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 = , Es =
(1) pHx26. 解：27. 解：＋Ex?Es0.059
0.312?0.2091 = + 0.059 =
= ＋0.088?0.2092 0.059 =
＋?0.017?0.2093 = 0.059 =
[A-] = L , ΦSCE =
E = ΦSCE － Φ2H+/H2
= － lg[H+
] [H+
] = k[HA]a[A?] = k0.01a0.01 = － lgk0.01a0.01 ka = - k-5
a = ×10
+
+ CrO2?4 = Ag2CrO4
+]2
=
Ksp[CrO2? 4] pHx = pHs (2) pHx (3) pHx [HA] = L , E = lg 2Ag [Ag E??SCE??Ag2CrO4/Ag - = - [ +
0.059Ksp2lg()] 2?2[CrO4] lgKspKsp-10
= - , = ×10 2?2?[CrO4][CrO4]2?4 [CrO
1.1?10?12-3
] = = ×10(mol/L) ?106.93?10-3
28. 解：pBr = 3 , aBr- = 10mol/L pCl = 1 , aCl- = 10mol/L 百分误差 =
-1
KBr?,Cl??aCl?aBr?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=×10
即aBr?由10mol/L变为×10mol/L 相差 - = pBr单位
29. 解：
-3
-3-3
-3
-1
-3