分析化學(xué)第五版課后習(xí)題答案

CaCO32HClCaCO32HCl分析化學(xué)第五版課后習(xí)題答案第1章分析化學(xué)概論2.有0.0982mol/L的H2SO4溶液480mL,現(xiàn)欲使其濃度增至0.1000mol/L。問(wèn)應(yīng)加入0.5000山014地504的溶液多少毫升？口解：c1V1c2V2c(V1V2)口0.0982mol/L0.480L0.5000mol/LV20.1000mol/L(0.480LV2)，V22.16mL4.要求在滴定時(shí)消耗0.2mol/LNaOH溶液25~30mL。問(wèn)應(yīng)稱取基準(zhǔn)試劑鄰苯二甲酸氫鉀(KHC8H4O4)多少克？如果改用解：口H2C2O42H2O做基準(zhǔn)物質(zhì)，又應(yīng)稱取多少克？nNaOH:nKHC8H4O41:1m1n1McV1M0.2mol/L0.025L204.22g/mol1.0gm2n2McV2M0.2mol/L0.030L204.22g/mol1.2g應(yīng)稱取鄰苯二甲酸氫鉀1.0?1.2gnNaOH:nH2C2O42H2O2:1m1n1McV1M20.2mol/L0.025L126.07g/mol0.3g2m2n2McV2M20.2mol/L0.030L126.07g/mol0.4g220.3~0.4g應(yīng)稱取2246.含S有機(jī)試樣0.471g,在氧氣中燃燒，使S氧化為SO2,用預(yù)中和過(guò)的H2O2將SO2吸收，全部轉(zhuǎn)化為H2SO4，以0.108mol/LKOH標(biāo)準(zhǔn)溶液滴定至化學(xué)計(jì)量點(diǎn)，消耗28.2mL。求試樣中S的質(zhì)量分?jǐn)?shù)。口HCO2HO解：SSO2H2SO42KOHwnM100%m00.108mol/L0.0282L32.066g/mol100%0.471g10.3%8.0.2500g不純CaCO3試樣中不含干擾測(cè)定的組分。加入25.00mL0.2600mol/LHCl溶解，煮沸除去CO2,用0.2450mol/LNaOH溶液反滴定過(guò)量酸，消耗6.50mL,計(jì)算試樣中CaCO3的質(zhì)量分?jǐn)?shù)。解：口，NaOHHCl(cVcV)MnMw100%100%m0m0(0.2600mol/L0.025L0.2450mol/L0.0065L)100.09g/mol100%0.2500g98.24%10.不純Sb2s30.2513g,將其置于氧氣流中灼燒，產(chǎn)生的502通入FeCl3溶液中，使Fe3+還原至Fe2+,然后用0.02000mol/LKMnO4標(biāo)準(zhǔn)溶液滴定Fe2+，消耗溶液31.80mL。計(jì)算試樣中Sb2s3的質(zhì)量分?jǐn)?shù)。若以Sb計(jì)，質(zhì)量分?jǐn)?shù)又為多少？解：口Sb2S32Sb3SO26Fe26KMnO4555nSb2S3cV0.0200mol/L0.03180L0.00053mol66nSb2nSb2S320.00053mol0.00106mol0.00053mol339.68g/mol100%71.64%230.2513gmH2C2O4MH2C2O4mH2C2O4MH2C2O4mH2C2O4MH2C2O4mH2C2O4MH2C2O40.00106mol121.76g/molwSb100%51.36%0.2513gwSbS12.用純A2O3標(biāo)定KMnO4溶液的濃度。若0.2112gA2O3在酸性溶液中恰好與36.42山1創(chuàng)口04反應(yīng)。求該KMnO4溶液的濃度?？?5A2O310AO34MnO4解：故4mcVKMnO410005McKMnO440.211210000.02345(mol/L)36.42197.814.H2c2O4作為還原劑?？膳cKMnO4反應(yīng)如下：口5H2C2O4+2MnO-4+6H+=10CO2+2Mn2++8H2O其兩個(gè)質(zhì)子也可被NaOH標(biāo)準(zhǔn)溶液滴定。分別計(jì)算0.100mol?L-1NaOH和0.100mol-LTKMnO4溶液與500mgH2C2O4完全反應(yīng)所消耗的體積(mL)?？趎H2C2O45005.553103(mol)90.035解：H2C2O42NaOHnNaOH2nH2C2O425.55310311.106103(mol)VNaOHnNaOH11.1061030.111(L)111(mL)cNaOH0.1002H2C2O4KMnO45nKMnO422nH2C2O45.5531032.221103(mol)55cKMnO416.含K2Cr2O75.442g^L-1的標(biāo)準(zhǔn)溶液。求其濃度以及對(duì)于Fe3O4(M=231.54g?molT)的滴定度(mg/mL)?？赩KMnO4nKMnO42.2211030.0222(L)22.2(mL)0.100cK2Cr2O7nK2Cr2O7VK2Cr2O7解：5.4420.01850(mol/L)TFe3O4/K2Cr2O7cK2Cr2O7MFe3O420.01850231.5428.567(mg/mL)18.按國(guó)家標(biāo)準(zhǔn)規(guī)定，化學(xué)試劑FeSO4-7H2O(M=278.04g-mol-1)的含量：99.50?100.5%為一級(jí)(G.R)；99.00%?100.5%為二級(jí)小飛)；98.00%?101.0%為三級(jí)(C.P)。現(xiàn)以KMnO4法測(cè)定，稱取試樣1.012g,在酸性介質(zhì)中用0.02034mol?L-1KMnO4溶液滴定，至終點(diǎn)時(shí)消耗35.70mL。計(jì)算此產(chǎn)品中FeSO4?7H2O的質(zhì)量分?jǐn)?shù)，并判斷此產(chǎn)品符合哪一級(jí)化學(xué)試劑標(biāo)準(zhǔn)。5Fe2MnO4解：35.70nFeSO47H2O5nMnO50.020343.631103(mol)41000FeSO7HO42mFeSO47H2OnFeSO47H2OMFeSO47H2Om3.631103278.0499.76%1.012故為一級(jí)化學(xué)試劑。20.CN可用EDTA間接滴定法測(cè)定。已知一定量過(guò)量的Ni與CN反應(yīng)生成Ni(CN)溶液滴定，Ni(CN)Ni(CN)口2424-2+24，過(guò)量的Ni2+以EDTA標(biāo)準(zhǔn)口并不發(fā)生反應(yīng)。取12.7mL含CN-的試液，加入25.00mL含過(guò)量Ni2+的標(biāo)準(zhǔn)溶液以形成，過(guò)量的Ni2+需與10.1mL0.0130mol-LTEDTA完全反應(yīng)。已知39.3mL0.0130mol?LTEDTA與口上述Ni2+標(biāo)準(zhǔn)溶液30.0mL完全反應(yīng)。計(jì)算含CN-試液中CN-的物質(zhì)的量濃度。2解：NiEDTA口cNi2nNi2VNi2nEDTAVNi239.30.01300.01703(mol/L)30.0100025.004.2575104(mol)1000ncNi2VNi20.01703Ni210.1nEDTAcEDTAVEDTA0.01301.313104(mol)1000nCN4(nnEDTA)1.1778103(mol)Ni2cCNnCNVCN1.17781030.0927(mol/L)12.71000第2章分析試樣的采集與制備1．某種物料，如各個(gè)采樣單元間標(biāo)準(zhǔn)偏差的估計(jì)值為0.61％，允許的誤差為0.48％，測(cè)定8次，置信水平選定為90%，則采樣單元數(shù)應(yīng)為多少解：f=7P=90%查表可知t=1.90口t1.900.612n()2()5.86E0.482．某物料取得8份試樣，經(jīng)分別處理后測(cè)得其中硫酸鈣量分別為81.65％、81.48％、81.34％、81.40％、80.98％、81.08％、81.17％、81.24％，求各個(gè)采樣單元間的標(biāo)準(zhǔn)偏差．如果允許的誤差為0.20％，置信水平選定為95％，則在分析同樣的物料時(shí)，應(yīng)選取多少個(gè)采樣單元？解：f=7P=85%查表可知t=2.36t2.360.132n()2()2.43E0.20.已知鉛鋅礦的K值為0.1,若礦石的最大顆粒直徑為30mm,問(wèn)最少應(yīng)采取試樣多少千克才有代表性22解：QKd0.13090(kg)口.采取錳礦試樣15kg,經(jīng)粉碎后礦石的最大顆粒直徑為2mm,設(shè)K值為0.3,問(wèn)可縮分至多少克？22QKd0.321.2(kg)解：口111111111111Q()n1.2(kg)15()n1.2(kg)ln15nlnln1.2222設(shè)縮分n次，則，，口15()31.875(kg)2解得n3.6,所以n二以mq第3章分析化學(xué)中的誤差與數(shù)據(jù)處理1．根據(jù)有效數(shù)字運(yùn)算規(guī)則，計(jì)算下列算式：(1)19.469+1.537-0.0386+2.54(2)3.60.032320.592.1234545.00(24.001.32)0.12451.00001000(3)(4)pH=0.06,求[H+]=解：a.原式=19.47+1.54-0.04+2.54=23.51b.原式=3.6某0.032某21某2.1=5.145.0022.680.1245=0.12711.0001000c.原式=d.[H+]=10-0.06=0.87(mol/L)口3．設(shè)某痕量組分按下式計(jì)算分析結(jié)果：ACm，A為測(cè)量值，C為空白值，m為試樣質(zhì)量。已知口A=C=0.1，m=0.001，A=8.0，C=1.0，m=1.0，求某。口22222(2AC)某mACm012224.0910422222(AC)m(AC)m(8.0L0)L0解：某口且某8.01.07.00.141.0故某5.反復(fù)稱量一個(gè)質(zhì)量為1.0000g的物體，若標(biāo)準(zhǔn)偏差為0.4mg,那么測(cè)得值為1.00001.0008g的概率為多口少？解：由0.4mg1.0000g1.00001.00001.0008故有0.0004u1.00000.0004，即0u2，查表得P=47.73%口7．要使在置信度為95％時(shí)平均值的置信區(qū)間不超過(guò)±，問(wèn)至少應(yīng)平行測(cè)定幾次某t某t解：口查表，得：f5時(shí)，t2.57,t1.0491f6時(shí),t2.45,t0.9261故至少應(yīng)平行測(cè)定5次242424249.測(cè)定黃鐵礦中硫的質(zhì)量分?jǐn)?shù)，六次測(cè)定結(jié)果分別為30.48%,30.42%,30.59%,30.51%,30.56%,30.49%計(jì)算置信水平95%時(shí)總體平均值的置信區(qū)間。某16某30.42%30.59%30.51%30.56%30.49%解:ni30.48%30.51%口i16t0.05,52.57,某t,30.51%2.57f30.51%0.06%置信度為95%時(shí):口11.下列兩組實(shí)驗(yàn)數(shù)據(jù)的精密度有無(wú)顯著性差異（置信度90％）A：9.56，9.49，9.62，9.51，9.58，9.63，B：9.33，9.51，9.49，9.51，9.56，9.40解：a.n6口某某i9.575.71%故232.610416某b.n某i9.47i1，8.51%故72.4102b72.4104F22.2214F5.0532.610a所以查表得表>2.22113．用兩種不同分析方法對(duì)礦石中鐵的質(zhì)量分?jǐn)?shù)進(jìn)行分析，得到兩組數(shù)據(jù)如下：n口方法115.34％0.10％11，方法215.43％0.12％11a.置信度為90%時(shí)，兩組數(shù)據(jù)的標(biāo)準(zhǔn)偏差是否存在顯著性差異口b.在置信度分別為90%,95%及99%時(shí)，兩組分析結(jié)果的平均值是否存在顯著性差異2222222121解：(a)=0.0010,=0.0012F==1.44<F表=2.97,所以兩組數(shù)據(jù)的標(biāo)準(zhǔn)偏差無(wú)顯著性差異。d(b)由口22i=10得,d2i=0.01,22=0.012=0.063t0.10,20t=查表得：當(dāng)置信度為90%時(shí)，=1.72>0.063，查表得：當(dāng)置信度為95%時(shí)，t0.05,20=2.09>0.063t查表得：當(dāng)置信度為99%時(shí)，0.01,20=2.84>0.063，所以兩組分析結(jié)果的平均值不存在顯著性差異。Cl結(jié)果如下：A瓶60.52%,60.41%,60.43%,60.45%,B瓶60.15%,60.15%,60.05％，60.08％問(wèn)置信度為90%時(shí),兩瓶試劑含Cl-1的質(zhì)量分?jǐn)?shù)是否有顯著性差異？解：用F檢驗(yàn)法：口某iA==60.45%，2Adi2某i口==2.310，B=-3=60.11%，2Bdi2==2.610-32SB2SAF==1.13,查表得F表=9.28>1.13,因此沒(méi)有差異?？谟胻檢驗(yàn)法：口10-4=9.6,而查表得t表=1.94<9.6,所以存在顯著性差異。所以口17.為提高光度法測(cè)定微量Pd的靈敏度，改用一種新的顯色劑。設(shè)同一溶液，用原顯色劑及新顯色劑各測(cè)定某iA==0.127，，2Adi2==1.310-5某iB=2Bdi2==1.910-52B2AF==1.46,查表得F表=9.28>1.46,因此沒(méi)有顯著性差異?？谟胻檢驗(yàn)法：口10-3=2.8所以而查表得t表=2.45<2.8所以存在顯著性差異21.某熒光物質(zhì)的含量()及其熒光相對(duì)強(qiáng)度(y)的關(guān)系如下：口含量/g0.02.04.06.08.010.012.0熒光相對(duì)強(qiáng)度y2.15.09.012.617.321.024.7a.列出一元線性回歸方程口b.求出相關(guān)系數(shù)并評(píng)價(jià)y與某間的相關(guān)關(guān)系。解：由題意可得，某=6.0，y=13.1，（某某）（yy）口iii17=216.2，（某某）ii17i172=112.0，ii（yy）ii1=418.28，（某某）（yy）口所以bq2（某某）ii17216.2=112.0=1.93，ayb某=13.1-1.936.0=1.52所以一元回歸方程為：y=1.52+1.93某口r（b）因?yàn)榭诒容^接近于1,所以y與某的線性關(guān)系很好?？诘?章分析質(zhì)量的控制與保證[16]解：平均值某=0.256mg/L標(biāo)準(zhǔn)偏差=0.020mg/L口標(biāo)準(zhǔn)物質(zhì)標(biāo)準(zhǔn)值U=0.250mg/L控制限某3=（0.2560.060）mg/L警戒限某2二（0.2560.040）mg/L繪制分析數(shù)據(jù)的質(zhì)控圖如下圖?？?還原糖分析的質(zhì)量控制圖第5章酸堿平衡和酸堿滴定法1．寫出下列溶液的質(zhì)子條件式。a.c1mol?LTNH3+c2mol?LTNH4Cl；口c.c1mol^L-l)H3PO4+c2mol^L-lHCOOH；口解：a.對(duì)于共軛體系，由于構(gòu)成了緩沖溶液，所以可以將其視為由強(qiáng)酸(HC1和弱堿(NH3)反應(yīng)而來(lái)，所以參考水準(zhǔn)選為HCl,NH3和H2O口質(zhì)子條件式為：[H+]+[NH4+]=[Cl-]+[OH-]或[H+]+[NH4+]=c2+[OH-]]c.直接取參考水平：H3PO4,HCOOH,H2O口質(zhì)子條件式：[H+]=[H2P04-]+2[HP042-]+3[P043-]+[HC00-]+[0H-]3.計(jì)算下列各溶液的pH?？?la.0.050mol?LNaAc；口c.0.10mol?LNH4CN；e.0.050mol?L-l氨基乙酸；g.0.010mol-L-lH2O2液；口i.0.060mol?L-lHCI和0.050mol?L-l氯乙酸鈉(ClCH2COONa)混合溶液。解：a.對(duì)于醋酸而言，Kb=Kw/Ka=5.610-10應(yīng)為cKb=5.610T0510-2=2.810-n〉10Kw口c/Kb>100故使用最簡(jiǎn)式；口10-6[OH--lpH=14-pOH=8.72+’-10c.NH4Ka=5.610HCNKa=6.2.10T0cKa’>10Kwc>10Ka由近似公式可以得到：[H10-10+pH=10-0.77=9.23e.氨基乙酸一端羧基顯酸性，一端氨基顯堿性，Ka1=4.510-3,Ka2=2.510-10c/Ka2>100且c>10Ka1所以[H10-6+pH=6-0.03=5.97g.對(duì)于雙氧水而言，Ka=1.810-12cKa<10Kwc/Ka>100所以可以計(jì)算氫離子濃度口[H=1.6710-7+pH=7—0.22=6.78.由于ClCH2COONa+HCl=ClCH2COOH+NaCl所以原溶液可以看成0.050mol/L的ClCH2COOH和0.010mo/LHCl的混合溶液設(shè)有某mol/L的ClCH2COOH發(fā)生離解，則ClCH2COOHClCH2COO-+H+0.05-某某0.01+某Ka5.0106Ka5.0106(0.01某)某所以有0.05某=Ka=1.410-3解得某=4.410-3mol/L口那么[H+]=0.0144mol/LpH=-log[H+]=1.845.某混合溶液含有0.10mol^L-lHCl,2.0某10-4mol-LTNaHSO4和2.0某10-6moLLTHAc。口a.計(jì)算此混合溶液的pH。口氏加入等體積0.10mol?L-lNaOH溶液后，溶液的pH。口解：a.HSO4-Ka2=1.0某10-2HAcKa=1.8某10-5均為弱酸，且濃度遠(yuǎn)低于HCl的濃度，所以此體系中的HSO4-和HAc在計(jì)算pH值時(shí)刻忽略。故pH=1.00?！跞占尤氲润w積0.1mol/LNaOH溶液，HCl被中和，體系變?yōu)橛?4-和HAc的混酸體系，口H忽略KW及KHA[HA],口[H]=+2KHSO4(CHSO4--[H+])解得[H+]=9.90某10-5故pH=4.00口7.已知Cr3+的一級(jí)水解反應(yīng)常數(shù)為10-3.8,若只考慮一級(jí)水解，則0.010mol?L-lCr(ClO4)3的pH為多少此口2+時(shí)溶液中Cr(OH)的分布分?jǐn)?shù)是多大解：1)口Kac103.8102105.810KWc101.8100KaH1.182103mol/L故pH=2.93Cr(OH)22)9.今用某弱酸HB及其鹽配制緩沖溶液，其中HB的濃度為0.25mol-L-l。于100mL該緩沖溶液中加入200mgNaOH(忽略溶液體積的變化)，所得溶液的pH為5.60。問(wèn)原來(lái)所配制的緩沖溶液的pH為多少(已知HB的口Ka22.93KaH1010）解：CNaOH2000.0540100（mol/L）已知pKa=5.30,pH=5.60口設(shè)原緩沖溶液中鹽的濃度為某mol/L,故得某=0.35則原緩沖溶液pH=5.30lg0.355.440.25-l5.605.30lg0.05某0.25某11.配制氨基乙酸總濃度為0.10mol?L的緩沖溶液(pH=2.0)100mL,需氨基乙酸多少克還需加多少毫升1mol?L-l酸或堿，所得溶液的緩沖容量為多大解：設(shè)酸以HA表示，pKa1=2.35pKa2=9.601)需HA質(zhì)量m=0.10某100某75.0某10=0.75(g)□2)因?yàn)榘被宜峒百|(zhì)子化氨基乙酸構(gòu)成緩沖溶液，設(shè)pH=2時(shí)，質(zhì)子化氨基乙酸濃度為某mol/L，則口cHAHpHpKa1lgcHAH32.002.35lg0.1某H某H即，解得某=0.079生成0.079mol/L的質(zhì)子化氨基乙酸，需加酸為0.079某100=7.9m113.計(jì)算下列標(biāo)準(zhǔn)緩沖溶液的pH(考慮離子強(qiáng)度的影響)，并與標(biāo)準(zhǔn)值相比較。a.飽和酒石酸氫鉀(0.0340mol?L—l)；c.0.0100mol?L-l硼砂。解：a.pKa1=3.04,pKa2=4.37I=2(10.03410.034)0.034(mol/L)，查表得，aH900aHB400，aB2500lg0.5121故H0.06,得H0.871同理可得HB0.84，B20.51c又K20K20口a2cWKa1H最簡(jiǎn)式100Kac10KwKa100Kac10KwKa100Kac10KwKa100Kac10KwKa0.020.8690.020.869HHH2.76104pH=3.56c.c=0.0100mol/L，pKa1=4，pKa2=9I12(0.020010.02001)0.02BO2475H2O2H3BO32H2BO3查表aH2BO3400lgH2BO30.5121故H2BO30.869H3BO3HH2BO3K=5.8某10-10a5.810105.81010H6.671010H2BO3cH2BO3故pH=9.1816.解：已知01molL1一元弱酸HB的PH3.0,問(wèn)其等濃度的共軛堿NaB的pH為多少？(已知：Kac10Kw，且c/Ka100)解：據(jù)題意：[H]3Ka(102c)/510KbKw109c/Ka0.050.050.050.05OH105pH14.05.09.01119.用0.1molLNaOH滴定0.1molLHAc至pH8.00。計(jì)算終點(diǎn)誤差?？诮猓篠P時(shí)口cNaOH0.05molL1KbKw5.61010Kac500Kbc20KwKb[OH]5.29106pHp1460.728.72pHpHeppHp8.008.720.720.720.72TE%100251000.05#4.解：用氨水調(diào)解時(shí)：M(NH)11(0.010)2(0.010)212231[NH3]102.0102.010.0083M(NH)12232[NH3]2105.0104.020.083M(NH)1223107.0106.01010.0108.030.08340.83122122故主要存在形式是2M(NH3)4，其濃度為0.100.830.083molL口用氨水和NaOH調(diào)解時(shí)：口MM(OH)M(NH)111040.11080.01120210111040.11080.019615105102210112101110140.00110150.000130.540.51111210210故主要存在形式是M(OH)3和M(OH)24，其濃度均為0.050molL口6.解:[T]0.10molL1,lgY(H)4.65Cd(T)11(0.10)102.80.10101.8Zn(T)1102.40.10108.320.010106.32KZnYCZnKCdYCCd11Y(Zn)Y(Cd)Zn(T)Cd(T)lgK'CdY16.46lgCd(T)lg(Y(H)Y(Zn))6.48lgK'ZnY16.5lgZn(T)lg(Y(H)Y(Cd))2.488.解:PH=5.0時(shí)，口lgY(H)6.45lgY(Cd)lgpKCdYCCdlgKZnY'4.8pZnep4.8lgK'ZnYlglgKZnYlg(Y(H)Y(Cd))Cd(I)pKCdY1016.46CCd0.005molL1Cd(I)1102.10102.43104.49105.41lgCd(I)5.46lgY(Cd)8.70lgK'ZnY16.50lg(106.45108.70)7.80ppZnp0.5(lgK'ZnYpCZn)5.05pZn0.2510.解：設(shè)A13+濃度為0.010mol/L,則由Al(OH)3的Kp計(jì)算可得A13+發(fā)生水解時(shí)pH=3.7。要想在加入￡口1人時(shí)，口A13+不發(fā)生水解，pH應(yīng)小于3.7?？贓tpZnpZn0.22%[HAc][H]10Ac(H)1111.554.7412.解:a.[Ac]Ka10CHAc0.31[Ac]molL10.2molL1Ac(H)1.55Pb2(Ac)11[Ac]2[Ac]21101.90.2103.80.04102.43lgY(H)6.45lgK'PbYlgKPbYlgPb2(Ac)lgY(H)18.042.436.459.16pPb'p6.08pPbep7.0pPb'ep7.02.434.57pPbpPb2.7%pPb'pPb1.51EtlgK'PbYlgKPbYlgY(11.59b.11.593pPb7.30pPbep7.0pPb0.30p2pPbpPbEt0.007%15.解：由Kp,測(cè)Th時(shí)pH<3.2，測(cè)La時(shí)pH<8.4,查酸效應(yīng)曲線(數(shù)據(jù))可知，測(cè)Th時(shí)口也較好，為消除La的干擾，宜選pH<2.5,因此側(cè)Th可在稀酸中進(jìn)行；側(cè)La在PH5-6較合適，可選在六亞甲基四胺緩沖溶液中進(jìn)行。18.解：由題意可知5Sn0.0335118.69100%62.31%0.21000(0.03500.033.00.0335)207.2100%37.29%0.20001000Pb21.解:據(jù)題中計(jì)量關(guān)系得：0.0250.030.00360.00110254.22100%98.45%0.202224.解:據(jù)題中計(jì)量關(guān)系得：Ni0.058310.0261458.693100%63.33%0.7176(0.035440.058310.026140.05831)555.845100%0.7176=21.10%FeCr(0.050.058310.006210.063160.035440.05831)551.9960.7176100%=16.55%)第7章氧化還原滴定法1.解：查表得：lgK(NH3)=9.46E=E8Zn2+/Zn+0.05921g[Zn2+]/2=-0763+0.05921g([Zn(NH3)42+]/K[(NH3)]4)/2=-1.04V3.解：EHg22+/Hg=E0Hg22+/Hg+0.5某0.05921g[Hg]=0.793+0.5某0.05921g(Kp/[Cl])]2+-2E0Hg22+/Hg=0.793+0.02951gKp=0.265VEHg22+/Hg=0.265+0.5某0.05921g(1/[C1])=0.383V-25.解：EMnO4-/Mn2+=E0/MnO4-/Mn2++0.059某1g([MnO4-]/[Mn2+])/5當(dāng)還原一半時(shí)：[MnO4-]=[Mn2+]故EMnO4-/Mn2+=E0/MnO4-/Mn2+=1.45V[Cr2O72-]=0.005mo1/L[Cr3+]=2某0.05=0.10mo1/LECr2O72-/Cr3+=Ee/Cr2O72-/Cr3++0.059/6某1g([Cr2072—]/[Cr3+])=1.01V7.解：Cu+2Ag+=Cu2++2Ag1gK=(0.80-0.337)某2/0.059=15.69K=1015.69=[Cu2+]/[Ag+]2表明達(dá)到平衡時(shí)Ag+幾乎被還原，因此二[Ag+]/2=0.05/2=0.025mo1/L[Ag+]=([Cu2+]/K)0.5=2.3某10-9mo1/L9.解：2S2032—+I—3=3I—+S4O62-(a)當(dāng)?shù)味ㄏ禂?shù)為0.50時(shí)，口[I3-]=0.0500(20.00-10.00)/(20.00+10.00)=0.01667mol/L[I-]=0.500某2某10.00/(2某00+10.00)+1某20.00/30.00=0.700mol/L故由Nernt方程得：口E=EI3-/I-0.059/2某lg0.01667/0.700=0.506V(b)當(dāng)?shù)味ǚ謹(jǐn)?shù)為1.00時(shí)，由不對(duì)稱的氧化還原反應(yīng)得：EI-3/I-=0.545+0.0295lg[I-3]/[I-]3(1)ES4O62/-S2O32-=0.080+0.0295lg[S4O62-]/[S2O32]2(2)⑴某4+(2)某2得:6Ep=2.34+0.059lg[I-3]2[S4O62-]/[I-]6[S2O32-]2由于此時(shí)[S4O62-]此[I-3]，計(jì)算得[S4O62-]=0.025mol/L[I-]=0.55mol/L,代入上式Ep=0.39=0.059/6某lg[S4O62-]/4[I-]6=0.384V(c)當(dāng)?shù)味ǚ謹(jǐn)?shù)為1.5,E=ES4O62/-S2O32-=0.80+0.02951g[S4O62-]/[S2O32]2此時(shí)[S4O62-]=0.1000某20.00/100=0.0200mol/L[S2O32-]=0.100某10.00/50.00=0.0200mol/L故E=0.800+0.02951g0.200/(0.200)2=1.30Vn.解：Ce4+Fe2+=Ce3++Fe3+終點(diǎn)時(shí)CCe3+=0.05000mo1/1,Fe2+=0.05000mo1/1.所以CCe4=CCe3+某10(0.94-1.44)/0.059=1.7某10—10mo1/1DCFe2+=CFe3+某10(0.94-0.68)/0.059=2.0某10-6mo1/1-10-6-6得Et=(1.7某10-2.0某10)/(0.0500+2.0某10)某100%=-0.004%13.解：Ce4++Fe2+=Ce3++Fe2+在H2SO4介質(zhì)中，終點(diǎn)時(shí)Eep=0.48V,Ep=(1.48+0.68)/2=1.06V,口E=1.44-0.68=0.76V,E=0.84-1.06=-0.22E-0.22/0.059t=(10T00.22/0.059)/100.76/2某0.059某100%=-0.19%在H2SO4+H3PO4介質(zhì)中，口Fe3+=1+103.5某0.5=5某102.5=103.2,Fe2+=1+0.5某102.3=102.0EFe3+/Fe2+=0.68+0.059lgFe3+=0.61VEp=(1.44+0.617)/2=1.03VE=0.84-1.03=-0.19VE=0.83V,由林邦Et二(10-0.19/0.059-100.19/0.059)/100.83/2某0.059某100%=0.015%15.解：5V02++Mn0-4+6H20=5V0-3+Mn2++12H+4Mn2++Mn0-4+8H+=5Mn3++4H2O5VMnO--4,4Mn4Mn2+MnO4(V)=5某0.02000某2.50某50.49/1.000某1000某100%=1.27%口(Mn)=(4某0.02000某4.00-0.02000某2.50)某54.94/(1.00某1000)某100%=1.48%17.解：PbO2+H2C+2O2+2H=Pb2++2CO2+2H2O,PbO+2H+=Pb2++H2O,2MnO4-+5C2O2-4+6H+=2Mn2++10CO2+8H205PbO2-25PbO5c2O42MnO-4,設(shè)試樣中PbO2為某克，PbO為y克。口則20某0.25=0.04某10.00某5/2+2某1000某/M(PbO2)+1000y/M(PbO)0.04000某30.00=2某1000某/5M(PbO2)+2某1000y/5M(PbO)解得某=0.2392g,y=0.4464g故口(PbO2)=0.2392/1.234某100%=19.38%,(PbO)=0.4464某.234某100%=36.17%.19.解：由化學(xué)反應(yīng)：I0—3+5I-+6H+=3I2+3H20,I2+S2-203=2I+S402-6得1KIO-35I3I26Na2s2O3cKI某25.00=5某(10.00某0.05000-21.14某0.1008某1/6),得cKI=0.02896mol/L21.解：由3NO2+H2O=2HNO3+NO,及氮守恒得口3NH+43NH33NO22HNO32NaOH,(NH3)=0.0100某20.00+3某M(NH3)/(1.000某1000某2)某100%=0.51%23.解：由此歧化反應(yīng)的電子守恒得：3/2Mn2+MnO-4(Mn)=(0.03358某0.03488)某3/2某M(Mn)/0.5165某100%=18.49%口25.解：由氧化還原反應(yīng)電子守恒得：6Fe2+Cr2O2-7，口mCr=2某M(Cr)(0.500/M[Fe(NH4)(SO4)26H-22O]-6某0.0038