2018年安徽化学竞赛 无机 10第10章 沉淀平衡共56页共56张

1、无,机,化,学,第,10,章,沉淀平衡,主讲：杨,骏,化学化工学院,10-1,溶度积原理,一,溶解度,在一定温度下，达到溶解平衡时，一定,量的溶剂中含有溶质的质量，叫做溶解,度通常以符号,s,表示,对水溶液来说，通常以饱和溶液中每,100 g,水所含溶质质量来表示，即以,g /100g,水表示,二,溶度积常数,在一定温度下，将难溶电解质晶体放入,水中时，就发生,溶解,和,沉淀,两个过程,BaSO,4,的沉淀和溶解,在一定条件下，当溶解和沉淀速率相等时,便建立了一种动态的多相离子平衡，可表,示如下,K,sp,BaSO,4,= Ba,2,SO,4,2,K,sp,称为溶度积常数,简称为,溶度积,K,

2、sp,反映了难溶电解质溶解能力的大小,溶解,aq,SO,aq,Ba,s,BaSO,2,4,2,4,沉淀,对于一般沉淀反应,K,sp,A,m,B,n,= A,m,n,B,n,m,例,Fe(OH,3,的,K,sp,Fe(OH,3,= ,aq,mB,aq,nA,s,B,A,n,m,m,n,对于一般沉淀反应,K,sp,A,n,B,m,= A,m,n,B,n,m,例,Fe(OH,3,的,K,sp,Fe(OH,3,= ,K,sp,Fe(OH,3,= Fe,3,OH,3,aq,mB,aq,nA,s,B,A,n,m,m,n,三、溶解度和溶度积的关系,例,25,o,C,AgCl,的溶解度为,1.92,10,3,

3、g,L,1,求同温度下,AgCl,的溶度积,三、溶解度和溶度积的关系,例,25,o,C,AgCl,的溶解度为,1.92,10,3,g,L,1,求同温度下,AgCl,的溶度积,解,已知,M,r,AgCl)=143.3,L,mol,10,34,1,L,mol,3,143,10,92,1,S,5,3,AgCl(s,Ag,aq) + Cl,aq,平衡浓度,s s,K,sp,AgCl) =Ag,Cl,s,2,(1.34,10,5,2,1.8,10,10,例,25,o,C,已知,K,sp,Ag,2,CrO,4,1.1,10,12,求同,温下,Ag,2,CrO,4,的溶解度,例,25,o,C,已知,K,sp

4、,Ag,2,CrO,4,1.1,10,12,求同,温下,Ag,2,CrO,4,的溶解度,解：设,Ag,2,CrO,4,的溶解度为,s,mol/L,aq,CrO,aq,2Ag,s,CrO,Ag,4,4,2,2,平衡浓度,2s s,K,sp,Ag,2,CrO,4,Ag,2,CrO,4,2,(2s,2,s=4s,3,L,mol,10,5,6,4,10,1,1,4,K,S,5,3,12,3,sp,比较溶解度与溶度积的关系,分子式,溶度积,溶解度,mol/L,AgCl,1.8,10,10,1.3,10,5,AgBr,5.0,10,13,7.1,10,7,AgI,8.0,10,17,9.1,10,9,Ag

5、,2,CrO,4,1.1,10,12,6.5,10,5,相同类型的难溶电解质，其,K,sp,大的,s,也大,AB,型,A,2,B,或,AB,2,型,sp,K,S,3,sp,4,K,S,思考题：求,Ca,3,PO,4,2,的,s,与,K,sp,间的关系,思考题：求,Ca,3,PO,4,2,的,s,与,K,sp,间的关系,平衡浓度,3s 2s,K,sp,(3s,3,2s,2,5,sp,108,K,S,Ca,3,PO,4,2,3Ca,2,2PO,4,3,设,Ca,3,PO,4,2,的溶解度为,s,mol/L,四、溶度积原理,当,JK,sp,时,是为过饱和溶液,沉淀从溶液中析出,当,J=K,sp,时,

6、溶液为饱和溶液,饱和溶液与沉淀物平衡,当,JK,sp,时,是不饱和溶液,无沉淀析出,若体系中已有固体存在,沉,淀将溶解,aq,mB,aq,nA,s,B,A,n,m,m,n,m,n,n,m,B,c,A,c,J,沉淀,溶解平衡的反应商判据,即,溶度积规则,c (CrO,4,2,mol/L,c (Pb,2,Mol/L,例题,25时，晴纶纤维生产的某种溶液中,c(SO,4,2,为,6. 0,10,4,mol/L,若在,40.0 L,该溶液中，加入,0.010,mol/L,BaCl,2,溶液,10.0L,问是否能生成,BaSO,4,沉淀？如果有沉淀生成，问能生,成,BaSO,4,多少克？最后溶液中,c(

7、SO,4,2,是多少,1,4,4,2,4,L,m,ol,10,8,4,0,50,0,40,10,0,6,SO,c,1,3,2,L,mol,10,0,2,0,50,0,10,010,0,Ba,c,J=c(SO,4,2,c(Ba,2,4.8,10,4,2.0,10,3,9.6,10,7,J K,sp,1.1,10,10,所以有,BaSO,4,沉淀生成,解,反应前浓度,2.0,10,3,4.8,10,4,变化浓度,(4.8,10,4,x) -(4.8,10,4,x,平衡浓度,1.52,10,3,x,x,1.52,10,3,x,x = 1.1,10,10,1.52,10,3,x 1.52,10,3,x

8、 = 7.2,10,8,c(SO,4,2,= 7.2,10,8,mol/L,m(BaSO,4,= (4.8,10,4,7.2,10,8,50.0,233 = 5.6 g,aq,SO,aq,Ba,s,BaSO,2,4,2,4,X,很小,五、同离子效应,在难溶电解质溶液中加入与其含有,相同,离子,的易溶强电解质，而使难溶电解质,的,溶解度降低,的作用,例题,求,25时,Ag,2,CrO,4,在,0.01 mol,L,1,K,2,CrO,4,溶液中的溶解度,例题,求,25时,Ag,2,CrO,4,在,0.01 mol,L,1,K,2,CrO,4,溶液中的溶解度,解：设其溶解度为,x,mol,L,1,

9、a,CrO,aq,2Ag,s,CrO,Ag,2,4,4,2,初始浓度,mol,L,1,0 0.01,平衡浓度,mol,L,1,2x 0.01+x,2x,2,0.01+x)=K,sp,1.1,10,12,X,很小,0.01+x 0.01,X = 5.2,10,6,mol/L,纯水中,S = 6.5,10,5,mol/L,注意,制备沉淀过程中，应该用与沉淀含有相同离子的溶液洗,涤，而不是用蒸馏水,所谓完全沉淀,一般不超过某一限度，如,10,6,mol/L,即,认为该离子沉淀完全了,10-2,沉淀与溶解,一、沉淀的生成,生成沉淀的条件,J K,sp,方法：加沉淀剂,例,将等体积的,0.004mol/

10、L AgNO,3,和,0.004 moL/L K,2,CrO,4,混合时,有无红色,Ag,2,CrO,4,沉淀析出, (K,sp,为,1.1,10,12,10-2,沉淀与溶解,一、沉淀的生成,生成沉淀的条件,J K,sp,方法：加沉淀剂,例,将等体积的,0.004mol/L AgNO,3,和,0.004 moL/L K,2,CrO,4,混合时,有无红色,Ag,2,CrO,4,沉淀析出, (K,sp,为,1.1,10,12,a,CrO,aq,2Ag,s,CrO,Ag,2,4,4,2,J,0.002,2,x 0.002,8 x 10,9,1.1 x 10,12,K,sp,1,金属氢氧化物沉淀的生成

11、,K,sp,M,n,OH,n,若有氢氧化物沉淀生成,则,J = c,M,n,c(OH,n,K,sp,aq,nOH,aq,M,s,M,OH,n,n,n,n,sp,M,c,K,OH,c,开始沉淀,n,6,sp,10,K,OH,c,完全沉淀,例,计算欲使,0.01 moL/L Fe,3,开始沉淀和沉淀,完全时最低,pH,值,(Fe(OH,3,的,K,sp,为,4,10,38,例,计算欲使,0.01 moL/L Fe,3,开始沉淀和沉淀,完全时最低,pH,值,(Fe(OH,3,的,K,sp,为,4,10,38,解,J=c,Fe,3,c(OH,3,K,sp,开始沉淀时所需,pH,值,0.01c(OH,3

12、,K,sp,4,10,38,12,3,36,10,6,1,10,4,c(OH,pOH=-lg1.6,10,12,11.8 pH,2.2,沉淀完全时所需,pH,值,53,3,pH,10,4,3,10,10,4,10,K,OH,c,11,3,6,38,3,6,s,p,例题,在含有,0.10mol,L,1,Fe,3,和,0.10,mol,L,1,Ni,2,的溶液中，欲除掉,Fe,3,使,Ni,2,仍留在溶液中，应控制,pH,值为多少,已知,Ni(OH,2,的,K,sp,2.0,10,15,Fe(OH,3,的,K,sp,4.0,10,38,例题,在含有,0.10mol,L,1,Fe,3,和,0.10,

13、mol,L,1,Ni,2,的溶液中，欲除掉,Fe,3,使,Ni,2,仍留在溶液中，应控制,pH,值为多少,已知,Ni(OH,2,的,K,sp,2.0,10,15,Fe(OH,3,的,K,sp,4.0,10,38,分析,求出刚生成,Ni(OH,2,沉淀时所需,pH,值,求出使,Fe(OH,3,沉淀完全时所需,pH,值,控制,pH,值在二者之间即可,解,刚生成,Ni(OH,2,沉淀时所需,pH,值,L,mol,10,4,1,1,0,10,0,2,Ni,c,K,OH,c,7,15,2,sp,pOH = 6.85 pH = 7.15,Fe(OH,3,沉淀完全时所需,pH,值,53,3,pH,10,4,

14、3,10,10,4,10,K,OH,c,11,3,6,38,3,6,s,p,pH,值应控制在,3.537.15,之间,可用于物质分离,PbS Bi,2,S,3,CuS CdS Sb,2,S,3,SnS,2,As,2,S,3,HgS,2,难溶硫化物沉淀,例,某溶液中,Zn,2,和,Mn,2,的浓度皆为,0.1,mol/L,如果不断将,H,2,S,气体通入溶液中,使溶液中的,H,2,S,始终处于,饱和状态,将,H,离子浓度控制在什么范围可将,Zn,2,和,Mn,2,分开,已知,K,sp,ZnS)=2.93,10,25,K,sp,MnS)=4.65,10,14,K,a,H,2,S)= 5.7,10,

15、8,K,a,HS,1.2,10,15,硫化锌和硫化锰是同类，由溶度积常数,可知,ZnS,的溶解度小于,MnS,的溶解度,应控制,S,2,浓度,使,ZnS,沉淀完全，而不,生成,MnS,沉淀,由,H,2,S,的电离平衡推出,H,和,S,2,的相互,关系,求出,pH,值,分析,解,ZnS,沉淀完全时所需,pH,值,c(Zn,2,c(S,2,K,sp,2.93,10,25,10,6,c(S,2,2.93,10,25,c(S,2,2.93,10,19,H,2,S,2,H,2,SK,a1,K,a2,6.84,10,24,H,6.84,10,24,2.93,10,19,1/2,4.83,10,3,ZnS,

16、沉淀完全时,pH=2.31,MnS,开始沉淀时所需,pH,值,c(Mn,2,c(S,2,K,sp,4.65,10,14,0.1c(S,2,4.65,10,14,c(S,2,4.65,10,13,H,2,S,2,H,2,SK,a1,K,a2,6.84,10,24,H,6.84,10,24,4.65,10,13,1/2,3.83,10,6,MnS,开始沉淀时,pH=5.41,pH,值应控制在,2.315.41,范围内,利用多重平衡,MS(s)+2H,M,2,H,2,S,思考题,利用多重平衡,MS(s)+2H,M,2,H,2,S,MS,K,K,K,S,S,S,H,M,H,K,s,p,aHS,S,aH

17、,2,2,2,2,2,2,M,MS,K,K,K,1,0,H,2,s,p,aH,S,S,aH,2,S,H,M,MS,K,K,K,H,2,2,s,p,aHS,S,aH,2,2,课后思考题,3,分步沉淀,例,在含有,0.1 mol/L Cl,和,I,的溶液中,逐渐加入,AgNO,3,可能生成,AgCl,和,AgI,沉淀,已知,K,sp,AgCl)=1.8,10,10,K,sp,AgI)=9.3,10,17,1,哪个离子先沉淀,3,分步沉淀,例,在含有,0.1 mol/L Cl,和,I,的溶液中,逐渐加入,AgNO,3,可能生成,AgCl,和,AgI,沉淀,已知,K,sp,AgCl)=1.8,10,1

18、0,K,sp,AgI)=9.3,10,17,1,哪个离子先沉淀,解,AgCl,AgI,开始沉淀所需,c(Ag,Cl,开始沉淀,c(Ag,1.8,10,10,0.1= 1.8,10,9,I,开始沉淀,c(Ag,9.3,10,17,0.1=9.3,10,16,所以,I,先沉淀,2,当,Cl,开始沉淀时,I,有多大,此时,I,Cl,Ag,同时满足下列等式,Ag,Cl,初,K,sp,AgCl,Ag,I,K,sp,AgI,则,Cl,初,I,K,sp,AgCl) /K,sp,AgI,1.8,10,10,9.3,10,17,1.9,10,6,I,Cl,初,1.9,10,6,5.3,10,8,mol/L,当,

19、Cl,开始沉淀时,I,已沉淀完全,分步沉淀的顺序并不固定,它取决于两种因素,沉淀物的溶度积,被沉淀离子的浓度,如,K,sp,AgBr)=5,10,13,K,sp,AgSCN)=1,10,12,当,SCN,Br,时,但,SCN,1 mol/L, Br,0.01 mol/L,时,分步沉淀的顺序并不固定,它取决于两种因素,沉淀物的溶度积,被沉淀离子的浓度,如,K,sp,AgBr)=5,10,13,K,sp,AgSCN)=1,10,12,当,SCN,Br,时,AgBr,首先产生沉淀,但,SCN,1 mol/L, Br,0.01 mol/L,时，首,先产生沉淀的则是,AgSCN,二、沉淀的溶解,沉淀溶解

20、的条件,JK,sp,方法：加入试剂降低溶液中离子的浓度,1,酸碱溶解法,Mg(OH,2,沉淀能溶于,HCl,和,NH,4,Cl,中,Mg(OH,2,s,Mg,2,2OH,M(OH,n,沉淀能溶于,HCl,中,M(OH,n,s,M,n,nOH,弱酸盐溶于酸中,例,CaCO,3,溶于,HCl,中,CaCO,3,s)+2H,Ca,2,H,2,CO,3,2,a,1,a,sp,2,3,2,3,2,3,2,2,K,K,K,CO,CO,H,CO,H,Ca,K,难溶弱酸盐溶于酸的难易程度与,K,sp,有,关,与生成弱酸的,K,a,值有关,K,sp,越大或,K,a,越小,溶解反应越容易进行,例,使,0.01mo

21、l/L SnS,溶于,1L,盐酸中,求所需盐酸,的最低浓度。,K,sp,3.25,10,28,K,a,H,2,S)= 5.7,10,8,K,a,HS,1.2,10,15,例,使,0.01mol/L SnS,溶于,1L,盐酸中,求所需盐酸,的最低浓度。,K,sp,3.25,10,28,K,a,H,2,S)= 5.7,10,8,K,a,HS,1.2,10,15,解,6,23,28,2,a,1,a,sp,2,2,2,2,2,10,75,4,10,84,6,10,25,3,K,K,K,S,S,H,S,H,Sn,K,SnS(s)+2H,Sn,2,H,2,S,初始浓度,H,初,0 0,平衡浓度,H,初,0

22、.02 0.01 0.01,6,2,10,75,4,02,0,H,01,0,初,H,初,4.61 mol/L,例题,在,0.20L,的,0.50mol,L,1,MgCl,2,溶,液中加入等体积的,0.10mol,L,1,的氨水,溶液，问有无,Mg(OH,2,沉淀生成？为了,不使,Mg(OH,2,沉淀析出，至少应加入多,少克,NH,4,Cl(s,设加入,NH,4,Cl(s,后体,积不变,(K,sp,1.8,10,11,K,b,1.8,10,5,解,(1) c,0,Mg,2,0.25 moL/L,c,0,NH,3,0.050 moL/L,0.050 /K,b,500,OH,(0.050,1.8,1

23、0,5,1/2,9.4,10,4,mol/L,J= c,0,Mg,2,OH,2,0.25,9.4,10,4,2,2.25,10,7,K,sp,1.8,10,11,所以有,Mg(OH,2,沉淀析出,2,为了不使,Mg(OH,2,沉淀析出,J K,sp,c(OH,2,c,0,Mg,2,K,sp,c(OH,K,sp,c,0,Mg,2,1/2,8.5,10,6,mol/L,4,2,3,aq,OH,aq,NH,O(l,H,aq,NH,平衡浓度,0.050,8.5,10,6,c+ 8.5,10,6,8.5,10,6,0.050 c,5,6,10,8,1,050,0,c,10,5,8,c=0.11 mol/L,m(NH,4,Cl)=0.11mol/L,0.40,53.5g/mol = 2.4 g,此题,2,也可以用双平衡求解,此题,2,也可以用双平衡求解,l,O,2H,aq,2NH,aq,Mg,aq,2NH,s,Mg(OH,2,3,2,4,2,此题,2,也可以用双平衡求解,平衡,y 0.25 0.05,浓度,l,O,2H,aq,2NH,aq,Mg,aq,2NH,s,Mg(OH,2,3,2,4,2,L,mol,11,0,y,056,0,y,05,0,25,0,2,2,m(