终点二分之一时 aA?aHA
? Ka?aH??aA?aHA?aH??6.6?10
?55．解： E??SCE??左 ?0.244?3?左 ?左?E内参?E膜 ?E?Ag,AgCl?0.0592alClg??aF?内0.0592 lgaF?外 ?E?Ag,AgCl?0.0592lg0?.12.5?100.0592?l3g
1?10?1?4
?0.22?4?0.24V80.05?920.0592?lg2.5
10 E??SCE??左
?0.244?3??0.003V70.24
6．解：E??SCE??pH ?K?0.059p2H
?Es?K?0.059p2sHEx?K?0.059p2H xpH?sEx?Es0.05920.06?0.2180.0592 pHx?
? ?5.00
=2.33 7．解：E??SCE??Mg2?
0.0592lgaMg2? 20.0592,pMg ?K? 2 ??SCE?K? 31 / 47
?1?pMx?pMs?2?Ex?Es?
0.0592?2
??lg?1.1?5?6.571?0?2??0.41?20.05920.?275
?2?pMx?pMs?2?Ex?Es??液0.0592?2?
5 ??lg?1.1?2??0.41?20.2??7液5?1?0?
0.0592 ?1.939?3 ?液??0.00V2 ?3? pMx1?1.93933.7?8?38 70??.液1?333.7?8?78?0.137??0.0 026.64 pMx2 =1.9393+33.7838??0.137-0.002?=6.50
?pM范围在6.50—6.64之间
8．解：?1?分别溶剂法求Ki,j
lgKi,j? ? Ki,j=0.17
?i??jS
67?1130.059?2??0.7 61000?2?误差=KK?,Na??aNa?aK?0.1?7??11?100%
=1.?00?2?31.0?1010?100%
?1.7%9．解：E??F??SCE
?g ?K?0.0592clF? 32 / 47
?Es?K?0.059c2slgEx?K?0.059c2lgx
?0.21?70.05920.158 lgcscx
cx?10?4mol/ l10．解：?1? VNaOH 15.50 15.60 15.70 15.80
E 7.70 8.24 9.43 10.03
?E?V 5.4 11.9 6
?E?V22 65 -59
6565?59 ?Vep?15.50??15.70?15.60?? =15.56ml
(2)滴定至一半时的体积V半?12
V终?12?15.56?7.83ml
7.83?7.0014.00?7.00 滴定一半pH半为：pH半=5.47+（6.60-5.47）? =5.6 此时 aH??10?5.6
?2.51?10?6
而滴定至一半时：aA?aHA
? ?Ka?aH??aA?aHA?aH??2.5?110
?6 33 / 47
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。
110．解：对于可逆波：id?h2
1?id1.71?83.121
64.72?id?1.94?A
11.解：设50ml试液中浓度为Cxmoll
??4.0?KCx?2? 则：9.0?k?10Cx?0.5?1?10?10?0.5? 解得Cx=3.67?10试样Zn质量数为：
?4??? ??moll
3.67?10?4?50?100.5000?3?65.39?100%?0.24%
12. 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?10x?50?2Cx?3.94?102??4?????
moll
?58.693?115.7mg?L
?1 ?试样中Ni质量为：
Cx?501013．解：由可逆金属的极谱波方程得： E?E1?20.0592id?ilg ni 34 / 47
0.0592id?17.1???0.66?E?lg???11?2217.1??0.0592id?19.9?? ??0.7?1E1?lg???2
2219.9???0.0592id?20.0??1.7?1E?lg1??2220.0??从上式接得id=20?A
从（1）式或（2）式得E1=-0.64V
214．解：
??得?E?
由E1212?E1??20.0592nn0.05922C0.0592lgKC?x?lg?L?n4??E1?C20.05920.0592lgKPb2??x?lg??Yn0.05922??
代入数据：??E12??0.405?Clg?1.0?1018??1?lg?1.00?10?2?
=-879V
??15.解：以?E1?对lg?L?作图
?2?Ck'?
VR'VM?tR'tM?1.58?0.250.25?5.32??0.0592由?E1??K?x?lg?L?可求VM?Vm?tM?Fc?0.25?30?7.5ml
n?2?CV???7.5ml Cx?2/mol?L ??E1?2??/V ?2??10.000 0.0200 0.0600 0.100 0.300 0.500 -0.692 -1.083 -1.113 -1.128 -1.152 -1.170 lg??x? ? -1.699 -1.222 -1 -0.523 -0.301 Cx?2=0时 E1即为M?III?的半波电位。 E1=—0.692V n?3
22作图或求直线的方程得：
??2???E1?=-0.06066lg??x?—1.18677
?2?c 35 / 47