北京化工大学
仪器分析习题解答
董慧茹 编
2010年6月
第二章 电化学分析法习题解答
25. 解： pHs = 4.00 , Es = 0.209V
26. 解： 27. 解： pHx = pHs ＋Ex?Es0.059
(1) pHx1 = 4.00 + 0.312?0.2090.059 = 5.75
(2) pHx2 = 4.00 ＋0.088?0.2090.059 = 1.95
(3) pHx3 = 4.00 ＋?0.017?0.2090.059 = 0.17
[HA] = 0.01mol/L , E = 0.518V
[A-] = 0.01mol/L , ΦSCE = 0.2438V
E = ΦSCE － Φ2H+/H2
0.518 = 0.2438 － 0.059 lg[H+
] [H+
] =
k[HA]a? = k0.01[A]a0.01
0.518 = 0.2438 － 0.059 lgk0.01a0.01 lgka = - 4.647 k-5
a = 2.25×10
2Ag+ + CrO2?4 = Ag2CrO4
[Ag+]2
=
Ksp[CrO2? 4]
E??SCE??Ag2CrO4/Ag - 0.285 = 0.2438 - [0.799 +
0.059Ksp2lg()] 2?2[CrO4] lgKspKsp-10
= - 9.16 , = 6.93×10 2?2?[CrO4][CrO4]2?4 [CrO
1.1?10?12-3 ] = = 1.59×10(mol/L) ?106.93?10-3
28. 解：pBr = 3 , aBr- = 10mol/L pCl = 1 , aCl- = 10mol/L 百分误差 =
KBr?,Cl??aCl?aBr?-1
6?10?3?10?1×100 = ×100 = 60
10?3 因为干扰离子Cl-的存在，使测定的aBr- 变为： aBr?= aBr?＋KBr?.Cl?×aCl?= 10＋6×10×10=1.6×10
即aBr?由10mol/L变为1.6×10mol/L 相差3.0 - 2.8 = 0.2 pBr单位
29. 解：
-3
-3-3
-3
-1
-3