第2章+分析化学中的化学平衡.docx
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第2章+分析化学中的化学平衡.docx
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第2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(forreference)
*1.离子强度及活度系数,小化只要求了解公式ai=уi[i]
(1)I=1/2∑CiZi²CNa+=0.050mol/LCNO2-=0.050mol/L
I=1/2(0.050×1²+0.050×1²)=0.05mol/L
(分子的离子强度为0,即HNO2的I=0)
(2)CNH4+=0.066mol/LCSO4²ˉ=0.033mol/L
I=1/2(0.066×1²+0.033×2²)=0.099mol/L
(3)CHCOOˉ=0.100mol/LCNa+=0.100mol/L
I=1/2(0.100×1²+0.100×1²)=0.100mol/L
(4)I=1/2(0.05×1²+0.05×1²)=0.05mol/L
2.I=1/2∑CiZi2=1/2(0.05×12+0.05×12)=0.05mol/L
aH+°=0.9aNH+°=0.3
lgγH+=-0.512×1²×[0.05½/(1+3.28×0.9×0.05½)]=-0.069
γH+=0.85
lgγNH+=-0.512×1²×[0.05½/(1+3.28×0.3×0.05½)]=-0.0938
γNH+=0.81
NH4+=H++NH3Ka(I=0)=aH+/aNH+
Ka(I=0)=Ka(I=0.05)γH+/γNH+∴Ka(I=0.05)=5.2×10-10=10-9.28
3.Zn²+4CNˉ=Zn(CN)2-4
β4(I=0)=aZn(CN)4/(aZn·aCN)=β4(I=0.1)·γZn2+·(γCN-)4
lgβ4(I=0)=lgβ4(I=0.1)+lgγZn2++4lgγCN-aZn°=0.9aCN-°=0.3
lgγZn2+=-0.512×2²×[0.1½/(1+3.28×0.6×0.1½)]=-0.3991
γZn2+=0.399
lgγCN-=-0.512×1²×[0.1½/(1+3.28×0.3×0.1½)]=-0.123
γCN-=0.753
lgβ4(I=0)=16.7+(-0.3991)+4×(-0.123)=15.81
β4(I=0)=1015.81
4.Ksp(I=0)=Ksp(I=0.1)γAg+γCl-
lgγAg+=-0.512×1²×[0.1½/(1+3.28×0.3×0.1½)]=-0.123
aAg+°=0.9γAg+=0.75;aCl-°=0.3γCl-=0.75
lgKsp(I=0)=lgKsp(I=0.1)+lgγCl-+lgAg+=-9.49-0.123-0.123=-9.74
Ksp(I=0)=10-9.74=1.8×10-10
5.pKa1~pKa6:
0.9,1.6,2.07,2.75,6.24,10.34(I=0.1,书P370)
K1H=1/Ka6=1010.34,K2H=1/Ka5=106.24,K3H=1/Ka4=102.75,
K4H=1/Ka3=102.07,K5H=1/Ka2=101.6,K6H=1/Ka1=100.9
K1H=1010.34β1=K1H=1010.34
K2H=106.24β2=K1H·K2H=1016.58
K3H=102.75β2=K1H·K2H·K3H=1019.33
K4H=102.07β2=K1H·K2H·K3H·K4H=1021.4
K5H=101.6β2=K1H···K5H=1023.0
K6H=100.9β2=K1H···K6H=1023.9
6.
(1)Br2+2e-=2BrˉE1°=1.087V
I2+6H2O+10e-=2IO3ˉ+12H+E2°=1.20V
I2+5Br2+6H2O=2IO3ˉ+10Brˉ+12H+
lgK=10(1.087-1.20)/0.059=-19.15
∴K=10-19.15=7.1×10-20
(2)lgK=n(E1°-E2°)/0.059=2[-0.91-(-0.44)]/0.059=-15.93
∴K=10-15.93=1.17×10-16
(3)lgK=2[1.359-(-2.37)]/0.059=126.4
∴K=10126.4=2.57×10126
(4)lgK=6(1.23-1.695)/0.059=-47.29
∴K=10-47.29=5.13×10-48
(5)Ag++e-=AgE1°=0.799V
Ag(S2O3)23-+e-=Ag+2S2O32-E2°=0.017V
Ag++2S2O32-=Ag(S2O3)23-
lgK=(0.799-0.017)/0.059=13.25
∴K=1013=1.78×1013
(6)lgK=(0.185-0.518)/0.059=-5.64
∴K=10-5.64=2.29×10-6
7.
C6H5OHKa=10-9.95Kb=10-4.05β1H=1/Ka1=109.95
H2C2O4Ka1=10-1.22Kb2=10-12.78β2H=1/(Ka1·Ka2)=105.41
H2C2O4-Ka2=10-4.19Kb1=10-9.81β1H=1/Ka1=101.22
CH3COOHKa=10-4.74Kb=10-9.26β1H=104.74
N+H3OHKa=10-5.96Kb=10-8.04β1H=105.96
N+H3CH2COOHKa1=10-2.35Kb2=10-11.65β2H=1011.95
N+H3CH2COO-Ka2=10-9.60Kb1=10-4.40β1H=102.35
酸性:
H2C2O4>N+H3CH2COOH>H2C2O4->CH3COOH>N+H3OH>N+H3CH2COO->C6H5OH
8.β5=1012.86;β2=1016.34;β1=1018.0;β2=108.9;
β4=106.46;β1=1018.80;β3=1021.0;β1=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+]/([NH3][H+])=1/Ka=109.26
(2)OH-+H+=H2OKt=1/([OH-][H+])=1/Kw=1.0×1014
(3)HCO3-+H+=H2CO3Kt=[H2CO3]/([HCO3-][H+])=1/Ka1=106.38
(4)HS-+OH-=S2-+H2O
Kt=[H2O][S2-]/([HS-][OH-])=Ka2/Kw=10-14.15/10-14=10-0.15
(5)H3BO3+OH-=H2BO3-+H2O
Kt=[H2BO3-]/([H3BO3][OH-])=Ka1/Kw=10-9.24/10-14=104.76
10.[HAc]=δHAc·C=C·β1H[H+]/(1+β1H[H+])
=C·([H+]/Ka)/(1+[H+]/Ka)=C·[H+]/([H+]+Ka)
=0.10×10-7.00/(10-7.00+10-4.74)=5.5×10-4
[Ac-]=δAc-·C=C/(1+β1H[H+])=C·Ka/([H+]+Ka)
=0.1×10-4.74/(10-7.00+10-4.74)=0.0995
11.δ0=[A2-]/CA=Ka1·Ka2/([H+]2+Ka1[H+]+Ka1·Ka2)=0.39
δ1=[HA-]/CA=Ka1[H+]/([H+]2+Ka1[H+]+Ka1·Ka2)=0.60
δ2=[H2A]/CA=[H+]/([H+]2+Ka1[H+]+Ka1·Ka2)=0.001
[C2O42-]=CH2A·δ0=0.039mol/L
[HC2O4-]=CH2A·δ1=0.060mol/L
[H2C2O4]=CH2A·δ2=1.0×10-4mol/L
12.δNH3=Ka/(Ka+[H+])=10-9.26/(10-9.26+10-9.0)=10-0.45=0.355
δNH4+=[H+]/(Ka+[H+])=10-9.0/(10-9.26+10-9.0)=10-0.19=0.646
[NH3]=δ0·C=0.2×0.355=0.071mol/L
[NH4+]=δ1·C=0.2×0.646=0.129mol/L
13.δCu(NH3)+=β1[NH3]/(1+β1[NH3]+β2[NH3]²)
=105.93-4/(1+105.93-4+1010.86-8)=10-0.98=0.105
δCu(NH3)2+=β2[NH3]/(1+β1[NH3]+β2[NH3]²)=102.86/102.91=10-0.05=0.891
[Cu(NH3)+]=C·δ1=0.01×0.105=1.05×10-3mol/L
[Cu(NH3)2+]=C·δ2=0.01×0.891=8.91×10-3mol/L
14.Hg-Cl的lgβ1~lgβ4:
6.74,13.22,14.07,15.07
δ1=β1[Cl]/(1+β1[Cl]+…+β4[Cl]4)
=106.74-3/(1+106.74-3+1013.22-6+1014.07-9+1015.07-12)
=103.74/(1+103.74+107.22+105.07+103.07)=10-3.48
δ2=107.22/107.223=0.993
δ3=105.07/107.22=10-2.51δ4=103.07/107.22=10-4.52
∴主要以HgCl2型体存在
[HgCl2]=C·δ2=1.0×10-3×0.993=9.93×10-4mol/L
15.Al3+CF=1.05/(42.00×0.05)=0.5mol/L
OH-F-HαF(H)pKa=3.18pH=5.0
αAl(OH)αAl(F)∴[F-]=δ0·CF=Ka·CF/(Ka+[H+])
=10-3.18×0.5/(10-3.18+10-5)
=10-0.31=0.49
αAl(F)=1+β1[F]+…+β6[F]6=1018.27
∴αAl=1018.27+100.4–1=1018.27
∴[Al3+]=CAl/αAl=0.01/1018.27=10-20.27=5.37×10-21mol/L
16.MgNH4PO4→Mg2++NH4++PO43-
[Mg2+]=5.6×10-4mol/L
[NH4+]=C·δNH4+=5.6×10-4×10-9.7/(10-9.7+1014-4.74)=10-3.82
[PO43-]=C·δPO43-=5.6×10-4×10-12.36/(10-9.7+10-12.36)=10-5.91
Ksp=[Mg2+][NH4+][PO43-]
=5.6×10-4×10-3.82×10-5.91=10-12.88=1.0×10-13
17.AgI=Ag++I-[I-]=0.002×1000/100=0.02mol/L
[Ag+]=Ksp/[I-]=10-16.03/0.02=10-14.33
αAg(NH3)=CAg+/[Ag+]=0.02/10-14.33=1012.63
又αAg(NH3)=1+β1[NH3]+β2[NH3]²
=1+103.24[NH3]+107.05[NH3]²
∴[NH3]=102.79mol/L
18.1.00/[Al(OH)3·0.01]=1.00/78×10=0.1282mol/L=[Al(OH)4-]
[Al(OH)4-]/[OH-]=10[OH-]=0.1282/10=0.01282mol/L
1.0·V=0.01282·(100+V)+0.1282·(100+V)
∴V=16.42mL
19.Ksp-PbI2=10-8.15Ksp-AgI=10-16.03
PbI2沉淀时,[I-]=(Ksp/[Pb2+])½=(10-8.15/0.20)½=10-3.73
AgI沉淀时,[I-]=Ksp/[Ag+]=(10-16.03/0.01)½=10-14.03
∴AgI先沉淀
[Ag+]=Ksp/[I-]=10-16.03/10-3.73=10-12.30=5.0×10-13
20.提示:
(1)Ag++Cl-=AgCl↓
αCl0.1+s
s=[Ag+]+[AgCl]+[AgCl2-]+[AgCl32-]+[AgCl43-]
=[Ag+](1+β1[Cl-]+β2[Cl-]2+β3[Cl-]3+β4[Cl-]4)
=Ksp(1/[Cl-]+β1+β2[Cl-]+β3[Cl-]2+β4[Cl-]3)
(2)s(0.1+s)=Ksp
21.
(1)AgI→Ag++I-αAg(NH3)=1+β1[NH3]+β2[NH3]2
αAg(NH3)=1+103.24×1+107.05×1²=107.05
[Ag+]’[I-]=Ksp’=Ksp·αAg(NH3)=10-16.03×107.05=10-8.98
S2=Ksp’=10-8.98S=10-4.49=3.24×10-5mol/L
(2)AgI→Ag++I-Ag(S2O3)的lgβ1--lgβ3:
I-S2O32-8.82,13.46,14.15
αAg(I)αAg(S2O3)
αAg(I)=1+106.58-2+1011.74-4+1013.68-6=108.01
αAg(S2O3)=1+108.82-2+1013.46-4+1014.15-6=109.48
∴αAg=αAg(I)+αAg(S2O3)–1=108.01+109.48–1=109.48
[Ag+]’[I-+0.01]=Ksp’=Ksp·αAg=10-16.03×109.48=10-6.55
S(S+0.01)=Ksp’=Ksp·αAg=10-16.03×109.48=10-6.55
S=10-4.55=2.82×10-5mol/L
22.PbSO4→Pb2++SO42-Ksp=1.6×10-8;HSO4-:
Ka=10-2
αPb(Cl)=1+β1[Cl-]+β2[Cl-]2+β3[Cl-]3
=1+101.2×1.0+100.6×1.02+101.2×1.03=101.56
αSO4(H)=1+[H+]/Ka=1+1/10-2=102
[Pb2+]’[SO42-]’=S²=Ksp’=Ksp·αPb(Cl)·αSO4(H)
S=(1.6×10-8×101.56×102)½=10-2.12=7.6×10-3mol/L
23.Ag2S→2Ag++S2-αS(H)=1+β1[H]+β2[H]2=1019.03
S2SS+0.1Ksp=10-48.7
Ksp’=[2S]2(S+0.1)=Ksp·αS(H)
∴S=(2×10-48.7×1019.03/0.4)½=10-14.64=2.31×10-15
24.
(1)CuS→Cu2++S2-αS(H)=1+β1[H]+β2[H]2
SSS=1+1014.15-7+1014.15+6.88-14=107.40
Ksp’=S2=Ksp·αS(H)
∴S=(Ksp·αS(H))½
=(10-35.2+7.40)½=10-13.90=1.3×10-14mol/L
(2)Cu2S→2Cu++S2-Ksp=10-47.7
由
(1)知,αS(H)=107.40
[Cu+]2[S2-]’=(2S)2·S=Ksp’=Ksp·αS(H)=10-47.7+7.4=10-40.3
∴S=10-13.63=2.3×10-14mol/L
(3)(Ksp)1/3和水电离相近,
故不可忽略水的电离,改用无机算法。
Tl2S+H2O→2Tl++HS-+OH–
S2SSS
S2-+H+=HS-1/Ka2;H2O→H++OH–Kw;
Tl2S→2Tl++S2-Ksp
∴(2S)2·S·S=K总=Ksp·Kw/Ka2
4S4=10-20.3×10-14.0/10-14.15=10-20.15
∴S=6.5×10-6mol/L=10-5.19
25.αY(H)=1+β1[H+]+…+β6[H+]6
=1+[H+]/Ka6+[H+]2/(Ka6·Ka5)+…+[H+]6/(Ka6…Ka1)
=1+1010.34[H+]+1016.58[H+]2+1019.33[H+]3+1021.4[H+]4+1023.0[H+]5+1023.9[H+]6
pH=4.5αY(H)=1+105.84+107.58+105.83+103.40+100.5+10-3.1=107.59
pH=5.5αY(H)=1+104.9+105.58+102.83+10-0.6+10-4.5+10-9.1=105.66
pH=4.5[Y]/CY=δ0=1/107.59=10-7.59=10-5.59%=2.57×10-6%
pH=5.5δ0=1/105.66=10-5.66=10-3.66%=2.19×10-4%
26.
(1)Co2++Y4-=CoY[NH3]=10-1.30
αCo(NH3)=1+100.81+101.14+100.89+100.35+10-0.77+10-2.69=31.43=101.50
lgαY(H)=2.27lgK=16.31
∴K’=1016.31/(31.43×102.27)=1012.54=3.5×1012
(2)lgαY(H)=0.45
lgK’=lgK-lgαY(H)-lgαCo(NH3)=16.31-0.45-1.50=14.36
(3)[NH3]=10-0.30
αCo(NH3)=1+101.81+103.14+103.89+104.35+104.23+103.31=104.71
lgK’=16.31–2.27–4.71=9.33
(4)lgK’=16.31–0.45–4.71=11.15
27.
(1)lgKCdY=16.46
pH=5.0lgαY(H)=6.45
αCd(I)=1+β1[I-]+…+β4[I-]4
=1+102.10-1+103.43-2+104.49-3+105.41-4=1+101.10+101.43+101.49+101.41=101.99
lgKCdY’=lgKCdY-lgαY(H)-lgαCd(I)=16.46-6.45-1.99=8.02
KCdY’=108.02
(2)pH=7.0lgαY(H)=3.32
lgKCdY’=16.46-3.32-1.99=11.15KCdY’=1011.15
(3)αCd(I)=1+102.10-2+103.43-4+104.49-6+105.41-8
=1+100.1+10-0.57+10-1.51+10-2.59=100.41
lgKCdY’=16.46–6.45–0.41=9.60KCdY’=109.60
(4)lgKCdY’=16.46–3.32–0.41=12.
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