《现代基础化学》习题解答
第一章,习题答案,P44-46
1. E = -B/n2 (B= 2.18⨯ 10-18 J
∆E = -B/52 + B/22 = 4.58 ⨯ 10-19 J hv = ∆E, λν = c
λ = c ⨯ h ÷ ∆E = 434 nm 3.①②不存在;③④存在
4.Ni 3s2 3p6 3d8 4s2的四个量子数是:
3s 2: n= 3, l=0, m=0, ms = +1/2; n= 3, l=0, m=0, ms = -1/2
3p 6: n= 3, l=1, m=-1, ms = +1/2; n= 3, 1=1, m=-1, ms = -1/2; n= 3, 1=1, m=0, ms = +1/2
n= 3, 1=1, m=0, ms = -1/2; n= 3, 1=1, m=1, ms = +1/2; n= 3, 1=1, m=1, ms = -1/2 3d 8: n= 3, 1=2, m=-2, ms = +1/2; n= 3, 1=2, m=-2, ms = -1/2; n= 3, 1=2, m=-1, ms = +1/2
n= 3, 1=2, m=-1, ms = -1/2; n= 3, 1=2, m=0, ms = +1/2; n= 3, 1=2, m=0, ms = -1/2 n= 3, 1=2, m=1, ms = +1/2; n= 3, 1=2, m=2, ms = +1/2 4s 2: n= 4, 1=0, m=0, ms = +1/2; n= 4, 1=0, m=0, ms = -1/2
5.① 3d ②4f ③2s ④3d ⑤1s ⑥3p; 结果:⑤
8.① n>3;3d ②l = 1;2p
③m= -1 /0/+1; 3p
④ ms = +1/2 或-1/2, 4s 9.48Cd 4s 24p 64d 105s 2.
10.①有三个19K ;24Cr ;29Cu
②均第四周期, K: s区,,IA 族;Cr: d区VIB 族;Cu: ds区IB 族 11. (1) Ar: 1s22s 22p 63s 23p 6 (2)Fe: 1s22s 22p 63s 23p 64s 23d 6
(3) I: 1s22s 22p 63s 23p 63d 104s 24p 64d 105s 25p 5 ; (4) Ag: 1s22s 22p 63s 23p 63d 104s 24p 64d 105s 1
12.①(1)Ca: s区, 第四周期,IIA 族;(2) Cl: p区,第三周期,VIIA ;
(3) Ti: d区,第四周期,IVB ;(4) Hg: ds区, 第六周期, IIB族 ② Ca: +2,Cl: +7,Ti: +4,Hg: +2 ③ (1)
14.48;[Kr]4d105s 2, IIB族;48Cd
15.甲:3s 23p 5,VIIA ,非金属,电负性高;乙:3d 24s 2,IVB ,金属,电负性低。 16.①Cl ;②Cs ;③Be ;④Cl ;⑤Mn 19.22Ti 3+ [Ar]3d1 1个未成对;
2+4
24Cr [Ar]3d 4个未成对;
3+6
27Co [Ar]3d 4个未成对;
2+8
46Pd [Kr]4d 2个未成对;
3+1
58Ce [Xe]4f 1个未成对;
3+0
57La [Xe]5d 无未成对;
第二章习题答案,P81-82
1.① BeH 2: sp 杂化,直线型 ② SiH 4: sp3杂化,正四面体。 ③ BF 3: sp2杂化,平面三角形 ④ CO 2: sp杂化,直线型。 2.BF 3采用sp 2杂化;[BF4]-采用sp 3 3.H 2O sp3不等性杂化;BeH 2 sp杂化
4.H 2O sp3不等性杂化,104.5度,CH 4 sp3杂化,109.5度;CO 2 sp杂化,180度。 11
12.①乙醇高,有氢键
②甲、乙、丙醇分子结构相似,分子量依次增大,色散力递增,熔沸点依次升高。 ③丙三醇分子量大,熔沸点高, ④HF 有氢键,熔沸点高。
13.①CH 4,CCl 4,CI 4分子量递增,色散力增大,熔沸点依次升高。
②H 2O 有氢键,CH 4无氢键。
第三章习题答案,P119
1.NaF 〉 AgBr 〉 NH 3 〉O 2
离子晶体 过渡晶体 含氢键
+-++3.熔点随半径下降低升高,随离子电荷升高而升高。
4.①SiC 是原子晶体,熔点高。 ②水有氢键,熔点高。 ③KCl 离子晶体,熔点高。
④MgCl 2离子晶体,半径小熔点高。 ⑤KI 极化力小,熔点高。
5.①Sn 4+> Sn2+ > Fe2+ >Sr2+ ②S 2->O2->F-
第四章习题答案,P151
8.
9.
[1**********]5] ① [Ni(CN)4]2-为dsp 2杂化, [Ni(NH3) 4]2+为sp 3杂化,
② [Fe(CN)
6]4-为d 2sp 3杂化,无单电子;[FeF6]3-为sp 3 d2杂化,有五个单电子。
六 1. W=-P(V2-V 1)=-P2(P1V 1/P2-V 1)=-1×105(5×2×10-3/1-2×10-3)=-800J
2. ΔU=Q+W=Q+p(V2-V 1)=6.48+[-93.3(150-50)]=4.68+(-9.33)=-2.85kJ 5. ΔH=Qp =20×858=17.16×103J=17.16kJ
W=-Pout ΔV=-101325×20×607×10-6=-1230J= -1.230kJ 7. 2②-5/2③-① Δf H m °C 2H 2=[90.2×2-483.7×2.5-(-1255.6)]=226.75kJ/mol 8. 1/2[①+②+③-④] Δr H m °= 0.5[-234.7-184.6-30.59+285.85]= -127.02kJ/mol 9. Δc H m °=-1366.75kJ/mol;92/46=2mol, ΔrH m °=2×(-1366.75)= -2733.5kJ 12. Δf H m °C 4H 10=4×(-393.51)+5×(-285.85)-(-2878.51)]kJ/mol= -124.78kJ/mol
七 4. ① Δr G m °=Δr H m °-T Δr S m °= -91.92-298.15(-198.3)×10-3=-32.79kJ/mol
② Δr G m °=Δr H m °-T Δr S m °=181.42-298.15×.216=117.0kJ/mol ③ Δr G m °=Δr H m °-T Δr S m °= -890.36-298.15×(-0.243)= -817.9kJ/mol 6. ① Δr G m °=[(-569.6)+2×(-228.59)-(805.0)-4(-95.27)= 159.5kJ/mol ② Δr G m °= -394.38-(-137.3)-(×228.59)= -28.49kJ/mol ③ Δr G m °= 3(-394.38)-(-741.0)-3(-137.30)= -30.24kJ/mol 7. K 25°=exp[-(Δr H m °-T Δr S m °)/RT] =exp[-(-211.5×103-298.15×116.9)/(8.314×298.2)]==exp99.384=1.45×1043 K 100°=exp[-(-211.5×103-373.15×116.9)/(8.314×373.15)]=exp82.23=5.175×1035 8. Δr G m °= -RTlnK°= -8.314×3500ln8.28= -61510.7J/mol= -61.51kJ/mol 10. Δr G m °=37.22-298.15×0.09496=8.908kJ/mol>0; ←
T 1>Δr H m °/Δr S m °=37220/94.96= 319.95K Δr G m °=299.8-298.15×0.3633= 191.48kJ/mol>0; ←
T 2>Δr H m °/Δr S m °=299.8/0.3633= 825.2K 第一种好。 17. Δr G m °=-402-298.15×0.1891= -345.6kJ/mol
Δr G m °=Δr H m °-T Δr S m °
因Δr S m °
反应正向进行的最低温度:T=Δr H m °/Δr S m °=-402×103/-189.1= 2125.9K
八. 思
6. ① v=kCa C c 三级;② k=0.05/(0.01×0.01)=5×10(mol/L).s
③ v=5×104×0.53=6250 7. ① C 0/C=v0/v=1×10-5/3.26×10-6=3.067; k=(1/t)ln(C0/C)=(1/3600)ln3.067=3.11×10-4/s ② t 1/2=ln2 /k=0.693/k=0.693/3.11×10-4=2226.26s ③ C=v/k=10-5/3.11×10-4=0.03212mol/L
10. k=Ae-Ea/RT Ea=RT1T 2ln(k2/k1)/(T2-T 1)=(8.314×300×310×ln2)/10=53594J=53.59kJ 15. ① k=v/(CA 2C B )=2.5×10-3/0.013=2500(mol/L)-2.s -1 ② v=2500×0.0152×0.030=0.01688mol/L.s
16. ① Q=176-12=164kJ/mol; ② Ea +=176kJ/mol;Ea -=12kJ/mol
③ 在曲线最高处有活化络合物存在。④ 是放热反应 九
33238. 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+]=KCa /Cb =1.75×10-5; pH=4.76 ④[HAc]=0.05; [NaAc]=0.05; [H+]=KCa /Cb =1.75×10-5; pH=4.76 11. [H+]=KCa /Cb =10-4.5=3.16×10-5; CAcNa =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=Ka2/Kw =4.68×103; ②K=Ka2/Kw =6.31×106; ③K=Ka1/Kw =2.09×105; ④K=Ka2/Kw =1.51×105 16. 1.37×10-4×10/811.5=1.688×10-7; Ksp =(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
Cr 3+; [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=KZnS /KAg2S =2.5×10-22/6.3×10-50=4.0×1027; ⑵K=KZnS /KPbS =2.5×10-22/8.0×10-28=3.1×105;
⑶K=KPbCl2/KPbCrO4=1.7×10-5/2.8×10-13=6.1×107; 28. [Ac-]= KHAc [HAc]/ [H+]=1.75×103;
[Ag+]=Ksp /[Ac-]=1.94×10-3/1.75×10-3=1.3mol/L; 1.3×170=221g 31. [S2-]=K1K 2[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>Ksp,PbS (8×10-28); 有沉淀。 十.
1. n=0.0406; t=nZF/I=(0.0406X2X96485/5X60)=2.61min 4. ⑴ 2KMnO 4+5K2SO 3+3H2SO 4=6K2SO 4+2MnSO4+3H2O
⑵ NaBiO 3(s)+2MnSO4+16HNO3=2HMnO4+5Bi(NO3) 3+2Na2SO 4+NaNO3+7H2O ⑶ 4Zn+NO3-+10H+=4Zn2++NH4++3H2O ⑷ 3Ag+NO3-+4H+=3Ag++NO+2H2O
⑸ Cl 2+2OH-=Cl-+ClO-+H2O
⑹ 8Al+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+0.0296lgKsp 1/2=-0.179V
③E CuS/Cu=0.342+0.0296lgKsp =-0.700V
9. ① E=-0.0595V ② E=-0.414 ③ E=-0.17V④
θ
10. ① E =0.799+0.05916lg(1.12X10-12) 1/2=0.445
θ
②E =0.771+0.05916lg2.79X10-39/4.87X10-17=0.546V 11. 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(cKa ) 1/2-[-0.760+0.296lg0.1]=-0.17-(-0.790)=0.62V 12. E=0.799-(-0.76)+0.0296lg(0.12/x)=1.51V; [Zn2+]=x=0.452V 13. lgK sp =(0.126-0.140-0.22)*2/0.05916; Ksp =1.24X10-8 14. ①→②←③→
θθθθ
15. ①K =1.14X1041 ②K =0.134 ③K =1.86X10-5 ④K =2.14X1062
θθ
16. ①E =0.822V; E=-0.744V ②Cr 3+稳定
θθ
17. E BrO -/Be- =0.76; EBrO3-/Br-=0.52
18. (3X1.679+2x)/(3+2)=1.507V x=1.249V
19. ⑴ΔG=-ZFE=-6X96500X(1.23-1.36)X10-3=75.27kJ/mol
⑵E Cr2O72-/Cr3+=1.23+(0.05916/6)lg1214; ECl2/Cl-=1.36+0.0296lg12-2=1.30V E=1.38-1.3=0.08V θθ
20. ΔG =-ZFE=-2X96500X(0.222-0.151)X10-3=-13.7kJ
θθθ
21. E =0.236;ΔG =-45.54; K =9.40X107 十一.
θ
1. K 不=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
《现代基础化学》习题解答
第一章,习题答案,P44-46
1. E = -B/n2 (B= 2.18⨯ 10-18 J
∆E = -B/52 + B/22 = 4.58 ⨯ 10-19 J hv = ∆E, λν = c
λ = c ⨯ h ÷ ∆E = 434 nm 3.①②不存在;③④存在
4.Ni 3s2 3p6 3d8 4s2的四个量子数是:
3s 2: n= 3, l=0, m=0, ms = +1/2; n= 3, l=0, m=0, ms = -1/2
3p 6: n= 3, l=1, m=-1, ms = +1/2; n= 3, 1=1, m=-1, ms = -1/2; n= 3, 1=1, m=0, ms = +1/2
n= 3, 1=1, m=0, ms = -1/2; n= 3, 1=1, m=1, ms = +1/2; n= 3, 1=1, m=1, ms = -1/2 3d 8: n= 3, 1=2, m=-2, ms = +1/2; n= 3, 1=2, m=-2, ms = -1/2; n= 3, 1=2, m=-1, ms = +1/2
n= 3, 1=2, m=-1, ms = -1/2; n= 3, 1=2, m=0, ms = +1/2; n= 3, 1=2, m=0, ms = -1/2 n= 3, 1=2, m=1, ms = +1/2; n= 3, 1=2, m=2, ms = +1/2 4s 2: n= 4, 1=0, m=0, ms = +1/2; n= 4, 1=0, m=0, ms = -1/2
5.① 3d ②4f ③2s ④3d ⑤1s ⑥3p; 结果:⑤
8.① n>3;3d ②l = 1;2p
③m= -1 /0/+1; 3p
④ ms = +1/2 或-1/2, 4s 9.48Cd 4s 24p 64d 105s 2.
10.①有三个19K ;24Cr ;29Cu
②均第四周期, K: s区,,IA 族;Cr: d区VIB 族;Cu: ds区IB 族 11. (1) Ar: 1s22s 22p 63s 23p 6 (2)Fe: 1s22s 22p 63s 23p 64s 23d 6
(3) I: 1s22s 22p 63s 23p 63d 104s 24p 64d 105s 25p 5 ; (4) Ag: 1s22s 22p 63s 23p 63d 104s 24p 64d 105s 1
12.①(1)Ca: s区, 第四周期,IIA 族;(2) Cl: p区,第三周期,VIIA ;
(3) Ti: d区,第四周期,IVB ;(4) Hg: ds区, 第六周期, IIB族 ② Ca: +2,Cl: +7,Ti: +4,Hg: +2 ③ (1)
14.48;[Kr]4d105s 2, IIB族;48Cd
15.甲:3s 23p 5,VIIA ,非金属,电负性高;乙:3d 24s 2,IVB ,金属,电负性低。 16.①Cl ;②Cs ;③Be ;④Cl ;⑤Mn 19.22Ti 3+ [Ar]3d1 1个未成对;
2+4
24Cr [Ar]3d 4个未成对;
3+6
27Co [Ar]3d 4个未成对;
2+8
46Pd [Kr]4d 2个未成对;
3+1
58Ce [Xe]4f 1个未成对;
3+0
57La [Xe]5d 无未成对;
第二章习题答案,P81-82
1.① BeH 2: sp 杂化,直线型 ② SiH 4: sp3杂化,正四面体。 ③ BF 3: sp2杂化,平面三角形 ④ CO 2: sp杂化,直线型。 2.BF 3采用sp 2杂化;[BF4]-采用sp 3 3.H 2O sp3不等性杂化;BeH 2 sp杂化
4.H 2O sp3不等性杂化,104.5度,CH 4 sp3杂化,109.5度;CO 2 sp杂化,180度。 11
12.①乙醇高,有氢键
②甲、乙、丙醇分子结构相似,分子量依次增大,色散力递增,熔沸点依次升高。 ③丙三醇分子量大,熔沸点高, ④HF 有氢键,熔沸点高。
13.①CH 4,CCl 4,CI 4分子量递增,色散力增大,熔沸点依次升高。
②H 2O 有氢键,CH 4无氢键。
第三章习题答案,P119
1.NaF 〉 AgBr 〉 NH 3 〉O 2
离子晶体 过渡晶体 含氢键
+-++3.熔点随半径下降低升高,随离子电荷升高而升高。
4.①SiC 是原子晶体,熔点高。 ②水有氢键,熔点高。 ③KCl 离子晶体,熔点高。
④MgCl 2离子晶体,半径小熔点高。 ⑤KI 极化力小,熔点高。
5.①Sn 4+> Sn2+ > Fe2+ >Sr2+ ②S 2->O2->F-
第四章习题答案,P151
8.
9.
[1**********]5] ① [Ni(CN)4]2-为dsp 2杂化, [Ni(NH3) 4]2+为sp 3杂化,
② [Fe(CN)
6]4-为d 2sp 3杂化,无单电子;[FeF6]3-为sp 3 d2杂化,有五个单电子。
六 1. W=-P(V2-V 1)=-P2(P1V 1/P2-V 1)=-1×105(5×2×10-3/1-2×10-3)=-800J
2. ΔU=Q+W=Q+p(V2-V 1)=6.48+[-93.3(150-50)]=4.68+(-9.33)=-2.85kJ 5. ΔH=Qp =20×858=17.16×103J=17.16kJ
W=-Pout ΔV=-101325×20×607×10-6=-1230J= -1.230kJ 7. 2②-5/2③-① Δf H m °C 2H 2=[90.2×2-483.7×2.5-(-1255.6)]=226.75kJ/mol 8. 1/2[①+②+③-④] Δr H m °= 0.5[-234.7-184.6-30.59+285.85]= -127.02kJ/mol 9. Δc H m °=-1366.75kJ/mol;92/46=2mol, ΔrH m °=2×(-1366.75)= -2733.5kJ 12. Δf H m °C 4H 10=4×(-393.51)+5×(-285.85)-(-2878.51)]kJ/mol= -124.78kJ/mol
七 4. ① Δr G m °=Δr H m °-T Δr S m °= -91.92-298.15(-198.3)×10-3=-32.79kJ/mol
② Δr G m °=Δr H m °-T Δr S m °=181.42-298.15×.216=117.0kJ/mol ③ Δr G m °=Δr H m °-T Δr S m °= -890.36-298.15×(-0.243)= -817.9kJ/mol 6. ① Δr G m °=[(-569.6)+2×(-228.59)-(805.0)-4(-95.27)= 159.5kJ/mol ② Δr G m °= -394.38-(-137.3)-(×228.59)= -28.49kJ/mol ③ Δr G m °= 3(-394.38)-(-741.0)-3(-137.30)= -30.24kJ/mol 7. K 25°=exp[-(Δr H m °-T Δr S m °)/RT] =exp[-(-211.5×103-298.15×116.9)/(8.314×298.2)]==exp99.384=1.45×1043 K 100°=exp[-(-211.5×103-373.15×116.9)/(8.314×373.15)]=exp82.23=5.175×1035 8. Δr G m °= -RTlnK°= -8.314×3500ln8.28= -61510.7J/mol= -61.51kJ/mol 10. Δr G m °=37.22-298.15×0.09496=8.908kJ/mol>0; ←
T 1>Δr H m °/Δr S m °=37220/94.96= 319.95K Δr G m °=299.8-298.15×0.3633= 191.48kJ/mol>0; ←
T 2>Δr H m °/Δr S m °=299.8/0.3633= 825.2K 第一种好。 17. Δr G m °=-402-298.15×0.1891= -345.6kJ/mol
Δr G m °=Δr H m °-T Δr S m °
因Δr S m °
反应正向进行的最低温度:T=Δr H m °/Δr S m °=-402×103/-189.1= 2125.9K
八. 思
6. ① v=kCa C c 三级;② k=0.05/(0.01×0.01)=5×10(mol/L).s
③ v=5×104×0.53=6250 7. ① C 0/C=v0/v=1×10-5/3.26×10-6=3.067; k=(1/t)ln(C0/C)=(1/3600)ln3.067=3.11×10-4/s ② t 1/2=ln2 /k=0.693/k=0.693/3.11×10-4=2226.26s ③ C=v/k=10-5/3.11×10-4=0.03212mol/L
10. k=Ae-Ea/RT Ea=RT1T 2ln(k2/k1)/(T2-T 1)=(8.314×300×310×ln2)/10=53594J=53.59kJ 15. ① k=v/(CA 2C B )=2.5×10-3/0.013=2500(mol/L)-2.s -1 ② v=2500×0.0152×0.030=0.01688mol/L.s
16. ① Q=176-12=164kJ/mol; ② Ea +=176kJ/mol;Ea -=12kJ/mol
③ 在曲线最高处有活化络合物存在。④ 是放热反应 九
33238. 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+]=KCa /Cb =1.75×10-5; pH=4.76 ④[HAc]=0.05; [NaAc]=0.05; [H+]=KCa /Cb =1.75×10-5; pH=4.76 11. [H+]=KCa /Cb =10-4.5=3.16×10-5; CAcNa =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=Ka2/Kw =4.68×103; ②K=Ka2/Kw =6.31×106; ③K=Ka1/Kw =2.09×105; ④K=Ka2/Kw =1.51×105 16. 1.37×10-4×10/811.5=1.688×10-7; Ksp =(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
Cr 3+; [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=KZnS /KAg2S =2.5×10-22/6.3×10-50=4.0×1027; ⑵K=KZnS /KPbS =2.5×10-22/8.0×10-28=3.1×105;
⑶K=KPbCl2/KPbCrO4=1.7×10-5/2.8×10-13=6.1×107; 28. [Ac-]= KHAc [HAc]/ [H+]=1.75×103;
[Ag+]=Ksp /[Ac-]=1.94×10-3/1.75×10-3=1.3mol/L; 1.3×170=221g 31. [S2-]=K1K 2[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>Ksp,PbS (8×10-28); 有沉淀。 十.
1. n=0.0406; t=nZF/I=(0.0406X2X96485/5X60)=2.61min 4. ⑴ 2KMnO 4+5K2SO 3+3H2SO 4=6K2SO 4+2MnSO4+3H2O
⑵ NaBiO 3(s)+2MnSO4+16HNO3=2HMnO4+5Bi(NO3) 3+2Na2SO 4+NaNO3+7H2O ⑶ 4Zn+NO3-+10H+=4Zn2++NH4++3H2O ⑷ 3Ag+NO3-+4H+=3Ag++NO+2H2O
⑸ Cl 2+2OH-=Cl-+ClO-+H2O
⑹ 8Al+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+0.0296lgKsp 1/2=-0.179V
③E CuS/Cu=0.342+0.0296lgKsp =-0.700V
9. ① E=-0.0595V ② E=-0.414 ③ E=-0.17V④
θ
10. ① E =0.799+0.05916lg(1.12X10-12) 1/2=0.445
θ
②E =0.771+0.05916lg2.79X10-39/4.87X10-17=0.546V 11. 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(cKa ) 1/2-[-0.760+0.296lg0.1]=-0.17-(-0.790)=0.62V 12. E=0.799-(-0.76)+0.0296lg(0.12/x)=1.51V; [Zn2+]=x=0.452V 13. lgK sp =(0.126-0.140-0.22)*2/0.05916; Ksp =1.24X10-8 14. ①→②←③→
θθθθ
15. ①K =1.14X1041 ②K =0.134 ③K =1.86X10-5 ④K =2.14X1062
θθ
16. ①E =0.822V; E=-0.744V ②Cr 3+稳定
θθ
17. E BrO -/Be- =0.76; EBrO3-/Br-=0.52
18. (3X1.679+2x)/(3+2)=1.507V x=1.249V
19. ⑴ΔG=-ZFE=-6X96500X(1.23-1.36)X10-3=75.27kJ/mol
⑵E Cr2O72-/Cr3+=1.23+(0.05916/6)lg1214; ECl2/Cl-=1.36+0.0296lg12-2=1.30V E=1.38-1.3=0.08V θθ
20. ΔG =-ZFE=-2X96500X(0.222-0.151)X10-3=-13.7kJ
θθθ
21. E =0.236;ΔG =-45.54; K =9.40X107 十一.
θ
1. K 不=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