部分习题答案

九

Conjugate Conjugate Substance Conjugate Conjugate

]=(0.025×-7)3]=4.68×23

3

8. ①[HCl]=0.025 pH=1.6

②[NH4Cl]=0.05; [H+]=(0.05×10-14/1.74×10-5)1/2=5.35×10-6; pH=5.27

③[HAc]=0.05; [NaAc]=0.05; [H+]=KC a/C b=1.75×10-5; pH=4.76

④[HAc]=0.05; [NaAc]=0.05; [H+]=KC a/C b=1.75×10-5; pH=4.76

11. [H+]=KC a/C b=10-4.5=3.16×10-5; C AcNa=0.45mol/L; 0.45×0.5×136=30.6g; 0.82×500/6=68.3ml

12.[H+]=1.75×10-5×0.02/0.8=4.38×10-7; pH=6.36

①[HAc]=(0.1×10/55)+0.1×5/55=0.027; [NaAc]=0.8-0.1×5/55=0.718;

[H+]=1.75×10-5×0.027/0.718=6.58×10-7; pH=6.18 ΔpH=0.18

②[HAc]=(0.1×10/55)-0.1×5/55=0.009; [NaAc]=(0.1×40/55)+0.1×5/55=0.736;

[H+]=1.75×10-5×0.009/0.736=2.148×10-7; pH=6.67 ΔpH=0.31

14.①K=K a2/K w=4.68×103; ②K=K a2/K w=6.31×106; ③K=K a1/K w=2.09×105; ④K=K a2/K w=1.51×105

16. 1.37×10-4×10/811.5=1.688×10-7; K sp=(3×1.688×10-7)3×(2×1.688×10-7)2=1.48×10-32

18. ①[Pb2+]=0.0050; [I-]=0.005; [Pb2+][I-]2=(0.0050)3=1.3×10-7>Ksp(9.8×10-9)；有沉淀。

②[Ba2+]=0.020; [CO32-]=0.30; [Ba2+][CO32-]=0.020×0.30=6.0×10-3>Ksp(2.58×10-9)；有沉淀。

③[Ag+]=0.010; [Cl-]=0.10; [Ag+][Cl-]=0.010×0.100=1.0×10-3>Ksp(1.77×10-10)；有沉淀。

④[OH-]=5.0×10-5; [Mg2+][OH-]2=10-7× (5.0×10-5)2=2.5×10-16

Cr3+; [OH-]=(6.3×10-31/0.02)1/3=3.2×10-10; pOH=9.5; pH=4.5; Cr3+先↓;

因：[OH-]=(6.3×10-31/2.0×10-6)1/3=6.8×10-9; pOH=8.2; pH=5.8>4.8; 无法满足要求。26．⑴K=K ZnS/K Ag2S=2.5×10-22/6.3×10-50=4.0×1027; ⑵K=K ZnS/K PbS=2.5×10-22/8.0×10-28=3.1×105;

⑶K=K PbCl2/K PbCrO4=1.7×10-5/2.8×10-13=6.1×107;

28. [Ac-]= K HAc [HAc]/ [H+]=1.75×103;

[Ag+]=K sp/[Ac-]=1.94×10-3/1.75×10-3=1.3mol/L; 1.3×170=221g

31. [S2-]=K1K2[H2S]/[H+]2=1.07×10-7×1.26×10-13×0.10/0.12=1.35×10-19

[Pb2+][S2-]=0.001×1.35×10-19=1.35×10-22>K sp,PbS(8×10-28); 有沉淀。

十.

1．n=0.0406; t=nZF/I=(0.0406X2X96485/5X60)=2.61min

4. ⑴2KMnO4+5K2SO3+3H2SO4=6K2SO4+2MnSO4+3H2O

⑵NaBiO3(s)+2MnSO4+16HNO3=2HMnO4+5Bi(NO3)3+2Na2SO4+NaNO3+7H2O

⑶4Zn+NO3-+10H+=4Zn2++NH4++3H2O

⑷3Ag+NO3-+4H+=3Ag++NO+2H2O

⑸Cl2+2OH-=Cl-+ClO-+H2O

⑹8Al+3NO3-+5OH-+18H2O=8[Al(OH)4]-+3NH3

5．(c) 正极半反应: 2H++2e=H2负极半反应: Pb+2Cl-=PbCl2+2e 氧化剂: H+,还原剂: Pb 原电池: -) Pb | PbCl2 | Cl- || H+ | H2 | Pt (+

6．

7．⑴⑸能共存，⑵⑶⑷不能共存

8．①E Cu2+/Cu=0.342+0.0296lg0.5=0.333V ②E=0.342+0.0296lgK sp1/2=-0.179V

③E CuS/Cu=0.342+0.0296lgK sp=-0.700V

4.①E=-0.0595V ②E=-0.414 ③E=-0.17V④

5.①Eθ=0.799+0.05916lg(1.12X10-12)1/2=0.445

②Eθ=0.771+0.05916lg2.79X10-39/4.87X10-17=0.546V

6.①E=0.342-[-0.760+0.0296lg0.001=0.342-(-0.849)=1.191V

②E=-0.126+0.0296lg0.1-[0.342+0.0296lg(6.3X10-36/0.1)=-0.156-(-0.670)=0.514V

③E=0.328

④E=0.05916lg(cK a)1/2-[-0.760+0.296lg0.1]=-0.17-(-0.790)=0.62V

7.E=0.799-(-0.76)+0.0296lg(0.12/x)=1.51V; [Zn2+]=x=0.452V

8.lgK sp=(0.126-0.140-0.22)*2/0.05916; K sp=1.24X10-8

9.①→②←③→

10.①Kθ=1.14X1041②Kθ=0.134 ③Kθ=1.86X10-5 ④Kθ=2.14X1062

11.①Eθ=0.822V; Eθ=-0.744V ②Cr3+稳定

12.EθBrO-/Be- =0.76; EθBrO3-/Br-=0.52

13.(3X1.679+2x)/(3+2)=1.507V x=1.249V

14.⑴ΔG=-ZFE=-6X96500X(1.23-1.36)X10-3=75.27kJ/mol

⑵E Cr2O72-/Cr3+=1.23+(0.05916/6)lg1214; E Cl2/Cl-=1.36+0.0296lg12-2=1.30V

E=1.38-1.3=0.08V

15.ΔGθ=-ZFEθ=-2X96500X(0.222-0.151)X10-3=-13.7kJ

16.Eθ=0.236；ΔGθ=-45.54;Kθ=9.40X107

十一.

1.K不θ=1.0X10-20

2.[Ag+]=9.92X10-8; [NH3]=0.3

3.[NH3]=2.25+0.2=2.25;

4.

5.⑴⑵⑶⑷⑸⑹

1.

①②③④⑤

⑴⑵⑶⑷⑸⑹