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1、1第二章 分析化学中的化学平衡第二章 分析化学中的化学平衡本章平衡常数方面基本在无机接触过，须灵活掌握以供今后计算使用；分布系数在计算浓度时非常有用，注意记清质子化与配合的区别联系，较易混淆；副反应系数和条件平衡常数在滴定计算，尤其是配合滴定中，非常有用，是本章又一重点及考点。作为本书的基础章节，虽难点不多，但需要熟练牢记，后面才不会卡住。书后习题解析（for reference）*1. 离子强度及活度系数，小化只要求了解公式ai =ii(1) I = 1/2CiZi CNa+ = 0.050 mol/L CNO2- = 0.050 mol/L I = 1/2(0.0501 + 0.0501)
2、 = 0.05 mol/L (分子的离子强度为0，即HNO2的I = 0)(2) CNH4+ = 0.066 mol/L CSO4= 0.033 mol/L I = 1/2(0.0661 + 0.0332) = 0.099 mol/L(3) CHCOO = 0.100 mol/L CNa+ = 0.100 mol/L I = 1/2(0.1001 + 0.1001) = 0.100 mol/L(4) I = 1/2(0.051 + 0.051) = 0.05 mol/L2. I = 1/2CiZi2 = 1/2(0.0512 + 0.0512 ) = 0.05 mol/L a H+= 0.9
3、a NH+= 0.3 lgH+ = -0.51210.05/(1+3.280.90.05)= -0.069 H+ = 0.85 lgNH+ = -0.51210.05/(1+3.280.30.05)= -0.0938 NH+ = 0.81NH4+ = H+ + NH3 Ka (I=0) = a H+/ a NH+Ka (I=0) = Ka (I=0.05)H+/NH+ Ka (I=0.05) = 5.210-10 =10-9.283. Zn + 4CN = Zn(CN)2-4 4(I=0) = aZn(CN)4/(aZnaCN)= 4(I=0.1)Zn2+(CN-)4 lg4(I=0) = l
4、g4(I=0.1) + lgZn2+ + 4lgCN- aZn= 0.9 a CN-= 0.3 lgZn2+ = -0.51220.1/(1+3.280.60.1)= -0.3991 Zn2+ = 0.399lgCN- = -0.51210.1/(1+3.280.30.1)= -0.123 CN- = 0.753lg4(I=0) = 16.7 + (-0.3991) + 4(-0.123)= 15.814(I=0) = 1015.814. Ksp(I=0) = Ksp(I=0.1)Ag + Cl- lgAg+ = -0.51210.1/(1+3.280.30.1)= -0.123 a Ag+=
5、 0.9 Ag+ =0.75 ; a Cl-= 0.3 Cl- = 0.75 lg Ksp(I=0) = lg Ksp(I=0.1) + lgCl- + lgAg+ = -9.49 - 0.123 - 0.123 = -9.74 Ksp(I=0) = 10-9.74 = 1.810-10 5. pKa1 pKa6 : 0.9 , 1.6 , 2.07 , 2.75 , 6.24 , 10.34 (I = 0.1,书P370) K1 H = 1/Ka6 = 1010.34 , K2 H = 1/Ka5 = 106.24 , K3 H = 1/Ka4 = 102.75 ， K4 H = 1/Ka3
6、 = 102.07 ， K5 H = 1/Ka2 = 101.6 ， K6 H = 1/Ka1 = 100.9 K1 H = 1010.34 1 = K1 H = 1010.34 K2 H = 106.24 2 = K1 HK2 H = 1016.58 K3 H = 102.75 2 = K1 HK2 HK3 H = 1019.33 K4 H = 102.07 2 = K1 HK2 HK3 HK4 H = 1021.4 K5 H = 101.6 2 = K1 HK5 H = 1023.0 K6 H = 100.9 2= K1 HK6H = 1023.96.(1) Br2 + 2e- = 2Br
7、 E1= 1.087 V I2 + 6H2O + 10e- = 2IO3+ 12H + E2= 1.20 V I2 + 5 Br2 + 6H2O = 2IO3+ 10 Br+ 12H + lgK = 10(1.087-1.20)/0.059 = -19.15 K = 10-19.15 = 7.110-20 (2) lgK = n(E1- E2)/0.059 = 2-0.91-(-0.44)/0.059 = -15.93 K = 10-15.93 = 1.1710-16 (3) lgK = 21.359-(-2.37)/0.059 = 126.4 K = 10126.4 = 2.5710126
8、(4) lgK = 6(1.23 - 1.695)/0.059 = -47.29 K = 10-47.29 = 5.1310-48 (5) Ag + + e- = Ag E1= 0.799 V Ag(S2O3)23- + e- = Ag + 2S2O32- E2= 0.017 V Ag + + 2S2O32- = Ag(S2O3)23-lgK = (0.799-0.017)/0.059 = 13.25 K = 10 13 = 1.7810 13 (6) lgK =(0.185-0.518)/0.059 = -5.64 K = 10-5.64 = 2.2910-6 7. C6H5OHKa = 1
9、0-9.95 Kb = 10-4.05 1H = 1/Ka1 = 10 9.95H2C2O4 Ka1 = 10-1.22 Kb2 = 10-12.78 2H = 1/(Ka1Ka2) = 10 5.41H2C2O4- Ka2 = 10-4.19 Kb1 = 10-9.81 1H = 1/Ka1 = 10 1.22CH3COOH Ka = 10-4.74 Kb = 10-9.26 1H = 10 4.74N+H3OH Ka = 10-5.96 Kb = 10-8.04 1H = 10 5.96N+H3CH2COOH Ka1 = 10-2.35 Kb2 = 10-11.65 2H = 10 11.
10、95N+H3CH2COO- Ka2 = 10-9.60 Kb1 = 10-4.40 1H = 10 2.35 酸性：H2C2O4 N+H3CH2COOH H2C2O4- CH3COOH N+H3OH N+H3CH2COO- C6H5OH8. 5 = 1012.86 ； 2 = 1016.34 ； 1 = 1018.0 ； 2 = 108.9 ； 4 = 106.46 ； 1 = 1018.80 ； 3 = 1021.0 ； 1 = 104.4 Cu-邻二氮菲EDTA柠檬酸乙酰丙酮NH3草酸酒石酸三乙醇胺9. (1) NH3 + H + = NH4+ Kt = NH4+ /(NH3H + )=
11、1/Ka = 109.26 (2) OH - + H + = H2O Kt = 1/(OH - H +)= 1/Kw = 1.01014 (3) HCO3- + H + = H2CO3 Kt = H2CO3/(HCO3- H + )= 1/Ka1 = 106.38 (4) HS - + OH - = S2-+ H2O Kt = H2OS2- /(HS - OH - )= Ka2/Kw = 10 -14.15 /10-14 = 10-0.15 (5) H3BO3 + OH - = H2BO3- + H2O Kt = H2BO3- /(H3BO3 OH - )= Ka1/Kw = 10 -9.24
12、 /10-14 = 104.76 10. HAc=HAc C = C1H H + /(1 + 1H H + ) = C(H + /Ka)/(1+H + /Ka)= CH + /(H + + Ka) = 0.1010-7.00 /(10-7.00 + 10-4.74 )= 5.510-4 Ac- =Ac - C = C /(1 + 1H H + )= CKa/(H + + Ka) = 0.110-4.74 /(10-7.00 + 10-4.74 )= 0.099511. 0 = A2- /CA = Ka1Ka2/(H + 2 + Ka1H + + Ka1Ka2)= 0.391 = HA- /CA
13、 = Ka1H + /(H + 2 + Ka1H + + Ka1Ka2)= 0.602 = H2A /CA = H + /(H + 2 + Ka1H + + Ka1Ka2)= 0.001C2O42- = CH2A0 = 0.039 mol/LHC2O4- = CH2A1 = 0.060 mol/LH2C2O4 = CH2A2 = 1.010-4 mol/L12. NH3 = Ka/(Ka + H + )= 10-9.26 /(10-9.26 + 10-9.0)= 10-0.45 = 0.355NH4+ = H + /(Ka + H + )= 10-9.0 /(10-9.26 + 10-9.0) = 10-0.19 = 0.646NH3= 0C = 0.20.355 = 0.071 mol/LNH4+ = 1C = 0.20.646 = 0.129 mol/L13. Cu(NH3)+ = 1NH3/(1 + 1NH3 + 2NH3)= 105.93-4/(1 + 105.93-4 + 1010.86-8)= 10-0.98 = 0.105Cu(NH3)2+ =2NH3/(1+1NH3+2NH3)
