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Principles of Instrumental Analysis, 6th ed. Chapter 23
 
 5
 0
CuBr 9
1.00 = 0.521 0.0592 log
5.2 10
E −−
×
 = 0.031 V 
 (b) SCE CuBr(sat’d), Br-(x M)|Cu 
 (c) Ecell = Eright – Eleft = 0.031 – 0.0592 log [Br–] – ESCE 
 = 0.031 + 0.0592 pBr – 0.244 
 pBr = (Ecell – 0.031 + 0.244)/0.0592 = (Ecell + 0.213)/0.0592 
 (d) pBr = (–0.095 + 0.213)/0.0592 = 1.99 
23-15. (a) Ag3AsO4(s) 3Ag+ + AsO4
3– Ksp = [Ag+]3[AsO4
3–] = 1.2 × 10–22
 Ag+ + e– Ag(s) E0 = 0.799 V 
 
3
4
sp
[AsO ]1 0.0592 = 0.799 0.0592log = 0.799 log
[Ag ] 3
E
K
−
+− − 
 When [AsO4
3–] = 1.00, E = 
3 4
0
Ag AsOE
 
3 4
0
Ag AsO 22
0.0592 1.00 0.799 log
3 1.2 10
E −= −
×
 = 0.366 V 
 (b) SCE Ag3AsO4(sat’d), AsO4
3–(x M)|Ag 
 (c) Ecell = 0.366 – 3
4 S
0.0592 log[AsO ] 
3
E− − CE 
 4 4
0.0592 0.0592 0.366 + pAsO 0.244 = 0.122 + pAsO
3 3
= − 
 cell
4
( 0.122) 3pAsO = 
0.0592
E − × 
 (d) 4
(0.247 0.122) 3pAsO = 
0.0592
− × = 6.33