天津大学无机化学课件第三章酸碱反应和沉淀反应

*,无机化学,*,2007-5-27,第一节水的解离反响和溶液的酸碱性,无机化学多媒体电子教案,第三章 酸碱反响和沉淀反响,第一节,水的解离反响,和溶液的酸碱性,2024/9/30,2007-5-27,3-1-1,酸碱的电离理论,3-1-1,酸碱的电离理论,酸碱理论,阿仑尼乌斯,(Arrhenius),电离理论,富兰克林,(Franklin),溶剂理论,布朗斯泰德,(Bronsted),质子理论,路易斯,(Lewis),电子理论,皮尔逊,(Pearson),软硬酸碱理论,2024/9/30,2007-5-27,3-1-1,酸碱的电离理论,3-1-1,酸碱的电离理论,阿仑尼乌斯酸碱电离理论,酸,：,在水溶液中解离产生的阳离子全部是,H,+,的化合物,碱,：在水溶液中解离产生的阴离子全部是,OH,-,的化合物,酸,碱,中和反应的实质：,H,+,和,OH,-,结合生成,H,2,O,酸碱的相对强弱可根据在水溶液中解离,出来的,H,+,和,OH,-,程度来衡量,2024/9/30,2007-5-27,3-1-2 水的解离反响和溶液的酸碱性,3-1-2 水的解离反响和溶液的酸碱性,水的解离反响,纯水或稀溶液中,H,2,O(l),H,+,(aq) + OH,-,(aq),c,(H,+,)/,c,c,(OH,-,)/,c,=,K,w,K,w,水的离子积,K,w,(298.15K)=1.0,10,-14,K,w,与温度有关,2024/9/30,2007-5-27,溶液的酸碱性和,pH,pH=-lg ,c,(H,+,)/,c,溶液的酸碱性,1.0,10,-7,1.0,10,-7,1.0,10,-7,c,(OH,-,)/,molL,-1,1.0,10,-7,c,(H,+,)/,molL,-1,碱性,中性,酸性,溶液酸碱性,c,(H,+,)/,c,c,(OH,-,)/,c,=,1.0,10,-14,=,K,w,pH=p,K,w,-pOH=14-pOH,pOH=-lg ,c,(OH,-,)/,c,碱性,中性,酸性,溶液酸碱性,7,=7,Kw , 浓度不很小, 可忽略水的解离,一元弱酸溶液中,c,(H,+,),的计算,一元弱电解质溶液中离子浓度的计算,若,(,c,/,c,)/,K,a,500,x,500,0.100-,x,x,则,x,=1.34,10,-3,c,(NH,4,),/,c,c,(OH,-,),/,c,xx,c,(,NH,3,H,2,O,),/,c, 0.100-,x,K,b,= = =1.8,10,-5,+,2024/9/30,2007-5-27,例,c,(OH,-,)=1.34,10,-3,molL,-1,pH = 14-2.87 =,11.13,计算,0.100molL,-1,氨水溶液中的,c,(OH,-,),、,pH,和氨水的解离度,c,(OH,-,) 1.34,10,-3,c,0.100,=,100%=,100%,=,1.34%,所以,pOH = -lg(1.34,10,-3,/c,),= 2.87,2024/9/30,2007-5-27,3-2-4,多元弱酸的分步解,3-2-4,多元弱酸的分步解离,多元弱酸在水溶液中,分步解离,如,H,2,S H,+,+ HS,-,HS,-,H,+,+ S,2-,K,a(1),= =1.1,10,-7,c,(H,+,)/,c,c,(HS,-,)/,c,c,(H,2,S)/,c,K,a(2),= =1.3,10,-13,c,(H,+,)/,c,c,(S,2-,)/,c,c,(HS,-,)/,c,分级解离常数逐级减小,多元弱酸的强弱主要取决于,K,a(1),2024/9/30,2007-5-27,例,常温、常压下H2S的溶解度为,0.10molL-1，计算H2S饱和溶液中c(H+)、,c(S2-)和H2S的解离度,解：,H,2,S H,+,+ HS,-,平衡浓度,/(molL,-1,) 0.10-,x x x,K,a(1),= = =1.1,10,-7,c,(H,+,)/,c,c,(HS,-,)/,c,xx,c,(H,2,S)/,c,0.10-,x,(,c,/,c,)/,K,a,(1),=0.10/(1.1,10,-7,)500, 0.10-,x,0.1,x,= 1.1,10,-4,c,(H,+,)=,1.1,10,-4,molL,-1,2024/9/30,2007-5-27,例,常温、常压下H2S的溶解度为,0.10molL-1，计算H2S饱和溶液中c(H+)、,c(S2-)和H2S的解离度,解：,HS,-,H,+,+ S,2-,K,a(2),= =1.3,10,-13,c,(H,+,)/,c,c,(S,2-,)/,c,c,(HS,-,)/,c,因为,K,a(2),500,所以,0.10-,x,0.10, 0.10+,x,0.10,解：,NH,3,H,2,O NH,4,+ OH,-,平衡浓度,/(molL,-1,) 0.10-,x,0.10+,x x,+,2024/9/30,2007-5-27,例,在0.10molL-1NH3H2O溶液中, 参加,固体NH4Cl,使其浓度为0.10molL-1, 计算,溶液中c(H+)、NH3H2O的解离度。,解：,NH,3,H,2,O NH,4,+ OH,-,平衡浓度,/(molL,-1,) 0.10-,x,0.10+,x x,+,c,(OH,-,)=1.8,10,-5,molL,-1,1.0,10,-14,1.8,10,-5,c,(H,+,) = molL,-1,=5.6,10,-10,molL,-1,1.8,10,-5,0.10,=,100%=,0.018%,未加,NH,4,Cl,的,0.10molL,-1,NH,3,H,2,O,溶液 ,1.34%,即,由于同离子效应，使,降低,利用同离子效应, 调节溶液的酸碱性,可控制弱电解质溶液中有关离子的浓度，,进展金属离子的别离和提纯。,2024/9/30,2007-5-27,3-2-6,缓冲溶液,3-2-6,缓冲溶液,演示实验,：,HOAc NaOAc,混合溶液,缓冲作用原理：如,HOAc+NaOAc,混合液,HOAc,H,+,+,OAc,-,大量 极小量 大量,外加适量酸,(H,),，平衡向左移动,外加适量碱,(OH,-,),，平衡向右移动,NaOAc,抵消,H,+,; HOAc,抵消,OH,-,缓冲作用：使溶液,pH,值基本保持不变的作用。具有缓冲作用的溶液，称为缓冲溶液,2024/9/30,2007-5-27,3-2-6,缓冲溶液,演示实验,：,HOAc NaOAc,混合溶液,缓冲作用原理：如,HOAc+NaOAc,混合液,HOAc,H,+,+,OAc,-,大量 极小量 大量,外加适量酸,(H,),，平衡向左移动,外加适量碱,(OH,-,),，平衡向右移动,NaOAc,抵消,H,+,; HOAc,抵消,OH,-,缓冲作用：使溶液,pH,值基本保持不变的作用。具有缓冲作用的溶液，称为缓冲溶液,HOAc+NaOAc混合溶液能抵抗少量外,来酸、碱及少量水的稀释作用，而保,持溶液pH值根本不变，为缓冲溶液。,2024/9/30,2007-5-27,常见的缓冲溶液,如弱酸与弱酸盐、弱碱与弱碱盐等混合液，,多元弱酸组成的两种不同酸度的盐混合液,缓冲溶液,p,K,i,pH,值范围,HOAc-NaOAc,4.76,3.65.6,NH,3,H,2,O-NH,4,Cl,4.76,8.310.3,NaHCO,3,-Na,2,CO,3,10.33,9.211.0,KH,2,PO,4,-K,2,HPO,4,7.20,5.98.0,H,3,BO,3,-Na,2,B,4,O,7,9.2,7.29.2,2024/9/30,2007-5-27,缓冲溶液,pH,值的计算,例,试计算含,0.10mol,L,-1,HOAc,、,0.10molL,-1,NaOAc,溶液的,pH,值。,解：,HOAc H,+,+ OAc,-,平衡浓度,/(molL,-1,) 0.10-,x,x,0.10+,x,K,a,= =1.8,10,-5,x,(0.10+,x,),0.10-,x,因为,(,c,/,c,)/,K,a,=0.10/(1.8,10,-5,)500,所以,0.10-,x,0.10, 0.10+,x,0.10,pH=,4.74,x,=1.8,10,-5,c,(H,+,)=,1.8,10,-5,molL,-1,=1.8,10,-5,0.10,x,0.10,2024/9/30,2007-5-27,例,(2) 在含0.100molL-1HOAc、0.100molL-1,NaOAc溶液中参加HCl, 使c(Cl-) = 0.001,molL-1，计算溶液的pH值。,HOAc H,+,+ OAc,-,K,a,= =1.8,10,-5,x,(0.099+,x,),0. 101-,x,平衡浓度,/(molL,-1,) 0.101-,x,x,0.099+,x,x,=1.8,10,-5,c,(H,+,)=,1.8,10,-5,molL,-1,pH=,4.74,=1.8,10,-5,0.099,x,0. 101,解,:,反应,H,+,+ OAc,-,HOAc,0.001 0.001 0.001,HOAc+NaOAc,混合溶液能抵抗少量外来酸的作用，而保持溶液,pH,值不变。,2024/9/30,2007-5-27,例,(3) 在含0.100molL-1HOAc、0.100,molL-1NaOAc溶液中参加KOH, 使,c(K+)=0.001molL-1，计算溶液的pH值。,HOAc H,+,+ OAc,-,K,a,= =1.8,10,-5,x,(0.101+,x,),0. 099-,x,平衡浓度,/(molL,-1,) 0.099-,x,x,0.101+,x,x,=1.8,10,-5,c,(H,+,)=,1.8,10,-5,molL,-1,pH=,4.74,=1.8,10,-5,0.101,x,0. 099,解,:,反应,OH,-,+ HOAc,OAc,-,+ H,2,O,0.001 0.001 0.001,HOAc+NaOAc,混合溶液能抵抗少量外来碱的作用，而保持溶液,pH,值不变。,2024/9/30,2007-5-27,例,(4) 在含0.10molL-1HOAc、0.10molL-1,NaOAc溶液中参加H2O, 使溶液稀释10倍,计算溶液的pH值。,解：,HOAc H,+,+ OAc,-,平衡浓度,/(molL,-1,) 0.010-,x,x,0.010+,x,K,a,= =1.8,10,-5,x,(0.010+,x,),0. 010-,x,c,(H,+,)=,1.8,10,-5,molL,-1,pH=,4.74,x,=1.8,10,-5,=1.8,10,-5,0.010,x,0. 010,HOAc+NaOAc,混合溶液能抵抗少量水的稀释作用，而保持溶液,pH,值不变。,2024/9/30,2007-5-27,例,(5)欲配制250ml, pH=5.00的缓冲溶液，,问在12.0mL 6.00molL-1HOAc溶液中,应参加固体NaOAc3H2O 多少克,解：,c,(HOAc)=(12.06.00)/250 molL,-1,=0.288 molL,-1,pH=5.00,c,(H,+,)=10,-5,molL,-1,HOAc H,+,+ OAc,-,平衡浓度,/(molL,-1,) 0.288-10,-5,10,-5,x+,10,-5,K,a,= =1.8,10,-5,10,-5,(,x,+,10,-5,),0. 288-10,-5,x,=,0.518,molL,-1,应参加NaOAc3H2O,(250/1000)0.518136=17.6 (g),2024/9/30,2007-5-27,第三节盐类的水解反响,无机化学多媒体电子教案,第三章 酸碱反响和沉淀反响,第三节,盐类的水解反响,2024/9/30,caikeying,2007-5-27,3-3-1 水解反响和水解常数,3-3-1 水解反响和水解常数,强碱弱酸盐的水溶液,水解反应式：,OAc,-,H,2,O,HOAc,OH,-,如,NaOAc,Na,+,+ OAc,-,H,2,O OH,-,+ H,+,水的解离,+,HOAc,水解反响,显碱性,2024/9/30,2007-5-27,强酸弱碱盐水溶液,水解反应式：,NH,4,H,2,O,NH,3,H,2,O,H,+,+,显酸性,+,NH,3,H,2,O,如,NH,4,Cl,Cl,-,+ NH,4,H,2,O H,+,+ OH,-,+,盐的水解反应,:,盐的组分离子与水解离出来的,H,+,或,OH,-,结合生成弱电解质的反应,2024/9/30,2007-5-27,弱酸弱碱盐,水溶液的酸碱性视生成弱酸,、,弱碱的,K,i,而定,A,+,+ B,-,+ H,2,O HB + AOH,K,a,(HB),K,b,(AOH),水溶液显,酸性,如,NH,4,F NH,4,+ F,-,+H,2,O,NH,3,H,2,O+HF,+,K,a,(HB),K,b,(AOH),水溶液显,中性,如,NH,4,OAc,NH,4,+ OAc,-,+H,2,O,NH,3,H,2,O+HOAc,+,如,NH,4,CN NH,4,+CN,-,+H,2,O NH,3,H,2,O + HCN,K,a,(HB),K,h(2),通常只须考虑第一步水解,2024/9/30,2007-5-27,如FeCl3的水解反响式,Fe,3+,+ H,2,O Fe(OH),2+,+,H,+,由于,K,h(1),K,h(2),K,h(3),所以一般只考虑一级水解,3-3-2,分步水解,Fe(OH),2+,+ H,2,O Fe(OH),2,+,H,+,Fe(OH),2,+ H,2,O Fe(OH),3,+,H,+,2024/9/30,2007-5-27,3-3-3,盐溶液,pH,的近似计算,3-3-3,盐溶液,pH,的近似计算,例,计算,0.10molL,-1,NaOAc,溶液的,pH,和水解度,解：,OAc,-,+H,2,O HOAc,+,OH,-,平衡浓度,/(molL,-1,),0.10-,x,x x,0.10-,x,0.10,x,=7.5,10,-6,c,(OH,-,)=7.5,10,-6,molL,-1,pH=14-lg(7.5,10,-6,)=,8.88,K,h,= = = =5.6,10,-10,K,w,1.0,10,-14,K,a,(HOAc),1.8,10,-5,xx,0.10,-x,7.5,10,-6,0.10,h,=,100%=7.5,10,-3,%,2024/9/30,2007-5-27,3-3-4,影响盐类水解度的因素,3-3-4,影响盐类水解度的因素,水解离子的本性,水解产物,弱酸或弱碱越弱，则水解,程度越大。,盐溶液,(0.1molL,-1,),水解产物,K,a,h,/%,pH,NaOAc,HOAc+OH,-,1.810,-5,0.0075,8.9,NaCN,HCN+OH,-,6.210,-10,1.4,11.2,Na,2,CO,3,HCO,3,+OH,-,4.710,-11,4.2,11.6,-,2024/9/30,2007-5-27,3-3-4,影响盐类水解度的因素,水解离子的本性,水解产物是弱电解质，且为难溶或为易,挥发气体，则水解程度很大或完全水解,Al,2,S,3,+6H,2,O,2,Al(OH),3,+ 3H,2,S,SnCl,2,+ H,2,O,Sn(OH)Cl,+ HClSbCl,3,+ H,2,O,SbOCl,+ 2HCl,Bi(NO,3,),3,+ H,2,O,BiONO,3,+ 2HNO,3,2024/9/30,2007-5-27,3-3-4,影响盐类水解度的因素,盐溶液浓度、温度,一般来说，盐浓度越小，温度越高，,盐的水解度越大。,盐溶液酸度,降低,溶液的,pH,，可增大阴离子的水解度,升高,溶液的,pH,，可增大阳离子的水解度,2024/9/30,2007-5-27,3-3-5,盐类水解的抑制和利用,3-3-5,盐类水解的抑制和利用,易水解盐溶液的配制,为抑制水解，必须将它们溶解在相应的碱或酸中,配制SnCl2、SbCl3溶液,应先参加适量HCl,如,Sn,2+,+ H,2,O Sn(OH),+,+ H,+,Sn(OH)Cl,+,Cl,-,加入,HCl,2024/9/30,2007-5-27,易水解盐溶液的配制,配制,SnCl,2,、,SbCl,3,溶液,应先加入适量,HCl,加入,HCl,如,Sb,3+,+ H,2,O Sb(OH),2+,+ H,+,SbOCl,+,Cl,-,+,H,2,O,Sb(OH),2,+,+ H,+,-H,2,O,(,氯化氧锑,),2024/9/30,2007-5-27,易水解盐溶液的配制,配制,Bi(NO,3,),3,溶液,应先加入适量,HNO,3,(,硝酸氧铋,),Bi,3+,+ H,2,O Bi(OH),2+,+ H,+,BiONO,3,+,NO,3,-,+,H,2,O,Bi(OH),2,+,+ H,+,-H,2,O,加入,HNO,3,2024/9/30,2007-5-27,易水解盐溶液的配制,配制,Na,2,S,溶液,应先加入适量,NaOH,S,2-,+ H,2,O HS,-,+ OH,-,+,H,2,O,H,2,S,+ OH,-,加入,NaOH,2024/9/30,2007-5-27,利用盐类水解进展离子的别离和提纯,如除去溶液中的,Fe,2+,、,Fe,3+,2Fe,2+,+ H,2,O,2,+ 2H,+,2Fe,3+,+ 2H,2,O,1.,加入氧化剂,(,如,H,2,O,2,),，使,Fe,2+,Fe,3+,2.,降低酸度，调节溶液,pH=34,，促使,Fe,3+,水解，生成,Fe(OH),3,.,加热，促使,Fe,3+,水解，生成,Fe(OH),3,2024/9/30,