-,=,x,=1.8,10,-5,mol/L,HAc,=0.1,x,=,0.1 mol/L,例,4:,计算,0.10 mol/L H,2,SO,4,溶液的,H,3,O,+,、,HSO,4,-,和,SO,4,2-,?,已知,HSO,4,-,的,K,a,=1.2,10,-2,解,:H,2,SO,4,+H,2,O,H,3,O,+,+HSO,4,-,(,完全电离,),HSO,4,-,(aq)+H,2,O(l)H,3,O,+,(aq)+SO,4,2-,(aq),起始,0.1,0.1,0,转化,x x x,平衡,0.1,x,0.1+,x x,K,a,=,x,(0.1,+x,)/(0.1,x,),x,2,+0.112,x,0.0012=0,x,=9.8510,-3,(mol/L),H,3,O,+,=0.1+,x,=0.11(mol/L),SO,4,2-,=,x,=9.8510,-3,(mol/L),HSO,4,-,=0.1,x,=0.090(mol/L),HSO,4,-,的水溶液只能作为弱酸，不能作为弱碱！,弱碱（,B,）和强碱（,MOH,）的混合溶液,如果,K,b,比较小，,B,电离产生的,OH,-,也比较小，可忽,略,(,强,碱浓度不是太小,),，因此,OH,-,=C,MOH,（碱性溶液,中水自电离产生的,OH,-,忽略）,如果,K,b,比较大，,B,电离产生的,OH,-,也比较大，不能忽略。碱性溶液中水自电离产生的,OH,-,忽略,B(aq,)+H,2,O(l)OH,-,(,aq,)+,BH,+,(aq,),平衡：,C,B,x,x+C,MOH,x,5.2.,多元弱酸（弱碱）的电离平衡,多元弱酸：电离出,2,个或,2,个以上的,H,3,O,+,，且部分电离,多元弱碱：电离出,2个或2个以上的,OH,-,，且部分电离,多元弱酸：,H,2,CO,3,、,H,2,C,2,O4,、,H,3,PO,4,多元弱碱：,Na,2,S,、,Na,3,PO,4,二元弱酸的电离平衡：,H,2,A,H,2,A(aq)+H,2,O(l),H,3,O,+,(aq),+HA,-,(,aq,),Ka,1,=H,3,O,+,HA,-,/H,2,A H,2,A,的一级解离常数,HA,-,(,aq,)+H,2,O(l),H,3,O,+,(aq),+A,2-,(aq),Ka,2,=H,3,O,+,A,2-,/HA,-,H,2,A,的二级解离常数,H,2,O(l)+H,2,O(l),H,3,O,+,(aq),+OH,-,(,aq,),K,w,=H,3,O,+,OH,-,物料平衡：,C,H2A,=H,2,A+HA,-,+A,2-,电荷平衡：,H,3,O,+,=HA,-,+2A,2-,+OH,-,等式右边三项的物理意义：,第,1,项：一级电离的贡献,第,2,项：二级电离的贡献,第,3,项：水自电离的贡献,一般,Ka,1,Ka,2,，差,3,4,个数量级,例：,H,2,CO,3,，,Ka,1,=4.3,10,-7,，,Ka,2,=4.8,10,-11,二元弱酸作为一元弱酸来处理（忽略二级电离）！,C,H2A,400K,a1,，近似式：,C,H2A,400K,a1,，最简式：,H,2,A(aq)+H,2,O(l)H,3,O,+,(aq)+HA,-,(,aq,),起始,C,H2A,0 0,转化x x x,平衡,C,H2A,x,x x,K,a1,=,x,2,/(C,H2A,x,),C,H2A,x,=H,2,A C,H2A,得到最简式：,推导：,例,5:,计算,0.1 mol/L,的,H,2,C,2,O,4,溶液中各离子的浓度,?,已知,:K,a1,=5.9,10,-2,K,a2,=6.4,10,-5,解：,C,H2C2O4,400,K,a1,H,2,C,2,O,4,(aq)+H,2,O(l)H,3,O,+,(aq)+HC,2,O,4,-,(aq)(1),平衡,C,x,x x,HC,2,O,4,-,=H,3,O,+,=0.0528(mol/L),H,2,C,2,O,4,=c,H,3,O,+,=0.1,0.0528=0.0472(mol/L),OH,-,=,K,w,/H,3,O,+,=1.89,10,-13,(mol/L),HC,2,O,4,-,(aq)+H,2,O(l)H,3,O,+,(aq)+C,2,O,4,2-,(aq)(2),K,a2,=H,3,O,+,C,2,O,4,2-,/,HC,2,O,4,-,=C,2,O,4,2-,C,2,O,4,2-,=K,a2,=6.4,10,-5,(mol/L),纯二元弱酸（,H,2,A,）水溶液中的,A,2-,=K,a2,!,(1)+(2),得到,H,2,C,2,O,4,总的电离反应：,H,2,C,2,O,4,(aq)+2H,2,Ol)2H,3,O,+,(aq)+C,2,O,4,2-,(aq),H,2,C,2,O,4,总的电离反应平衡常数：,K=K,a1,K,a2,=H,3,O,+,2,C,2,O,4,2-,/H,2,C,2,O,4,问题：,H,3,O,+,=2C,2,O,4,2-,（,X,，不成立,）,三元弱酸的电离平衡：,H,3,A,H,3,A(aq)+H,2,O(l),H,3,O,+,(aq),+H,2,A,-,(aq),K,a1,=H,3,O,+,H,2,A,-,/H,3,A H,3,A,的一级解离常数,H,2,A,-,(aq)+H,2,O(l),H,3,O,+,(aq),+HA,2-,(aq),K,a2,=H3,O,+,HA,2-,/H,2,A,-,H,3,A,的二级解离常数,HA,2-,(aq)+H,2,O(l)H,3,O,+,(aq)+A,3-,(aq),K,a3,=H,3,O,+,A,3-,/HA,2-,H,3,A,的三级解离常数,C,H3A,Ka,2,K,a3,三元弱酸作为一元弱酸来处理（忽略二级电离和三级电离）！,H,3,A(aq)+H,2,O(l)H,3,O,+,(aq)+H,2,A,-,(aq),起始,C,H3A,0 0,转化,x x x,平衡,C,H3A,x,x x,K,a1,=,x,2,/(C,H3A,x,),C,H3A,x,=H,3,A C,H3A,得到最简式：,推导：,H,3,A(aq)+H,2,O(l)H,3,O,+,(aq)+H,2,A,-,(aq),平衡,0.1,x,x x,问题：计算,0.1 mol/L,的,H,3,A,水溶液中各物种的浓度？,HA,2-,=K,a2,A,3-,=K,a2,K,a3,/H,3,O,+,二元弱碱的电离平衡：,Na,2,S,Na,2,S,2Na,+,+S,2-,S,2-,(aq)+H,2,O(l),OH,-,(,aq,),+HS,-,(,aq,),K,b1,=OH,-,HS,-,/S,2-,S,2-,的一级解离常数,K,b1,=,K,w,/K,a2,HS,-,(,aq,)+H,2,O(l),OH,-,(,aq,),+H,2,Saq),K,b2,=OH,-,H,2,S/HS,-,S,2-,的二级解离常数,K,b2,=,K,w,/K,a1,K,a1,K,a2,分别为,H,2,S,的一级解离常数和二级解离常数！,一般,K,b1,K,b2,二元弱碱作为一元弱碱来处理（忽略二级电离）！,C,S2-,400K,b1,，近似式：,C,S2-,400K,b1,，最简式：,S,2-,(aq)+H,2,O(l)OH,-,(aq)+HS,-,(aq),起始,C,S2-,0 0,转化,x x x,平衡,C,S2-,x,x x,K,b1,=,x,2,/(C,S2-,x,),C,S2-,x,=S,2-,C,S2-,得到最简式：,推导：,例,6,：算,0.1 mol/L Na,2,S,溶液中,OH,-,、,S,2-,、,HS,-,、,H,3,O,+,和,H,2,S,的浓度,?,已知,:H,2,S,的,K,a1,=9.1,10,-8,K,a2,=1.1,10,-12,S,2-,(aq)+H,2,O(l)OH,-,(aq)+HS,-,(aq),平衡,C,x,x x,K,b1,=K,W,/K,a2,=9.110,-3,Na,2,S,2Na,+,+S,2-,C,=0.1 K,b2,K,b3,三元弱碱作为一元弱碱来处理（忽略一级电离和二级电离）！,C,K,w,/K,a1,K,a1,K,a2,K,w,酸性,K,a2,K,w,/K,a1,K,a1,K,a2,K,w,/K,a1,酸性,Na,2,HPO,4,水溶液：,Na,2,HPO,4,2Na,+,+HPO,4,2-,HPO,4,2-,+H,2,O,H,3,O,+,+PO,4,3-,K,a,=K,a3,=1,10,-13,HPO,4,2-,+H,2,O,H,2,PO,4,-,+OH,-,K,b,=K,w,/K,a2,=1.6,10,-7,K,a3,K,w,/K,a2,碱性,H,2,CO,3,的,K,a1,=,4.3,10,-7,，,K,a2,=5.6,10,-11,NaHCO,3,水溶液,:K,a2,K,w,/K,a1,碱性,H,3,O,+,计算公式的推导：,H,2,A,是二元弱酸（K,a1,，,K,a2,），,NaHA,的浓度,C,NaHA,Na,+,+HA,HA,-,的初始浓度为,C,HA,-,+H,2,O,H,3,O,+,+A,2-,K,a,=K,a2,HA,-,+H,2,O,H,2,A,+OH,-,K,b,=K,w,/K,a1,H,2,O+H,2,O H,3,O,+,+OH,-,K,w,=H,3,O,+,OH,-,物料平衡：,C=HA,-,+,H,2,A,+,A,2-,电荷平衡：,H,3,O,+,+Na,+,=HA,-,+2A,2-,+OH,-,Na,+,=C=HA,-,+H,2,A+A,2-,代入电荷平衡式得：,H,3,O,+,+H,2,A,=,A,2-,+,OH,-,(*),由,K,a,=K,a2,=H,3,O,+,A,2-,/HA,-,得：,A,2-,=K,a2,HA,-,/H,3,O,+,由,K,b,=,K,w,/K,a1,=OH,-,H,2,A/HA,-,=K,w,H,2,A/(H,3,O,+,HA,-,),得：,H,2,A=H,3,O,+,HA,-,/K,a1,或直接由,K,a1,的定义可得：,H,2,A=H,3,O,+,HA,-,/K,a1,将,A,2-,的表达式，,H,2,A,的表达式,以及,OH,-,=,K,w,/H,3,O,+,代入等式,(*),可得：,C,HA,-,HA,-,(C,HA,-,不是太小,),精确表达式：,近似式,（K,a2,C 20K,w,，,C 20K,a1,）,（,K,a2,C 20K,w,）,近一步近似：,（,C 20K,a1,）,最简式,例,7,：计算,0.10 mol/L NaHCO,3,溶液中各组分的浓度？,已知：,H,2,CO,3,的,K,a1,=4.3,10,-7,、,K,a2,=5.6,10,-11,NaHCO,3,Na,+,+HCO,3,-,Na,+,=0.10 mol/L,HCO,3,-,的初始浓度为,0.10 mol/L,(1)HCO,3,-,+H,2,O,H,3,O,+,+CO,3,2-,K,a,=K,a2,(2)HCO,3,-,+H,2,O,OH,-,+H,2,CO,3,Kb,=,K,w,/K,a1,物料平衡：,C=HCO,3,-,+CO,3,2-,+H,2,CO,3,(3),由,(1),得：,K,a2,=CO,3,2-,H,3,O,+,/HCO,3,-,于是，,CO,3,2-,=K,a2,HCO,3,-,/H,3,O,+,由,(2),得：,K,w,/K,a1,=H,2,CO,3,OH,-,/HCO,3,-,=H,2,CO,3,K,w,/(H,3,O,+,HCO,3,-,),于是，,H,2,CO,3,=H,3,O,+,HCO,3,-,/K,a1,或者直接由,K,a1,的定义得,H,2,CO,3,=H,3,O,+,HCO,3,-,/K,a1,将,CO,3,2-,H,2,CO,3,的表达式代入,(3),得：,CO,3,2-,=K,a2,HCO,3,-,/H,3,O,+,=1.1,10,-3,(mol/L),H,2,CO,3,=H,3,O,+,HCO,3,-,/K,a1,=1.1,10,-3,(mol/L),OH,-,=,K,w,/H,3,O,+,=2.0,10,-6,(mol/L),问题：,NaH,2,PO,4,水溶液的浓度,C,，,H,3,O,+,=?,NaH,2,PO,4,Na,+,+H,2,PO,4,-,H,2,PO,4,-,+H,2,O,H,3,O,+,+HPO,4,2-,K,a,=K,a2,H,2,PO,4,-,+H,2,O,H,3,PO,4OH,-,K,b,=K,w,/K,a1,H,3,PO,4,K,a1,K,a2,K,a3,问题：,Na,2,HPO,4,水溶液的浓度,C,，,H,3,O,+,=?,Na,2,HPO,4,2Na,+,+HPO,4,2-,HPO,4,2-,+H,2,O,H,3,O,+,+PO,4,3-,K,a,=K,a3,HPO,4,2-,+H,2,O,H,2,PO,4,-,+OH,-,K,b,=K,w,/K,a2,近似式,（,K,a3,C 20K,w,，,C 20K,a2,）,（,K,a3,C 20K,w,）,近一步近似：,（,C 20K,a2,）,最简式,解,:NaHSO,4,Na,+,+HSO,4,-,(完全电离,),HSO,4,-,(aq)+H,2,O(l)H,3,O,+,(aq)+SO,4,2-,(aq),平衡,0.1,x,x x,例,8,：,P.324/,例,12,2,HSO,4,-,的水溶液只能作为弱酸，不能作为酸式盐！,5.4.,氨基酸及其等电点,COOH,：羧基；,NH,2,：胺基,氨基酸,K,a2,：,NH,3,+,的解离常数,K,a1,：,COOH,的解离常数,+H,2,O,H,3,O,+,+,K,a,=K,a2,(K,a2,:NH,3,+,的解离常数,),+H,2,O,OH,-,+,推导：,K,b,=,K,w,/K,a1,(K,a1,:COOH,的解离常数,),氨基酸相当于酸碱两性物质！,取最简式：,氨基酸的等电点,I.P.,（,Isoelectric,Point):,电中性的氨基酸水溶液的,pH,值,例,9,：计算,0.10 mol/L,氨基乙酸溶液的,pH,值。,已知：,K,a1,=4.5,10,-3,、,K,a2,=1.7,10,-10,I.P.=-lgH,3,O,+,=(-lgK,a1,-,lgK,a2,)/2=(pK,a1,+pK,a2,)/2,pH=-lgH,3,O,+,=-(lgK,a1,+,lgK,a2,)/2,=-(lg(4.510,-3,)+lg(1.710,-10,)/2=6.06,在,pH 6.06,的水溶液中，氨基乙酸带负电荷,有些氨基酸的侧基,-R,上带有能给出质子的官能团，如：,这些氨基酸的质子化产物是三元酸,H,3,A,+,，它们有三个离解常数：,K,a1,，,K,a2,和,K,a3,(K,a1,K,a2,K,a3,),。因为这些氨基酸的第一级离解产物就是不带电荷的两性品种,H,2,A,，所以这类氨基酸的等电点也是：,I.P.=(pK,a1,+pK,a2,)/2,半胱氨酸,H,2,A+H,2,O,H,3,O,+,+HA,-,K,a,=K,a2,H,2,A+H,2,O,OH,-,+H,3,A,+,推导：,取最简式：,K,b,=,K,w,/K,a1,P351/,表,12-9,：半胱氨酸,I.P.=(pK,a1,+pK,a2)/2=(1.96+8.18)/2=5.07,I.P.=-lgH,3,O,+,=(pK,a1,+pK,a2,)/2,有一些氨基酸的侧基,-R,上带有能接受质子的,N,原子，如,:,这些氨基酸可以在水溶液中接受二个质子，它们的质子化产物也是三元酸,H,3,A,2+,，也有三个离解常数：,K,a1,，,K,a2,和,K,a3,(K,a1,K,a2,K,a3,),。但是,H,3,A,2+,的第一级离解产物是带正电荷的两性品种,H,2,A,+,，第二级离解产物才是不带电荷的两性品种,HA,。所以它们的等电点是：,I.P.=(pK,a2,+pK,a3,)/2,赖氨酸,HA+H,2,O,H,3,O,+,A,-,K,a,=K,a3,HA+H,2,O,OH,-,+H,2,A,+,推导：,取最简式：,K,b,=,K,w,/K,a2,P351/,表,12-9,：赖氨酸,I.P.=(pK,a2,+pK,a3,)/2=(8.95+10.53)/2=9.74,I.P.=-lgH,3,O,+,=(pK,a2,+pK,a3,)/2,作业：,P334,：,2,(HF,K,a,=6.6,10,-4,;HNO,2,K,a,=4.610,-4,),P346,：,1,，,3,，,P347:,4,3,：,H,2,CO,3,，,K,a1,=4.3,10,-7,，,K,a2,=4.8,10,-11,4,：,P319/,表,12-4,H,3,PO,4,的,K,a1,=7.5,10,-3,、,K,a2,=6.2,10,-8,、,K,a3,=2.0,10,-13,