仪器分析习题解答

北京化工大学

仪器分析习题解答

董慧茹 编

2005年2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

(1) 由pH―V图查得 V终点 = 9.10ml

由ΔpH/ΔV―V图查得 V终点 = 9.10ml (2) 计算[HA]

0.1000?9.10[HA] = = 0.09100 (mol/L)

10.00(3) 化学计量点的pH值

pH Δ2pH/ΔV2 6.80 +42.5 9.10 -40.0

9.1?6.8pH终点 = 6.80 + ×42.5 = 7.98

42.5?40.0