'31. 解：lgKCu= 18.8 – 6.6 = 12.2

化学计量点 [Cu']?2?spcCu?'KCuY0.02/2?7.1 ?1012.210 即 pCu计 = 7.1 pCu终 = pCut = 8.8

⊿pCu = 8.8 – 7.1 = 1.7

Et?10?pM?10-?pMcCu计K'CuY?101.7?10?1.710?2.0?12.2?100%?0.04%

由计算结果可知，终点的相对误差为0.04%＜0.1%，故选用PAN指示剂可行。

32. 解： αY(Ca)=1+10-2+10.7=108.7

αY = αY(H) + αY(Ca) -1= 106.8 + 108.7 – 1 ≈ 108.7

'?lgKPbY?lg?= 18.0 – 8.7 = 9.3 (1) lgKPbY[Pb']?spcPb?'KPbY0.010?10?5.7 9.310即 pPb计= 5.7 = pY'计 = pCaY计

[Pb2+]计 = [CaY]计 = 10-5.7(mol?L-1)

(2)

⊿pPb =7.0 – 5.7=1.3

Et?10?pM?10-?pMcPb计K'PbY?101.3?10?1.310?2.0?9.3?100%?0.4%

[PbY]10?2.0?4.3-1[CaY]终?[Y']终???10 (mol?L) '?7.0?9.3[Pb]KPbY10

33. 解： (1) 六次甲基四胺介质

'lgKPbY?lgKPbY?lg?Y(H)=18.0 – 6.6=11.4

spcPb10?4?7.7??10 '11.4KPbY10[Pb']?即 pPb计 = 7.7 ⊿pPb=7.0 – 7.7= – 0.7

Et?10?pM?10-?pMcPb计K'PbY?10?0.7?100.710?4.0?11.4?100%??0.1%

(2) HAc-NaAc介质

?Pb(Ac)=1+[Ac]β1+[Ac]

2

β2= 1+10-1+1.9+10-2+3.3 =101.5

'lgKPbY?lgKPbY?lg?Y(H)?lg?Pb(Ac)=18.0 – 6.6 –1.5 = 9.9

spcPb10?4??10?7.0 '9.9KPbY10[Pb']? 即 pPb'计= 7.0 pPb'终 = 7.0 – 1.5 = 5.5 ⊿pPb'=5.5 – 7.0 = – 1.5

Et?10?pM'?10-?pM'cPbspK'PbY?10?1.5?101.510?4.0?9.9?100%??3.5%

结论：应选用六次甲基四胺

34. 解：用EDTA滴定Zn2+ 时，存在的各种反应：

Zn2+ + YOH-Zn(OH)+ ?Zn(OH)ZnYNH3Zn(NH3)2+ ?Zn(NH3)H+HY ?Y(H) Y(H) = 0.45，

lg? M(OH) = 2.4，

由于是NH3 – NH4Cl溶液，其pH = 10。查表得，lg?lgKZnY = 16.5。

[NH3]?cNH3?NH3[OH?]10?4?0.1??0.1??4?0.085 (mol?L?1) ??5[OH]?Kb10?1.8?10?Zn(NH)?1?102.27?0.085?104.61?(0.085)2?107.01?(0.085)3?109.06?(0.085)4?6.65?104

3?Zn??Zn(OH)??Zn(NH)?1?102.4?6.65?104?6.65?104

3'lgKZnY?lgKZnY?lg?Y?lg?Zn?16.5?0.45?lg6.65?104?11.23

sp'spKMY)≥6，本题中cM金属离子被准确滴定的条件是lg(cM = 0.1/2 = 0.05，故

sp'lg(cZnKZnY)= lg0.05 + 11.23 = 9.93＞6

所以，可用EDTA准确滴定0.1000mol ? L-1 的Zn2+。

35. 解：用EDTA滴定Zn2+ 时，存在的各种反应：

Zn2+ + YH+HY ?Y(H)ZnYCa2+CaY ?Y(Ca)

终点时， [Ca2+]=0.020/2 = 0.01 (mol?L-1)

αY(Ca) =1+KCaY[Ca2+]=1+1010.7×0.01 =1+108.7 ≈108.7 αY =αY(H) +αY(Ca) – 1=105.5 +108.7 – 1≈108.7

'lgKZnY?lgKZnY?lg?Y?16.5?8.7?7.8

sp'spKMY)≥6，本题中cM金属离子被准确滴定的条件是lg(cM = 0.020/2 = 0.010，故

sp'lg(cZnKZnY)= – 2.0 + 7.8= 5.8 < 6

所以不能用EDTA准确测定Zn2+ 。

36. 解：查表得lgKBiY=27.94，lgKPbY=18.04，Pb(OH)2的Ksp=10-14.93 (1)判断能否分布滴定 ∵CBi=CPb

⊿lgK=lgKBiY – lgKPbY = 27.94 - 18.04 = 9.90 > 6 ∴可能在Pb2+存在下滴定Bi3+

(2) pH=1时，lgαY(H) =18.01，lgαBi(OH) = 0.1 αY(N) =1+KPbY[Pb] = 1+1018.04 × 10-2.00 = 1016.04 αY= αY(H) +αY(N) -1 =1018.01 + 1016.04 -1≈ 1018.01 lgK′BiY=lgKBiY-lgαY- lgαBi = 27.94 - 18.01 - 0.1 = 9.83 lgCBiK′BiY = -2.00 + 9.83 = 7.83＞6 ∴pH=1时可准确滴定Bi3+ (3)(a)计算滴定Pb2+的最高酸度

lgαY(H)= lgKPbY – 8 = 18.04 – 8 = 10.04，查表得pH=3.25

(b)计算滴定Pb2+的最低酸度.

Ksp10?14.93?6.46[OH]???10[Pb2?]0.01?pH=7.54，因此可在pH=4～7的范围内滴定Pb2+ (c)计算最适酸度.CPb,sp=1/3×0.020 =10-2.18 当pH=5时，pPbep=7， lgαY(H) = 6.45 lgK’PbY= lgKPbY - lgαY = 18.04 - 6.45 = 11.59

pPbsp= (pCPb,sp+lgK’PbY)/2 = (2.18+11.59)/2 = 6.9，与pPbep十分接近. 结论:滴定Pb2+的最佳pH范围为5.0

37. 解：用EDTA滴定Pb2+ 时，存在的各种反应：

Pb2++ YH+HY ?Y(H)PbYAl3+AlY ?Y(Al)F-AlF63- ?Al(F)

αAl(F)=1+10-2.0+6.1+10-4.0+11.2+10-6.0+15.0+10-8.0+17.7+10-10.0+19.4+10-12.0+19.7 =109.9

[Al]?Y(Al)=1+[Al

3+

spcAl?Al(F)10?2.0 ?9.9?10?11.9 (mol?L-1)10]KAlY = 1+10-11.9+16.3 =104.4

?Y=?Y(H)+ ?Y(Al) – 1= 104.4 + 105.5 – 1 = 105.5

'lgKPbY?lgKPbY?lg?Y?18.0?5.5?12.5

[Pb']?spcPb10?2?7.3

??10'KPbY1012.5 即 pPb计=7.3

⊿pPb=7.6 – 7.3 = – 0.3

Et?10?pM'?10-?pM''cPbspKPbY?10?0.3?100.310?2.0?12.5?100%??8?10?4%