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