发布时间 : 星期日 文章分析化学课后习题答案更新完毕开始阅读
?Es?K?0.0592lgcsEx?K?0.0592lgcx
lgcs0.217?0.158? cx0.0592?4 cx?10mol/l
10.解:?1? VNaOH E
?E 6 ?V?2E 2 65 -59
?V ?Vep?15.50??15.70?15.60?? =
65
65?5911V终??15.56?7.83ml 227.83?7.00 滴定一半pH半为:pH半=+()?
14.00?7.00(2)滴定至一半时的体积V半? = 此时 aH??10?5.6?2.51?10?6
而滴定至一半时:aA??aHA ?Ka?aH?gaA?aHA?aH??2.51?10?6
第十章 极谱分析法
3.答:极谱分析是微量分析方法,测定依次在电极发生的物质的量的多少。 4.答:底液,即含有支持电解质 ,除氧剂,络合剂及极大抑制剂的溶液。
10.解:对于可逆波:id?h
12?id83.1? 11.7164.7212?id?1.94?A
11.解:设50ml试液中浓度为Cxmoll
???4.0?KCx??2?? 则:9.0?kg10Cx?0.5?1?10
??10?0.5?? 解得Cx=3.67?10试样Zn质量数为:
?4moll
3.67?10?4?50?10?3?65.39?100%?0.24%
0.500012. 解:50ml中溶液Cx
???id1?46.3?8.4?KCx??2?? 则:id?68.4?8.4?K(C?0.5?2.3?10
x2??50??Cx?3.94?10?4moll
?试样中Ni质量为:
2?
Cx?50?58.693?115.7mggL?1 1013.解:由可逆金属的极谱波方程得: E?E1?20.0592id?ilg ni0.0592id?17.1???0.66?E?lg?1??1?2217.1??0.0592id?19.9??lg2?? ??0.71?E1??2219.9???0.0592id?20.0??1.71?E?lg1??2220.0??从上式接得id=20?A 从(1)式或(2)式得E1=
214.解:
0.05920.0592lgKC?xglg?L?2C2nn0.05920.05924??得E1?E1?lgKPb2??xglg?Y??2C2 nn代入数据:0.05920.0592E1??0.405?lg?1.0?1018??1glg?1.00?10?2?2C22由E1?E1????????
=-879V
??15.解:以?E1?对lg?L?作图
?2?Ck'?VR'tR'1.58?0.25???5.32VMtM0.25??0.0592由?E1??K?xglg?L?可求VM?Vm?tMgFc?0.25?30?7.5ml
n?2?CV?7.5ml
Cx?2/molgL?1 ???E1?/V ?2?2?lg??x??
-1
Cx?2=0时 E1即为M?III?的半波电位。 E1=— n?3
22作图或求直线的方程得:
??2?x?E1?=lg????—
?2?c斜率 ?xg0.0592??0.06066 3?3得x?3 ?络合物组成:Mx 截距 K?E1—
20.0592lgKc= 3得Kc=1.2?10
16.解: 在处,只有Fe?III????Fe?II?,得到Fe?II?,浓度等于Fe?III?的浓度, 即:CF?II?'?CF?III?
ee25 由尤考方程得:
12.5?KCF?III? ①
e在处Fe?III????Fe?0?。溶液中发生还原反应Fe?II?的总浓度为CF?II?+CF?II?'
ee?=2KCFe?II?'?CFe?II?
即:=KCF?II??CF?III?
ee即: =KCF?II??CF?III? ② ee①
??????? — 得:
②
CFe?III?CFe?II??CFe?III?5? 1?12.55? 156?
CFe?III?CFe?II?