天津大学无机化学课件第三章酸碱反应和沉淀反应

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1、*,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

2、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,

3、+,)/,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,-,)/,mol·L,-1,<1.0,×,10,-7,1.0,×,10,-7,>1.0,×,10,-7,c,(H,+,)/,mol·L,-1,碱性,中性,酸性,溶液酸碱性,,[,

4、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,<7,pH,pH,越小，溶液酸性越强,,pH,越大，溶液碱性越强,2024/9/30,,,2007-5-27,,,,,,例１,0.10 mol·L-1HOAc溶液中, c(H+)=1.34×10-3 mol·L-1, pH=,,例,2,0.10 mol·L-1NH3·H2O溶液中, c(OH-) = 1.32×10-3 mol·L-1, pH=,,,

5、,,,pH=-lg[,c,(H,+,)/,c,Θ,]=-lg[1.34,×,10,-3,]=,2.87,pH=14-pOH=14-(-lg[,c,(OH-)/,c,Θ,]),,=14+lg(1.32×10,-3,)=,11.12,2024/9/30,,,2007-5-27,,,,,,c,(H,+,)/(mol·L,-1,),1,,,,10,-1,10,-2,10,-3,10,-4,10,-5,10,-6,10,-7,10,-8,10,-9,10,-10,10,-11,10,-12,10,-13,10,-14,,pH,0 1 2,,3,,4 5,,,,6 7 8,,,,9 1

6、0 11 12 13 14,酸性增强　中性　碱性增强,,,,,,溶液的酸碱性和,pH,pH能否 14,2024/9/30,,,2007-5-27,,,,,酸碱指示剂,能在一定,pH,范围内保持一定颜色的某些,,有机弱酸或弱碱,,,甲基橙,红,←,,橙,→,,黄,3.1 ~ 4.4,,,酚酞,,无色←,,粉红,,→,红,8.0 ~ 10.0,pH,试纸在不同的,pH,溶液中显不同的颜色,,,,,,石蕊,红,←,紫,,→,蓝,,5.0 ~ 8.0,2024/9/30,,,2007-5-27,,第二节弱电解质的解离反响,,,,无机化学多媒体电子教案,第三章 酸碱反响和沉淀反响,第二节,,弱

7、电解质的,,解离反响,2024/9/30,,,2007-5-27,,3-2-1,解离平衡和解离常数,,,,3-2-1,解离平衡和解离常数,一元弱酸,HA(aq) H,+,(aq) + A,-,(aq),弱电解质在水中局部解离, 存在解离平衡,[,c,(H,+,)/,c,][,c,(A,-,)/,c,],,[,c,(HA)/,c,],K,i,,(HA)=,,,,,c,(H,+,)·,c,(A,-,),,,c,(HA),简化为,K,i,,(HA)=,,解离平衡,K,i,,——,标准解离常数,,K,a,、,K,b,——,分别表示弱酸、弱碱标准解,,,离常数,,,,,2024/9/30

8、,,,2007-5-27,,,,,K,i,,,越小,,,弱电解质解离越困难,,,电解质越弱,,弱电解质：一般,K,i,≤,10,-4,,,中强电解质：,K,i,=10,-2,~10,-3,,,解离常数,,K,i,,表示,弱电解质解离限度大小的特性常数,,,,K,i,与,浓度无关,，与,温度有关,,,由于温度对,K,i,,影响不大,,,室温下一般可不考虑其影响,,,,,,2024/9/30,,,2007-5-27,,,,,,,例,试计算,298.15K,、标准态下,K,a,(HOAc),值,,解：,HOAc H,+,+ OAc,-,,∆,f,G,m,/(kJ·mol,-1,) -

9、396.46 0 -369.31,,∆,r,G,m,=[(-369.31)-(-396.46)]kJ·mol,-1,,,=27.15 kJ·mol,-1,,,-∆,r,G,m,-27.15,×,1000,,,RT,8.314,×,298.15,ln,K,= = =,-,10.95,,,K,a,(HOAc)=,1.8,×,10,-5,,,,,提示：根据热力学性质计算。,2024/9/30,,,2007-5-27,,,,,稀释定律,,一元弱酸,HA(aq) H,+,(aq) + A,-,(aq),

10、起始浓度,c,,,平衡浓度,c,-,c,α,c,α,,c,α,,[,c,(H,+,)/,c,][,c,(A,-,)/,c,],,[,c,(HA)/,c,],K,a,(HA)=,,,,,,=( ),,α,2,c,,1-,α,,c,,,K,i,,≈( ),α,2,,cc,,K,i,,c,/,c,,α,≈,,若,(,c,/,c,)/,K,i,,≥,500, 1-,α,≈,1,,,,α,与,,c,有关，,K,i,,与,,c,无关,,,,,,,2024/9/30,,,3-2-3,弱酸或弱碱溶液中离子浓度计算,,,,,3-2-3,弱酸或弱碱溶液中离子浓度计算,假设Ki >>Kw ,

11、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,<<,c,,,c,-x,≈,c,,,,K,a,=,[,x,],2,,,[,c-x,],x = c,(H,+,),≈,cK,a,,,,,,,,一元弱酸,HA H,+,+ A,-,,c-x x x,2024/9/30,,,2007-5-27,,,,,,3-2-3,弱酸或弱碱溶液中离子浓度计算,,,一元弱碱溶液中,c,(H,+,),的计算,,c,(OH,-,),,,c,pOH= -lg

12、,,一元弱碱,BOH B,+,+ OH,-,,同理,c,(OH,-,),≈,cK,b,,,,c,(OH,-,),,,c,,pH=14 +lg,,,,,,2024/9/30,,,2007-5-27,,,,,,例,计算,0.100mol·L,-1,氨水溶液中的,c,(OH,-,),、,pH,和氨水的解离度,解：,NH,3,·H,2,O NH,4,+,+ OH,-,平衡浓度,/(mol·L,-1,) 0.100-,x x x,,,,,,c,(OH,-,)=,1.34,×,10,-3,,mol·L,-1,(,c,/,c,)/,K,b,(NH

13、,3,·H,2,O),=0.100/(1.8,×,10,-5,)>500,,,,0.100-,x,≈,x,则,x,=1.34,×,10,-3,,[,c,(NH,4,),/,c,][,c,(OH,-,),/,c,],x·x,,,[,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,mol·L,-1,pH = 14-2.8

14、7 =,11.13,,,,,,,计算,0.100mol·L,-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,,α,=,×,100%=,×,100%,,=,1.34%,,所以,pOH = -lg(1.34,×,10,-3,/c,Θ,),,= 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,+,+

15、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,

16、常温、常压下H2S的溶解度为,,0.10mol·L-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,/(mol·L,-1,) 0.10-,x x x,,,,,K,a(1),= = =1.1,×,10,-7,[,c,(H,+,)/,c,][,c,(HS,-,)/,c,],x·x,,[,c,(H,2,S)/,c,],,0.10-,x,(,c,/,c,)/,K,a,(1),=0.10/(1.1,,1

17、0,-7,)>500, 0.10-,x,≈0.1,,,,x,= 1.1,×,10,-4,,c,(H,+,)=,1.1,×,10,-4,,mol·L,-1,,,,,2024/9/30,,,2007-5-27,,,,,,例,常温、常压下H2S的溶解度为,,0.10mol·L-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,-,)/

18、,c,],,,,因为,K,a(2),<<,K,a(1),所以,c,(HS,-,)≈,c,(H,+,),,,,c,(S,2-,)≈,K,a(2),c,≈1.3,×,10,-13,mol·L,-1,,,,K,a(1),,c,/,c,,α≈,,1.1,×,10,-7,,,,0.10,= =,0.11%,二元弱酸溶液中,:,c,(A,2-,)≈,K,a(2),·,c,,,2024/9/30,,,2007-5-27,,,,,H,2,S H,+,+ HS,-,,K,a(1),=,,[,c,(H,+,)/,c,][,c,(HS,-,)/,c,],,[,c,(

19、H,2,S)/,c,],,,,,K,a(2),=,,[,c,(H,+,)/,c,][,c,(S,2-,)/,c,],,[,c,(HS,-,)/,c,],,,,,c,(H,+,),与,c,(S,2-,),的关系,HS,-,H,+,+ S,2-,,两式相加,H,2,S,,2H,+,+ S,2-,K,a,=,K,a(1),·,K,a(2),,,,,[,c,(H,+,)],2,[,c,(S,2-,)],,c,(H,2,S)(,c,),2,,=,K,a(1),·,K,a(2),=1.4,×,10,-20,,,常温下,, H,2,S,饱和溶液中,,,c,(H,2,S)=0.10,mol·L,-1,[,c,

20、(H,+,)],2,c,(S,2-,) = 1.4,×,10,-21,(mol·L,-1,),3,,,,,,调节,H,2,S,溶液的酸度，可控制,c,(S,2-,),,2024/9/30,,,2007-5-27,,3-2-5,解离平衡的移动　同离子效应,,,,,3-2-5,解离平衡的移动　同离子效应,,同离子效应,——,在弱电解质溶液中，加入,,含有相同离子的易溶强电解质，使弱电解,,质解离度降低的现象,HOAc H,+,+ OAc,-,,,NaOAc,→,,Na,+,+ OAc,-,平衡向左移动,,,,,,演示实验,: HOAc,溶液,+,甲基橙,再参加 NaOAc,甲基橙,红

21、,←,,橙,→,,黄,3.1 ~ 4.4,2024/9/30,,,2007-5-27,,,,,,例,在0.10mol·L-1NH3·H2O溶液中, 参加,,固体NH4Cl,使其浓度为0.10mol·L-1, 计算,,溶液中c(H+)、NH3·H2O的解离度,,x,=1.8,×,10,-5,,c,(OH,-,)=,1.8,×,10,-5,,mol·L,-1,,K,b,= =1.8,×,10,-5,x,(0.10+,x,),,0.10-,x,0.10,x,,0.10,=1.8,×,10,-5,,因为,(,c,/,c,)/,K,b,=0.10/(1.8

22、,×,10,-5,)>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,/(mol·L,-1,) 0.10-,x,0.10+,x x,+,2024/9/30,,,2007-5-27,,,,,,例,在0.10mol·L-1NH3·H2O溶液中, 参加,,固体NH4Cl,使其浓度为0.10mol·L-1, 计算,,溶液中c(H+)、NH3·H2O的解离度。,,,,,,解：,NH,3,·H,2,O NH,4,+ OH,-,平衡浓度,/(mol·L,-1,)

23、 0.10-,x,0.10+,x x,+,c,(OH,-,)=1.8,×,10,-5,mol·L,-1,1.0,×,10,-14,,,1.8,×,10,-5,,c,(H,+,) = mol·L,-1,=5.6,×,10,-10,mol·L,-1,1.8,×,10,-5,,0.10,α,=,×,100%=,0.018%,未加,NH,4,Cl,的,0.10mol·L,-1,NH,3,·H,2,O,溶液 ＝,1.34%,,,即,由于同离子效应，使,α,降低,,α,,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

24、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-

25、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。

26、,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,4.76,3.6~5.6,NH,3,·H,2,O-NH,4,Cl,4.76,8.3~10.3,NaHCO,3,-Na,2,CO,3,,10.33,9.2~11.0,KH,2,PO,4,-K,2,HPO,4,,7.20,5.9~8.0,H,3,BO,3,-Na,2,B,4,O,7,,9.2,7.2~9.2,,,,,,2024/9/30,,,2007-5-27,,,,,,,缓冲溶液,p

27、H,值的计算,,,例,,,试计算含,0.10mol,·,L,-1,HOAc,、,0.10mol·L,-1,,NaOAc,溶液的,pH,值。,解：,HOAc H,+,+ OAc,-,平衡浓度,/(mol·L,-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,

28、×,10,-5,,c,(H,+,)=,1.8,×,10,-5,,mol·L,-1,,=1.8,×,10,-5,,0.10,x,,0.10,,,,,2024/9/30,,,2007-5-27,,,,,,,,例,,(2) 在含0.100mol·L-1HOAc、0.100mol·L-1,,NaOAc溶液中参加HCl, 使c(Cl-) = 0.001,,mol·L-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,/(mol·L,-1,)

29、 0.101-,x,,x,0.099+,x,x,=1.8,×,10,-5,,c,(H,+,)=,1.8,×,10,-5,,mol·L,-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0.100mol·L-1HOAc、0.100,,mol·L-1NaOAc溶液中参加KOH, 使,

30、,c(K+)=0.001mol·L-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,/(mol·L,-1,) 0.099-,x,,x,0.101+,x,x,=1.8,×,10,-5,,c,(H,+,)=,1.8,×,10,-5,,mol·L,-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

31、 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0.10mol·L-1HOAc、0.10mol·L-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,/(mol·L,-1,) 0.010-,x,,x,0.010+,x,,K,a,= =1.8,×,10,-5,x,(0.010+,x,),,0. 010-,x,c,(H

32、,+,)=,1.8,×,10,-5,,mol·L,-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 6.00mol·L-1HOAc溶液中,,,应参加固体NaOAc·3H2O 多少克,解：,c,(HOAc)=[(12.0×6.00)/250] mol·L,-1,,,=0.288 mol·L,-1,,,pH

33、=5.00,c,(H,+,)=10,-5,mol·L,-1,,,,HOAc H,+,+ OAc,-,平衡浓度,/(mol·L,-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,mol·L,-1,应参加NaOAc·3H2O,,(250/1000)×0.518×136=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

34、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,+,

35、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,,+

36、,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,b,(AOH),水溶液显,碱性,,,+,强酸强碱盐不水解,,,溶液呈中性,2024/9/30,,,2007-5-27,,,,水解常数,,水解常数,,OAc,-,＋,H,2,O,,HOAc,＋,OH,-,K,h,=,,[,c,(HOAc/,c,][,c,(OH,-,),/,c,,],,[,

37、c,(OAc,-,),/,c,,],,,,[,c,(HOAc/,c,][,c,(OH,-,)/,c,][,c,(H,+,)/,c,],,,[,c,(OAc,-,)/,c,,][,c,(H,+,)/,c,,],,,,,,,K,h,= =,K,w,,K,a,,,,即,一元弱酸盐,,一元弱碱盐,,一元弱酸弱碱盐,K,w,,K,a,,,,K,h,=,,K,w,,K,b,,,,K,h,=,,,K,w,,K,a,K,b,,,,,K,h,=,,,,,,,2024/9/30,,,2007-5-2

38、7,,,,,,,K,h,,值越大，相应盐的水解限度越大,,,水解度,——,表示盐的水解程度,盐水解部分的浓度,,盐的开始浓度,水解度,(,h,)=,×,100%,,K,h,,与盐的浓度无关,,,h,与盐的浓度有关,,,,,,2024/9/30,,,2007-5-27,,3-3-2,分步水解,,,,3-3-2,分步水解,多元弱酸盐或多元弱碱盐水解是分步的,如,S,2-,+ H,2,O HS,-,+,OH,-,HS,-,+ H,2,O H,2,S + OH,-,,,K,w,1.0,×,10,-14,,K,a(2),(H,2,S),,1.3,×,10,-13,,,

39、K,h(1),= = = 7.7,×,10,-2,,,,K,w,1.0,×,10,-14,,K,a(1),(H,2,S),,1.1,×,10,-7,,,K,h(2),= = =9.1,×,10,-8,,,,K,h(1),、,K,h(2),——,分级水解常数,,,,,,,,K,h(1),>>,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

40、 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,0.10mol·L,-1,NaOAc,溶液的,pH,和水解度,解：,OAc,-,+H,2,O HOAc,+

41、,OH,-,平衡浓度,/(mol·L,-1,),0.10-,x,,x x,,0.10-,x≈,0.10,x,=7.5,×,10,-6,c,(OH,-,)=7.5,×,10,-6,mol·L,-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,,,,x·x,,,0.10,-x,,,,,,7.5,×,10,-6,,0.10,h,=,×,100%=7

42、.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,,,(0.1mol·L,-1,),水解产物,K,a,h,/%,pH,NaOAc,HOAc+OH,-,1.8×10,-5,,0.0075,8.9,NaCN,HCN+OH,-,6.2×10,-10,1.4,11.2,Na,2,CO,3,,HCO,3,+OH,-,4.7×10,-11,4.2,11.6,,-,2024/9/30,,,2007-5-27,,,,,,3

43、-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

44、,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

45、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

46、,,,,,,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=3~4,，促使,,,Fe,3+,水解，生成,Fe(OH),3,↓,３,.,加热，促使,Fe,3+,水解，生成,Fe(OH),3,↓,,,,,2024/9/30,