常用酸碱溶液质子条件式

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,*,*,*,常用酸碱溶液的质子条件,式,物质,零水准,质子条件式,(,PCE,),HA,HA,H,2,O,H,+,=A,-,+OH,-,A,-,A,-,H,2,O,H,+,+HA=OH,-,*,HA,-,(,两性物,),HA,-,H,2,O,H,2,A+H,+,=A,2,-,+OH,-,H,2,A,H,2,A,H,2,O,H,+,=HA,-,+2A,2,-,+OH,-,H,2,O,H,2,O,H,+,=OH,-,1,2,1.,一元弱酸,(,HA,),的,H,+,的计算,质子条件式,:,H,+,=A,-,+OH,-,若将 代入精确表达式,展开得一元三次方程,H,+,3,+,K,a,H,+,2,-(,K,a,c,a,+,K,w,)H,+,-,K,a,K,w,=0,代入平衡关系式,精确表达式：,3,若,:,K,a,c,20,K,w,(10,-12.7,),忽略,K,w,(,即忽略水的酸性,),HA=,c,a,-A,-,=,c,a,-(H,+,-OH,-,),c,a,-H,+,得近似式,:,展开,:,H,+,2,+,K,a,H,+,-,c,a,K,a,=0,若,20,K,w,K,a,/,c,=10,-1.26,/0.20=10,-0.56,2.510,-3,故应用近似式,:,解一元二次方程,:,H,+,=10,-1.09,则,pH=1.09,如不考虑酸的离解,(,用最简式,:,pH=0.98,),则,E,r,=29%,7,2.,一元弱碱,(A,-,),溶液,的,OH,-,计算,质子条件,:,H,+,+HA=OH,-,引入平衡关系,:,OH,-,=,精确表达式,:,若,:,K,b,c,b,20,K,w,(10,-12.7,),忽略,K,w,(,水的碱性,),近似式,:,若,0.05,(,即,K,b,/,c,b,K,a,2,HA,-,c,(HA),(,若,p,K,a,3.2,则,5%,),又,:,K,a,2,c,(HA,-,)20,K,w,得,近似式,:,精确表达式,:,12,近似式,:,如果,c,(HA,-,)/,K,a,1,20,则“,1,”,可略,得,最简式,:,(,若,c,/,K,a,1,20,K,a,2,c,20,K,w,即使,K,a,1,与,K,a,2,相近,也可使用最简式计算,),13,14,15,例,3,计算,0.050molL,-1,NaHCO,3,溶液,的,pH,.,解,:,p,K,a,1,=6.38,p,K,a,2,=10.25,p,K,a,=3.87 3.2,HA,-,c,(HA,-,),K,a,2,c,=10,-10.25,0.050=10,-11.55,20,K,w,c,/,K,a,1,=0.050/10,-6.38,=10,5.08,20,故可用最简式,:,H,+,=10,-8.32,molL,-1,pH=8.32,16,例,4,计算,0.033molL,-1,Na,2,HPO,4,溶液,的,pH,.,p,K,a,1,p,K,a,3,:2.16,7.21,12.32,解,:,K,a,3,c,=10,-12.32,0.033=10,-13.80,K,w,c,/,K,a,2,=0.033/10,-7.21,=10,5.73,20,故,:,(,若用最简式,pH=9.77,E,r,=22%),pH=9.66,17,氨基乙酸,(,甘氨酸,),NH,2,CH,2,COOH,NH,3,+,CH,2,COOH NH,3,+,CH,2,COO,-,NH,2,CH,2,COO,-,p,K,a,1,=2.35 p,K,a,2,=9.78,H,2,A,+,H,+,A,-,A,-,氨基乙酸,氨基乙酸,氨基乙酸,阳离子 双极离子 阴离子,等电点时,:H,2,A,+,=A,-,-H,+,-H,+,+H,+,+H,+,甘氨酸溶液的,pH,与其等电点时的,pH,(,即,pI,),是否相等,?,18,19,蛋白质基本结构单位,氨基酸,氨基酸,结构式,p,K,a,等点电,pH,甘氨酸,Glycine,H,H-,C-COO,-,+,NH,3,2.34,9.60,5.97,丝氨酸,Serine,H,HO-CH,2,-C-COO,-,+,NH,3,2.21,9.15,5.68,丙氨酸,Alanine,H,CH,3,-C-COO,-,+,NH,3,2.34,9.69,6.02,20,氨基酸,氨基酸,结构式,p,K,a,等点电,pH,缬氨酸,Valine,CH,3,H,CH,3,-CH-C-COO,-,+,NH,3,2.32,9.62,5.97,谷氨酸,GlutamicAcid,H,HOOC,-CH,2,-CH,2,-,C-COO,-,+,NH,3,2.19,4.25,9.67,3.22,赖氨酸,Lysine,H,H,3,N,+,-C,4,H,8,-,C-COO,-,+,NH,3,2.18,8.95,10.53,9.74,21,蛋白质的等电点,(,pI,),鱼精蛋白,12.0 12.4,胰岛素,5.30 5.35,细胞色素,C,9.7,血清甲种,球蛋白,5.1,肌红蛋白,7.0,血清蛋白,4.80 4.85,血清丙种,球蛋白,6.3 7.3,胃蛋白酶,2.71 3.00,血清乙种,球蛋白,5.6,丝蛋白,2.0 2.4,22,4.,弱碱,(,A,-,),+,弱酸,(,HB,),溶液的,pH,(,K,a,(HA),K,a,(HB),质子条件,:,H,+,+HA=B,-,+OH,-,精确表达式,:,近似式,:,最简式,:,若,c,(HB)=,c,(A,-,),(,与两性物相同,),23,例,5,计算,0.10molL,-1,Ac,-,+0.20molL,-1,H,3,BO,3,溶液的,pH.,解：,K,a,(HAc,)=10,-4.76,K,a,(H,3,BO,3,)=10,-9.24,pH=6.85,24,25,26,5.,多元弱酸,(,碱,),溶液的,pH,计算,以二元酸,(H,2,A),为例,质子条件,:H,+,=HA,-,+2A,2-,+OH,-,精确式：,若,/H,+,0.05,则该项可略,.,近似式,：,以下与一元酸的计算方法相同,.,大部分多元酸均可按一元酸处理,忽略第二步及以后各步的离解,.,27,28,6.,强酸,(,HCl,),、,强碱,(,NaOH,),溶液,的,pH,计算,强酸,(,HCl,),质子条件,:H,+,=,c,(HCl,)+OH,-,一般情况,:H,+,=,c,(HCl,),强碱,(,NaOH,),质子条件,:H,+,+,c,(NaOH,)=OH,-,一般情况,:OH,-,=,c,(NaOH,),c,(NaOH,)10,-6,molL,-1,则可以忽略水的碱性,.,29,30,7.,强碱,(,NaOH,)+,弱碱,(A,-,),的,pH,计算,质子条件,:,H,+,+HA+,c,(NaOH,)=OH,-,近似式,:,忽略弱碱的离解,:,最简式：,OH,-,c,(NaOH,),31,例,6,计算,20.00 ml 0.1000molL,-1,HA(,p,K,a,=7.00)+20.04 ml 0.1000molL,-1,NaOH,的,pH.,解方程得,解,：,先按简化式，,弱碱,A,解离的,OH,不能忽略,，,32,33,34,35,36,37,38,39,40,41,42,8.,强酸,(,HCl,)+,弱酸,(HA),的,pH,计算,:,质子条件,:,H,+,=,c,(HCl,)+A,-,+OH,-,近似式,:,忽略弱酸的离解,:,最简式,：,H,+,c,(HCl,),H,2,SO,4,为何体系？,43,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,.,44,9.,两弱酸,(HA+HB),溶液,H,+,=,两,弱碱溶液类似处理,.,45,以,HA,、,A,-,、,H,2,O,为零水准,H,+,+HA,A,=OH,-,+,A,-,HA,c,a,=HA,HA,+,A,-,HA,即,A,-,HA,=,c,a,-HA,HA,，,代入上式,c,a,=HA,A,+HA,HA,+H,+,-OH,-,=HA+H,+,-OH,-,可得,HA=,c,a,H,+,+OH,-,10.,共轭酸碱体系,(HA+A,-,),的,pH,计算,46,类似地，将,c,b,=A,-,A,+HA,A,代入质子条件式，可得,A,-,=,c,b,+H,+,-OH,-,代入平衡关系,得精确式：,47,酸性,略去,OH,-,(,近似式,),碱性,略去,H,+,:,(,近似式,),若：,c,a,OH,-,-H,+,，,c,b,H,+,-OH,-,则得最简式：,计算方法,:,(1),先按最简式计算,H,+,.,(2),再将,OH,-,或,H,+,与,c,a,、,c,b,比较,看忽略是否合理,.,48,例,7,计算以下溶液的,pH,(a)0.040 molL,-1,HAc,0.060 molL,-1,NaAc,解,(a),先按最简式,:,c,a,H,+,c,b,H,+,结果合理,pH=4.94,49,(b),0.080molL,-1,二氯乙酸,0.12molL,-1,二氯乙酸钠,解,(b),先用最简式,应用近似式,:,解一元二次方程,：,H,+,=10,-1.65,molL,-1,pH=1.65,50,例,8,在,20.00mL 0.1000molL,-1,HA(,K,a,=10,-7.00,),溶液中,加入,0.1000molL,-1,NaOH,溶液,19.96mL,计算,pH,.,解,:,c,a,=10,-4.00,molL,-1,c,b,=10,-1.30,molL,-1,先按最简式,OH,-,=10,-4.30,molL,-1,应用近似式,:,解一元二次方程,:OH,-,=10,-4.44,molL,-1,pH=9.56,51,酸碱溶液,H,+,计算小结,一,.,一元弱酸,(,碱,):,HA,(A,-,),多元弱酸,(,碱,):,H,n,A,(A,n,-,),二,.,两性物质,:,HA,-,类两性物质,:,A,-,+HB,三,.,混合酸碱,:,强,+,弱、弱,+,弱,四,.,共轭酸碱,:,HA+A,-,判断条件,确定用近似式或最简式,.,先用,最简式计算,再看是否合理,.,52,复杂酸碱体系溶液中,酸碱形态,的计算,例,9,10mL 0.20molL,-1,HCl,溶液,与,10mL0.50molL,-1,HCOONa,和,2.010,-4,molL,-1,Na,2,C,2,O,4,溶液混和,计算溶液中的,C,2,O,4,2-,.,已知,:,p,K,a,(HCOOH,)=3.77;,H,2,C,2,O,4,的 为,1.25,4.29,53,解,(,方法一,),混和后发生了酸碱中和反应,可忽略,Na,2,C,2,O,4,消耗的,H,+,.,c,(HCOOH)=0.10 molL,-1,c,(HCOO,-,),=,0.15 molL,-1,C,2,O,4,2-,=,c,(C,2,O,4,2-,),x,(C,2,O,4,2-,),=3.210,-5,(molL,-1,),54,(,方法二,),混和后不考虑发生反应,Zero Level:,HCl,，,HCOO,-,，,C,2,O,4,2,-,，,H,2,O,PCE:,H,+,+HCOOH+HC,2,O,4,-,+2H,2,C,2,O,4,=OH,-,+,c,(HCl,),混和后,c,(HCl,)=0.10 molL,-,1,c,(HCOO,-,)=0.25 molL,-,1,解得,:,H,+,=10,-,3.95,molL,-,1,C,2,O,4,2,-,=3.210,-,5,molL,-,1,55,3.4,酸碱缓冲溶液,3.4.1,缓冲容量,使,1L,溶液的,pH,增加,dpH,单位时所需强碱,d,b,mol,使,1L,溶液的,pH,减少,dpH,单位时所需强酸,d,a,mol,或是,56,质子条件,:,H,+,+,b,=A,-,+OH,-,HA-A,-,体系：,HA(,c,)+OH,-,(,b,),根据,和,换算得,:,57,强酸,:,H,+,=,2.3H,+,=,2.3,c,(HCl),(pH11),HA A,-,:,HA A,-,=,2.3,c,x,1,x,0,1.,c,c,增大则,增大,2.,H,+,=,K,a,，,即,pH,=,p,K,a,时,max,=,0.575,c,3.,缓冲范围,:p,K,a,-,1,pH,p,K,a,+,1,共轭体系缓冲容量的特点：,58,2.88,4.76,p,K,a,H,+,0.1molL,-,1,HAc,0.1molL,-,1,Ac,-,8.88,OH,-,HAc,+,Ac,-,0.1,molL,-,1,HAc,溶液的,-pH,曲线,59,p,K,a,-,1,x,1,=0.9,x,0,=0.1,=0.36,max,p,K,a,+,1,x,1,=0.1,x,0,=0.9,=0.36,max,p,K,a,-,2,x,1,=0.99,x,0,=0.01,=0.04,max,p,K,a,+,2,x,1,=0.01,x,0,=0.99,=0.04,max,60,3.4.2,缓冲溶液的选择,1.,有较大的缓冲能力,:,c,较大,(,0.01,1molL,-1,);,pHp,K,a,即,c,a,:,c,b,1:1,4.5,6,8,10,HAc,NaAc,:,p,K,a,=4.76 (pH,4,5.5,),NH,4,Cl NH,3,:,p,K,b,=4.75 (pH ),(CH,2,),6,N,4,(CH,2,),6,N,4,H,+,:,p,K,b,=8.87,(pH,),2.,不干扰测定,:,例如,EDTA,滴定,Pb,2+,时,不用,HAc,-Ac,-,控制,pH.,61,3.4.3,缓冲溶液的配制方法,1.,按比例加入,HA,和,A,(,NaAc+HAc,NH,4,Cl+NH,3,);,2.,溶液中,H,+,大,加过量,A,-,溶液中,OH,-,大,加过量,HA;,3.,溶液中有,HA,可加,NaOH,中和一部分,溶液中有,A,-,可加,HCl,中和一部分,.,缓冲溶液,浓度的计算,方法有,:,缓冲容量法,、,缓冲溶液,pH,计算式法,、及,摩尔分数法,.,62,配制缓冲溶液的计量方法,缓冲容量法,:(,引入酸碱均适用,),1.,根据引入的酸,(,碱,),量求,2.,根据所需,pH,求总浓度,c,：,=2.3,c,x,0,x,1,3.,按,x,0,与,x,1,的比值配成总浓度为,c,的缓冲溶液,.,例,10,欲制备,200,mL,pH=9.35,的,NH,3,-NH,4,Cl,缓冲溶液，且使该溶液在加入,1.0,mmol,的,HCl,或,NaOH,时，,pH,的改变不大于,0.12,单位，需用多少克,NH,4,Cl,和多少毫升,1.0 molL,-1,氨水？,63,解：,又,m,(NH,4,Cl)=,c,x,(NH,4,+,),V,M,V,(NH,3,)=,c,x,(NH,3,),V,/,c,(NH,3,),64,缓冲溶液,pH,计算式法,(,确定引入酸或碱时,),:,1.,按所需,pH,求,c,a,/,c,b,:,2.,按引入的,酸,(,碱,),量,c,a,(,c,b,),及,pH,变化限度计算,c,a,和,c,b,:,3.,据,c,a,、,c,b,及所需溶液体积配制溶液,.,65,摩尔分数法,(,确定引入酸或碱时,),：,1.,分别计算,pH,0,和,pH,1,时的,x,(HA),、,x,(A),2.,引入酸的量,c,(HCl,)=,x,(HA),c,总,(,引入碱的,量,c,(NaOH,)=,x,(A),c,总,),求出,c,总,3.pH,0,时,c,a,=,c,总,x,(HA),0,c,b,=,c,总,x,(A),0,66,常用缓冲物质,弱酸,p,K,a,弱酸,p,K,a,氨基乙酸,2.5 9.7,*硼砂,9.24,氯乙酸,2.7,*磷酸,2.0 6.9 11.7,甲酸,3.65,弱碱,p,K,b,丁二酸,(,琥珀酸,),4.00 5.28,*六次甲基四胺,8.74,*乙酸,4.65,三乙醇胺,7.9,柠檬酸,3.0 4.4 6.1,*三羟甲基甲胺,5.79,碳酸,6.3 10.1,*NH,3,4.63,表,中,打,*,者最为常用,.,67,p,K,a,:3,5,7,9,11,五种弱酸,c,0.1molL,-1,max,=,0.058,全域缓冲溶液,pH,图,68,常用标准缓冲溶液,(,用于,pH,计校准,),标准缓冲溶液的,pH,是,实验测定的,计算时应做活度校正,.,缓 冲 溶 液,pH(25,o,C),饱和酒石酸氢钾,(0.034 molkg,-1,),3.557,邻苯二甲酸氢钾,(0.050 molkg,-1,),4.008,0.025 molkg,-1,KH,2,PO,4,+,0.025 molkg,-1,Na,2,HPO,4,6.865,硼砂,(0.010 molkg,-1,),9.180,饱和氢氧化钙,12.454,69,习 题,3.7 3.8 3.9 3.10 3.14,70,71,72,73,74,75,76,77,78,79,80,81,82,