分析课后习题解答4章

分析化学课后习题解答-1-第4章酸碱滴定法1.根据教材上P56，公式4-4a、4-4b、4-4c（1）当pH=4.88时，[H+]=10-4.88mol/L，145.0710.0145.010101010)10()10(2157.519.419.488.4288.4288.40当pH=5.00时，[H+]=10-5.00mol/L，带入公式可求得δ0=0.109，δ1=0.702，δ2=0.189（2）当pH=4.88，c=0.01mol/L时[H2A]=0.01×0.145=1.45×10-3(mol/L)[HA-]=0.01×0.710=7.10×10-3(mol/L)[A-]=0.01×0.189=1.89×10-3(mol/L)2．pH=6.37[H+]=4.27×10-7mol/LpH=4.43[H+]=3.72×10-5mol/LpOH=3.28，pH=14-3.28=10.72[H+]=1.91×10-11mol/LpOH=9.86，pH=14-9.86=4.14[H+]=7.24×10-5mol/L[H+]=0.139mol/LpH=0.86[H+]=2.07×10-5mol/LpH=4.683．pKa=-lgKaKb=Kw/KaH3PO4Ka1=7.52×10-3;Ka2=6.23×10-8;Ka3=2.2×10-13Kb1=4.55×10-2;Kb2=1.61×10-7;Kb3=1.33×10-12H2C2O4Ka1=5.6×10-2;Ka2=5.42×10-5Kb1=1.85×10-10;Kb2=1.8×10-13苯甲酸Ka=6.28×10-5;Kb=1.59×10-10NH4+Ka=5.70×10-10;Kb=1.75×10-54．（1）[H+]=(0.1+0.01)/2=0.055mol/L，pH=1.26（2）[H+]=(0.1+0.001)/2=0.0505mol/L，pH=1.30（3）[H+]=(0.1+0.00001)/2=0.050005mol/L，pH=1.30（4）pH=1.00，[H+]=0.1mol/L；pH=14.00，溶液为碱，pOH=14-14.0=0，[OH-]=1mol/L，酸碱反应后溶液呈碱性，[OH-]=(1-0.1)/2=0.45，pOH=0.35，pH=14-0.35=13.65（5）pH=5.00，[H+]=1.00×10-5mol/L；pH=9.00，溶液为碱，pOH=14-9.00=5.00，[OH-]=1.00×10-5mol/L，[H+]=[OH-]，酸碱反应后溶液呈中性，pH=7.005．（1）NaH2PO4为两性物质，Ca=0.1mol/L，Ka1=7.52×10-3;Ka2=6.23×10-8。67.4)21.712.2(21)(2120/,202112aaaawaapKpKpHKCKKC（2）K2HPO4为两性物质，Ca=0.05mol/L，Ka2=6.23×10-8;Ka3=2.2×10-13。分析化学课后习题解答-2-131431314103820.052.2101.110200.052.21010[]0.1610(/)0.05116.23109.80aawaawaaCKKCKKHmolLCKpH6．（1）稀释前pH=-lg0.10=1.00；稀释后pH=-lg0.010=2.00（2）稀释前pOH=-lg0.10=1.00，pH=13.00；稀释后pOH=-lg0.010=2.00，pH=12.00（3）HAc为弱酸，稀释前Ca=0.10mol/L；Ka=1.75×10-588.2)/(1032.11075.110.0][500/;2035pHLmolKCHKCKKCaaaawaa稀释后Ca=0.010mol/L38.3)/(1018.41075.1010.0][500/;2045pHLmolKCHKCKKCaaaawaa（4）溶液组成为缓冲溶液，pKa=9.25，则稀释前25.91.01.0lg25.9][][lg43NHNHpKpHa稀释后25.901.001.0lg25.9][][lg43NHNHpKpHa7．（1）溶液组成为缓冲溶液，pKa=4.76，76.405.005.0lg76.4][][lgHAcAcpKpHa（2）溶液组成为弱碱，Cb=0.1mol/L，Ka=1.75×10-5，88.812.500.1400.1412.5)/(1056.71.01075.110][6514pOHpHpOHLmolCKKCKOHbawbb（3）溶液组成为两性物质，Ca=0.05mol/L，Ka1=1.12×10-3;Ka2=3.90×10-6，18.4)41.595.2(21)(2120/,202112aaaawaapKpKpHKCKKC8．（1）先求出弱酸的浓度21.41016.608518.0)10(][][)/(08518.05264.2210001001.12204.1aaaaaapKCHKKCHLmolC分析化学课后习题解答-3-（2）04.896.51496.5248.321010101010.0)10(][][awbaaaaKKKCHKKCH9．NaOH滴定氯乙酸，化学计量点时pH值应大于7，现滴定至pH＝4.4，说明还没到化学计量点，溶液中组成为氯乙酸和氯乙酸钠的缓冲溶液。假设未被中和的氯乙酸为x%，则根据缓冲溶液计算公式：56.2,4.41lg82.2lgxxxCCpKapHab10．NaOH滴定HCOOH至化学计量点，溶液组成为HCOONa。又因为CbKb≥20Kw，Cb/Kb≥500，则22.878.5)/(1068.11078.11005.0][6414pHpOHLmolKCOHbb可以选择酚酞作指示剂。11．计量点前0.1%，溶液是剩余的KOH，强碱。0.90.5)/(100.198.192002.002.0][5pHpOHLmolOH计量点时，为KCl+H2O，则pH=7.0计量点后0.1%，溶液是多余的HCl，强酸。0.5)/(100.102.202002.002.0][5pHLmolH滴定突跃为pH=5.0~9.0，选用酚酞、甲基红均可。12．（1）根据强碱滴定弱酸误差计算公式%0.1%100)10101004530.010(%)/(04530.070.2000.2570.201000.0.10][,10][%100)][][][(%100)][(%2.420.620.680.780.720.6TELmolCOHHKHHCOHCOHTEspaspHAsp则已知（2）化学计量点生成共轭碱，43.857.5141410101004530.0][57.52.414pOHpHKKCKCOHawbbb（3）根据（1）已知误差为-1.0%，设苯甲酸的起始浓度为c，则)/(08363.000.25%1)1000.070.20(1000.070.20Lmolcc分析化学课后习题解答-4-13．（1）HZ+NaOH=NaZ+H2O1.3371020.410900.0250.13NaOHNaOHHZNaOHHZHZHZHZHZNaOHVCmnmMMmnn（2）滴定剂加到8.24mL时，溶液的pH=4.30。此时溶液应是HZ+NaZ组成的缓冲溶液。51026.190.405093.001273.0lg30.4)/(01273.024.8500900.024.8)/(05093.024.8500900.0)24.820.41(lglgaabaabaabaKpKLmolCLmolCCCpHpKCCpKpH（3）化学计量点时，生成NaZ。76.824.5141410101004066.0][)/(04066.020.415020.410900.024.590.414pOHpHKKCKCOHLmolCawbbbSPb14．先求出与HCl反应的NaOH体积数，)(97.23030.000.42250.0mLCVCVVCVCnnNaOHHClHClNaOHHClHClNaOHNaOHHClNaOHH2A+2NaOH=Na2A+2H2O7.11810)97.200.38(3030.06300.0223212122222NaOHNaOHAHAHNaOHNaOHAHAHNaOHAHVCmMVCMmnn15．Na2B4O7+2HCl+5H2O=4H3BO3+2NaCl则4B∽Na2B4O7∽2HClHClOBNaBHClOHOBNaOBNannnnnn24;74227427422110（1）%44.93%100000.137.3811050.242000.0%100%10032110211010102742274227422742SMnSMnwOHOBNaHClOHOBNaOHOBNaOHOBNa分析化学课后习题解答-5-（2）%34.49%100000.137.2011050.242000.0%100%10032121742742742742SMnSMnwOBNaHClOBNaOBNaOBNa（3）%58.10%100000.18.101050.242000.02%1002%1003SMnSMnwBHClBBB16．NaAc稀释后浓度)/(043.070/)301.0(LmolCb中和所需要的HCl体积数)(301.0/)301.0(mLVHCl化学计量点时，生成HAc14.31025.71075.103.0][)/(03.03070043.07045pHKCHLmolCaaHAc超过化学计量点08.31024.81.01001.01.01001025.7][44pHHΔpH=3.14-3.08=0.06＜0.3，不适合滴定（见书中P72）17．弱酸被NaOH滴定一半时，溶液是由HX和NaX组成的缓冲溶液，而且[HX]=[X-]。00.5][][lg00.5lglgHXXCCpHpKCCpKpHabaaba弱酸滴定至化学计量点时，溶液由NaX组成。)/(05012.0101010][15.585.8141485.815.500.514LmolCCKKCKCOHpHpOHpHbbawbbb即[Na+]=0.05012mol/L。则mLVVVNaOHNaOHNaOH29.60)00.60(05012.01.0由于HX是一元弱酸，则nHX=nNaOH，%24.50%100900.000.751029.601000.0%100%100%100%100%3SMVCSMnSMnSmwHXNaOHNaOHHXNaOHHXHXHXHX18．过量的酸体积为)(46.5975.060.5mLV过量，则实际参与反应的酸体积为分析化学课后习题解答-6-20.00-5.46=14.54mL。CaCO3+2HCl=CaCl2+H2O+CO2↑%35.13%1002815.0441054.141175.0%100%37.30%1002815.009.1001054.