9武汉大学(第五版)定量分析化学课后习题答案

第9章重量分析法1．解：S0=[CaSO4]=β[Ca2+][SO42-]=β×Ksp=200×9.1×10-6=1.82×10-3mol/L非离解形式Ca2+的百分数为340[CaSO]1.821037.6%SSKsp水＋3．解：（1）2410.1mol/,0.10,2BaSOLNaClIciZi在中224224()5,()4,0.38,0.355BaSOaBaaSO＋＋查表得22402250104[][]2.8610(1.110)BaSOKspsBaSOKspKsp＋＋（2）14210.10.302BaSOmolLBaClICiZi在中，224()5,()4aBaaSO＋查表得2220.30lg0.512100.5909,0.2610.32850.30BaBa＋＋22420.30lg0.512100.6526,0.2210.32840.30BaSO＋2240102241.110[][](0.10)0.260.22BaSOKspBaSOss＋＋811.9210smolL5．解：210.10,2IciZi224()5,()4aBaaSO＋查表得，2240.38,0.355BaSO＋24221.0100.1250.071.010SO2224440101.110(0.01)BaSOBaSOSOKspaaSs＋716.4410smolL7．解：10,1.810AgClAgClKsp13,510AgBrAgBrKsp在同一溶液中，Ag只有一种浓度,AgClAgBrKspKspAgCl的溶解度大得多AgAgCl浓度由决定1051[]1.8101.3410AgClsAgKspmolL9．解：3CaCO已知沉淀在水中的主要离解平衡为：2323CaCOHOCaHCOOH＋233[][][]KspCaHCOOHs＋2233232[][][][][][][]COHKspKwKspCaHCOOHCOHKa＋914331122.910105.610KspKwsKa518.0210smolL51[]8.0210OHsmolL4.1,9.9pOHpH11．解：48.993315.142246.1382.8)O(SAg101002.31010101001.0101]Ag[Ag322＝）（＋）（＋－－cKSP＝9.3×10－17＝[Ag+][I－]＝010.01048.9ss=2.81×10－51molL13．解：混合后，23110001[]0.14.910150()BamolLMBa231450[]0.013.310150SOmolL剩余的2Ba=33(4.9103.3)150137.333.3mg100mL纯水洗涤时损失的4BaSO:251[]1.0510sBaKspmolL51.0510100233.40.245mg为＝100mL0.0101molL24HSO洗涤时121240.010[H]1.4110molLHSOmolL的24102224221.110[][](0.01)0.011.4110SOKaKspBaSOsssKa81844s=2.6510,2.6510100233.46.210molLBaSOmgmg－－损失数为：16．解：（1）4240.005[][]0.005[]HHHNHHFNHFHFHFHF41222228[][][][]5.84100.001[][](20.005)0.0005(0.01)2[]0.0005(0.010.56)1.5710FAgClHFKaHFHmolLKaCaFHKaKsp有沉淀生成(2)33.247.026AgNH1100.510(0.5)2.810[]AgcAg（）＝960.05[][]0.58.9102.810AgClAgClKsp有沉淀生成(3)0.059.26lg8.260.5pH＝261226214()5.74,[]1.8210[][]0.005(1.8210)1.6610MgOHpOHOHmolLMgOHKsp无沉淀生成19．解：22342234123423412342-71[][][()][()][()][]{1[][][][]}{1[][][][]}[]2.510sZnZnOHZnOHZnOHZnOHZnOHOHOHOHKspOHOHOHOHOHmolL－2－＋＋主要状态可由数值得22．解：(1)234()0.23512()MCrOFMPbCrO(2)422272(7)2.215()MMgSOHOFMMgPO(3)3424343[()]0.082662[(NH)12]MCaPOFMPOMoO(4)254343()0.0378322[(NH)12]MPOFMMPOMoO25．解：设CaC2O4为x,MgC2O4y=0.6240-x332424()()(0.6240)0.4830()()MCaCOMMgCOxxMCaCOMMgCO240.4773,%76.49%xgCaCO240.1467,%23.51%ygMgCO28．解：()107.86835.4530.58050.58051.4236()35.453MAgClMNaClNa22.988865Na解得31．解：设为xyFeO55.85160.543455.850.3801xyx0.38010.00680655.85x则0.0102030.0068062yx23FeO为34．解：AgCl:3.1431050＝0.035(mol·L-1)cNH3＝3/2＝1.5(mol·L-1)[Ag+]原＝2035.0＝0.0175(mol·L-1)[I-]原＝205.0＝0.025(mol·L-1)设混合后[Ag+]＝x/mol·L-1Ag++2NH3Ag(NH3)2+x1.5-2×(0.0175-x)0.0175-x≈1.5≈0.01752)5.1(0175.0x＝ß2＝107.40x＝3.1×10-10[Ag+][I-]＝3.1××0.025＝7.8×10-12AgIKsp有AgI沉淀生成。Copyright©WuhanUniversity.ALLRightsResverved.武汉大学版权所有