分析化学(高教第四版)习题及答案(第一章定量分析化学概论).doc

分析化学习题及答案（第四版）第一章：定量分析化学概论5.某试样中汉mgo约30%，用定量法测定时，fe有1%进入沉淀。若要求测定结果的相对误差小于0.1%，求试样中fe2o3允许的最高质量分数为多少？解：设 fe2o3的质量分数为 fe2o3,试样质量记为m样测定的相对误差= = 即 0.1% =fe2o3 2fe , nfe o = nfe fe o =,nfe o =mfe=nfe mfe=2nfe2o3 mfe=2mfe= 代入0.1%= 0.001= fe o =0.043=4.3%6. 某含cl-试样中含有0.10%br-,用agno3进行滴定时，br-与cl-同时被滴定，若全部以cl-计算，则结果为20.0%。求称取的试样为下列质量时，cl-分析结果的绝对误差及相对误差：a. 0.1000g, b. 0.5000g, c. 1.000g.解： 设 cl- 真实含量x,称取试样质量ms(g)。则 0.2000=x+ x=0.200-0.001=0.200-4.4 10-4=0.1996e=x-t绝对误差：4.4 10-4（即0.044%）相对误差：100%=0.22%7. 某试样中含有约5%的s，将s氧化为so42-，然后沉淀为baso4。若要求在一台灵敏度为0.1mg的天平上称量baso4的质量时可疑值不超过01.%，问必须称取试样多少克？解：设 必须称取x克 s - so42-baso432.066 233.39x 5% m baso 天平上的绝对误差0.1mg 2 (两次读取)相对误差= 0.1%= x= = =0.6g8. 用标记为0.1000 hcl标准溶液标定naoh溶液，求得浓度为0.1018 ,已知hcl溶液真实浓度为0.0999 ，标定过程中其他误差均小，可以不计，求naoh溶液的真实浓度。解：设 真实浓度为x， 则有oh-+h+=h2ohcl的相对误差应等于naoh的相对误差所以= x=0.1017 9. 称取纯金属锌0.3250g，溶于hcl后，稀释到250ml容量瓶中。计算zn2+溶液的浓度。解：czn2+= /(250 103)=0.01988 mol/l10. 有0.0982 的h2so4溶液480ml，现欲使其浓度增至0.1000 问应加入0.5000 的h2so4溶液多少毫升？解：设 加入x毫升 =0.1000解得 x=2.16ml11. 在500ml溶液中，含有9.21gk4fe(cn)6。计算该溶液的浓度及在以下反应中对zn2+的滴定度：（mkfe(cn)=368.35） 3zn2+2fe(cn)64-+2k+= k2zn3fe(cn)62解：（1）c= /（500 103）=0.0500（2）tzn/fe(cn)=c/1000 mzn=0.0500/1000 65.39=0.00490g/ml12. 要求在滴定时消耗0.2 naoh溶液2530ml。问应称取基准试剂邻苯二甲酸氢钾（khc8h4o4）多少克?如果该用h2c2o42h2o做基准物质,又应称取多少克?(m khc h o =204.126, mh c o 2h o=126.07)解: 反应2oh-+ h2c2o4=2h2o+ c2o42- (1) 0.2 25=1030.2 30=103x=1.01.2g (2) 0.2 25=1030.2 30=103x=0.30.4g13.欲配制na2c2o4溶液用于在酸性介质中标定0.02 的kmno4溶液，若要使标定时，两种溶液消耗的体积相近。问应配制多大浓度的na2c2o4溶液？配制100ml这种溶液应称取na2c2o4多少克？（m na c o =134.00）解：5 c2o42-+ 2mno4-+16h+=10co2+2mn2+8h2o(1) cc = cmn =0.02=0.05 m na2c2o4=134.00=0.7g14. 含s有机试样0.471g，在氧气中燃烧，使s氧化为so2,用预中和过的h2o2将so2吸收，全部转化为h2so4，以0.108 moll-1koh标准溶液滴定至化学计量点，消耗28.2ml。求试样中s的质量分数。解：sso2+ h2o2h2so4+2koh=h2o+ k2so4ms=ns ms=nh2so4 ms= = =0.104=10.4%15. 将50.00ml0.1000 ca(no3)2溶液加入到1.000g含naf的试样溶液中，过滤、洗涤。滤液及洗液中剩余的ca2+用0.0500 edta滴定，消耗24.20ml。计算试样中naf的质量分数。（mf=18.998,mnaf=41.998）解： ca2+2f-=caf mnaf=nnaf mnaf=2nca2+ mnaf= 2mnaf(50.00 0.1-0.05 24.20)/1000naf= =0.318316.0.2500g不纯caco3试样中不含干扰测定的组分。加入25.00ml 0.2600 hcl溶液，煮沸除去co2，用于0.2450 naoh溶液返滴过量酸，消耗6.30ml。计算试样中caco3的质量的分数。解：2h+ caco3=ca2+h2co3 2h+2oh-=2h2ocaco= = = =0.982417. 今含有mgso47h2o纯试剂一瓶，设不含其他杂质，但有部分失水变为mgso46h2o，测定其中mg含量后，全部按mgso47h2o计算，得质量分数为100.96%。试计算试剂中mgso46h2o的质量分数。解：设mgso46h2o质量分数 xmgso47h2o为1- x100.96%=1- x+ x x= = =0.1217若考虑反应,设含mgso47h2o为n1 molmgso46h2o为n2 mol样本质量为100g。n=n1+n2n 246.47=100.96n1 228.455+ n2 246.47=10018n1=0.96 n=0.253m(mgso46h2o)=n m mgso 6h o=0.053 226.45=12.18= =0.121818.不纯sb2s3 0.2513g，将其在氧气流中灼烧，产生的so2通入fecl3溶液中，使fe3+还原至fe2+，然后用0.02000 moll-1kmno4标准溶液滴定fe2+，消耗kmno4溶液31.80ml。计算试样中sb2s3的质量分数若以sb计，质量分数又为多少？（m sb s =339.7,msb=121.76）解：2 sb2s3+9o2=2sb2o3+6so2so2+2fecl3+2h2o=2fecl2+h2so4+2hcl5fe2+mno4-+8h+=mn2+5fe3+4h2osb2s3 3so2 6fecl3mno4-nsb s = nmn sb s = = =0.7164sb= =0.513519.已知在酸性溶液中,fe2+与kmno4反应时,1.00ml kmno4溶液相当于0.1117gfe，1.00mlkhc2o4h2c2o4溶液在酸性介质中恰好与0.20ml上述kmno4溶液完全反应。问需要多少毫升0.2000moll-1naoh溶液才能与上述1.00ml khc2o4h2c2o4溶液完全中和？解：5fe2+mno4-+8h+=mn2+5fe3+4h2o5khc2o4h2c2o4+4 kmno4+17h+=20co2 +4 mn2+9k+16 h2o5khc2o4h2c2o4+3oh-=k+2 c2o42-+3 h2otfe/= c = = =0.4000ckhc o h c o = =0.4 0.20=0.10 moll-1vnaoh= =1.5ml20.用纯as2o3标定kmno4溶液的浓度,若0.2112g as2o3在酸性溶液中恰好与36.42ml kmno4反应。求该kmno4溶液的浓度？解：as2o3+3h2o=h3 aso32mno4-+5 aso33-+6h+=5aso43-+2mn2+3h2oas2o3h3 aso3 mno4- nmn = nas o cmn =n mn /v= nas o /v=0.02345 moll-121.测定氮肥中nh3的含量。称取试样1.6160g,溶解后在250ml容量瓶中定容，移取25.00ml,加入过量naoh溶液，将产生的nh3导入40.00ml c( h2so4)=0.1020 moll-1的h2so4标准溶液吸收，剩余的h2so4需17.00ml, c(naoh)=0.09600 moll-1naoh溶液中和。计算氮肥的nh3的质量分数。解： +oh-= + h2so4= nh =(nnhm nh )/(1.6160 0.1)= m nh /(1.6160 0.1) (40.00 0.1020-17.00 0.09600)/1000=0.258022.称取大理石式样0.2303g，溶于酸中，调节酸度后加入过量(nh4)2c2o4溶液，使ca2+沉淀为cac2o4。过滤，洗净，将沉淀溶于稀h2so4中。溶解后的溶液用c( kmno4)=0.2012 moll-1kmno4标准溶液滴定，消耗22.30ml，计算大理石中caco3的质量分数。解：caco3-ca2+-cac2o4-h2c2o4- mno4-5h2c2o4+2mno4-+6h+=2mn2+10co2+8h2ocaco =ncacomcaco /m样= (n mn mcaco )/ m样 = (c mnvmno4- mcaco3)/( m样 1000)= =0.9750注：c( kmno4)=0.2012 kkmno =acknowledgements my deepest gratitude goes first and foremost to professor aaa , my supervisor, for her constant encouragement and guidance. she has walked me through all the stages of the writing of this thesis. without her consistent and illuminating instruction, this thesis could not havereached its present form. second, i would like to express my heartfelt gratitude to professor aaa, who led me into the world of translation. i am also greatly indebted to the professors and teachers at the department of english: professor dddd, professor ssss, who have instructed and helped me a lot in the past two years. last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. i also owe my sincere gratitude to my friends and my fellow classmates who gave me their help and time in listening to me and helping me work out my problems during the difficult course of the thesis. my deepest gratitude goes first and foremost to professor aaa , my supervisor, for her constant encouragement and guidance. she has walked me through all the stages of the writing of this thesis. without her consistent and illuminating instruction, this thesis could not havereached its present form. second, i would like to express my heartfelt gratitude to professor aaa, who led me into the world of translation. i am also greatly indebted to the professors and teachers at the department of english: professor dddd, professor ssss, who have instructed and helped me a lot in the past two years. last my thanks would go to my beloved