济南大学 分析化学课件 重量分析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,*,1,第五章 重量分析与沉淀滴定法,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,*,2,、盐效应：,当溶液中,存在强电解质时，会使沉淀溶解度增大，这一现象称为盐效应,原因,:,强电解质的存在,，溶液中离子浓度增大，,离子强度,亦相应,增大,，从而使,活度系数减小,。在一定温度下，,K,0,sp,为常数，当,M,+,、,A,-,减小时，必将引起,沉淀溶解度增大,。,2,沉淀的溶解度及影响因素,以,MA,型沉淀为例进行说明。,2,沉淀的溶解度及影响因素,一般：盐效应的影响较小，可不予考虑,当,s,大且,I,高时，应考虑盐效应,PbSO,4,在,Na,2,SO,4,溶液中的溶解情况,c,Na,2,SO4,(mol/L),0,0.001,0.01,0.02,0.04,0.100,0.200,s,PbSO4,10,-3,mol/L,0.15,0.024,0.016,0.014,0.013,0.016,0.023,3,、,酸效应,：溶液,酸度对沉淀溶解度,的影响,例如，,CaC,2,O,4,沉淀在溶液中存在如下平衡：,CaC,2,O,4,C,2,O,4,2-,+Ca,2+,HC,2,O,4,-,H,2,C,2,O,4,(1),当溶液,pH,已知时,解题思路：,s,2,沉淀的溶解度及影响因素,K,sp,A(H),pH,K,sp,=,Ca,2+,C,2,O,4,2-,Ca,2+,C,2,O,4,2-,C,2,O,4,(H),=,K,sp,C,2,O,4,(H),ss,=,s,2,例,2,：,计算,CaC,2,O,4,沉淀在下列两种情况下的溶解度：,不考虑酸效应；,pH=4.0,、,pH=2.0,时,考虑酸效应。已知,K,sp,=2.010,-9,，,H,2,C,2,O,4,：,K,a1,=5.910,-2,，,K,a2,=6.410,-5,解：,不考虑酸效应，其溶解度为：,s,=Ca,2+,=C,2,O,4,2-,=4.510,-5,(mol/L),考虑酸效应进行溶解度计算：,2,沉淀的溶解度及影响因素,Ca,2+,=,s,C,2,O,4,2-,+HC,2,O,4,-,+H,2,C,2,O,4,=C,2,O,4,=,s,K,sp,=,Ca,2+,C,2,O,4,2-,=,Ca,2+,C,2,O,4,2-,C,2,O,4,(H),=,K,sp,C,2,O,4,(H),=,s,2,2,沉淀的溶解度及影响因素,2,沉淀的溶解度及影响因素,pH=4.0:,pH=2.0:,K,sp,K,sp,例,2,：计算在,pH=4.0,，,C,2,O,4,2-,总浓度为,0.01mol/L,的溶液中，,CaC,2,O,4,的溶解度。（,K,sp,=2.010,-9,）,解：同离子效应和酸效应同时存在,Ca,2+,=,s,K,sp,=,Ca,2+,C,2,O,4,2-,C,2,O,4,s,结论：同离子效应起主导作用，使溶解度降低。,低于纯水中溶解度,2,沉淀的溶解度及影响因素,=0.01+,s,=Ca,2+,C,2,O,4,2-,C,2,O,4,(H),=,K,sp,C,2,O,4,(H),=,s,(0.01+,s,),0.01,s,C,2,O,4,2-,+HC,2,O,4,-,+H,2,C,2,O,4,=C,2,O,4,CaF,2,Ca,2+,+2F,-,H,+,HF,:,例,3,：计算在,pH=3.0,的溶液中,CaF,2,的溶解度。（,K,sp,=10,-10.57,）,K,sp,=,Ca,2+,F,-,2,2,沉淀的溶解度及影响因素,s,s,2,s,=,s,(2,s,),2,=4,s,3,=Ca,2+,F,-,2,2,F(H),=,K,sp,2,F(H),弱酸盐的,弱酸根,在溶液中存在着酸效应，确定溶液的,pH,是一关键。,MS,M,2+,+S,2-,H,+,HS,-,:,(2),当溶液,pH,未知时,溶液,pH,的主要取决于,2,沉淀的溶解度及影响因素,水的解离,弱酸根的解离,金属离子的解离（一般忽略）,2,沉淀的溶解度及影响因素,确定体系,pH,的方法：,求出,原始溶解度,s,*,若,n,s,*10,-7,则溶液的,pH,主要,取决于弱酸根的解离,M,m,A,n,mM,n,+,+,nA,m,-,s,m,s,n,s,A,m-,+H,2,O HA,(m-1)-,+OH,-,n,s,n,s,n,s,例,4,：,计算,CuS,在水中的溶解度。,已知：,CuS,的,K,sp,=10,-35.2,H,2,S:,K,a1,=10,-6.88 ,K,a2,=10,-14.15,解,：,确定,pH,溶液 的,pH,主要 由水的解离来确定,pH=7.0,确定溶解度,CuS,=Cu,2+,+S,2-,K,sp,=Cu,2+,S,2-,=,K,sp,S(H,),2,沉淀的溶解度及影响因素,=,s,s,2,沉淀的溶解度及影响因素,例,5:,MnS,在纯水中的溶解度。,已知：,K,sp,=10,-9.7,H,2,S:,K,a1,=10,-6.88,K,a2,=10,-14.15,解：,确定,pH,溶液的,pH,应由,S,2-,水解来确定,假设,S,2-,第一步水解是完全的,OH,-,=,s,MnS,+H,2,O Mn,2+,+HS,-,+OH,-,s,s,s,K,=Mn,2+,HS,-,OH,-,=,s,3,2,沉淀的溶解度及影响因素,2,沉淀的溶解度及影响因素,分析：,OH,-,=10,-3.2,H,+,=10,-10.8,S(H),=1+10,14.15-10.8,+10,14.15+6.88-21.6,=1+10,3.4,+10,-0.3,S,2-,:HS,-,:H,2,S=1:10,3.4,:10,-0.3,以,HS,-,为优势，可见前面假设第一步水解完全是正确的。,例,6:,计算,Ca,3,(PO,4,),2,在纯水中的溶解度。已知：,K,sp,=10,-28.7,H,3,PO,4,:,K,a1,=10,-2.12,K,a2,=10,-7.20,K,a3,=10,-12.36,解：,确定,pH,Ca,3,(PO,4,),2,3Ca,2+,+2PO,4,3-,n,s,*10,-7,溶液,pH,的由,PO,4,3-,的水解决定,假设第一步水解完全,Ca,3,(PO,4,),2,+2H,2,O 3Ca,2+,+2 HPO,4,2-,+2OH,-,s 3s 2s 2s,2,沉淀的溶解度及影响因素,K,=Ca,2+,3,HPO,4,2-,2,OH,-,2,2,沉淀的溶解度及影响因素,=(3,s,),3,(2s),2,(2s),2,=432s,7,=,Ca,2+,3,HPO,4,2-,2,OH,-,2,PO,4,3-,2,H,+,2,/PO,4,3-,2,H,+,2,=,K,sp,K,w,2,/,K,a3,2,=10,-28.70-14.0,2+12.362,=10,-31.98,Ca,3,(PO,4,),2,+2H,2,O 3Ca,2+,+2 HPO,4,2-,+2OH,-,s 3s 2s 2s,问题,:,计算在纯水中,CaF,2,的溶解度,?,（,K,sp,=10,-10.57,）,解：,溶液的,pH,应由,F,-,水解来确定,CaF,2,+2H,2,O Ca,2+,+2HF,+2OH,-,s,2,s,2,s,K,=Ca,2+,HF,2,OH,-,2,=16,s,5,=10,-32.21,错误的解答,H,2,S:,K,a,1,=10,-6.88,K,a,2,=10,-14.15,HF:,K,a,=10,-3.18,H,3,PO,4,:,K,a1,=10,-2.12,K,a2,=10,-7.20,K,a3,=10,-12.36,H,2,CO,3,:,K,a,1,=10,-6.38,K,a,2,=10,-10.25,可见,F,-,是极弱的碱,第一步水解不能完全,CaF,2,+2H,2,O Ca,2+,+2HF,+2OH,-,s,2,s,2,s,CaF,2,:,s,=,s,1,+,s,=10,-5.29,+10,-6.68,=10,-5.29,