分析化学各章节习题(含答案) 下载本文

0.042?0.032?0.022?0.012标准偏差:s??0.03%

4?1变异系数:1-5 解:

s0.03%??0.2% x20.06%x?24.87%?24.93%?24.90%

2RE?24.90%?25.05%??0.6%

25.05%相对相差:

24.93%?24.87%?0.2%

24.90%1-6 解:(1)用Q值检验法:

Q?12.47?12.42?0.36?Q表?0.73

12.47?12.33∴%应保留。

(2) 用4d检验法:

12.33?12.34?12.38?12.42%?12.37@.04?0.03?0.01?0.05d?%?0.03%

44d?0.12%x?xi?x?12.47%?12.37%?0.10%?4d∴%应保留。 1-7 解:解法一:

x?21.62% s?0.03%x??t?s 21.62%?21.42%n??4?13.3?t表?3.180.03%∴ 有系统误差。 解法二:

??x?tsn?21.62?3.18?0.034?(21.62?0.05)%

∵%不在平均值置信区间内,所以有系统误差。 1-8 解:

x1?96.4% x2?93.9%

s1 = % s2 = %

s2大0.92F?2??2?F表?9.12 ∴s1和s2间无显著性差异 2s小0.6t?96.4?93.94?5?6?t表?2.37

0.64?5即两组数据平均值有显著性差异,有系统误差,即温度对测定结果有影响。 1-9解:%为可疑值 (1)用Q值检验法:

Q?30.12?30.49?0.77?Q表?0.76

30.60?30.12∴%应舍弃。 (2)用4d检验法:

30.60%?30.52%?30.49%?30.540.06%?0.02%?0.05%d??0.044d?0.16%x?xi?x?30.12%?30.54%?0.42%?0.16%∴%应舍弃。

x?30.54%0.062?0.022?0.052s?%?0.06%

3?1ts2.92?0.06??x??30.54??(30.54?0.10)%n31-10 解:

(1) 用返滴定法测定某组分含量,测定结果按下式计算:

0.1023mol?L-1?(0.02500L-0.01921L)?106.0g?mol-1w?

0.5123g计算结果应以三位有效数字报出。 (2) 已知pH=,c(H) = 10

+

-5

(pH=为两位有效数字)

(3) pH =

第二章 滴定分析法

2-1 答:分析纯NaCl试剂若不作任何处理就用以标定AgNO3溶液的浓度,结果会偏高,原因是NaCl易吸湿,使用前应在500~600C条件下干燥。若不作上述处理,则NaCl因吸湿,称取的NaCl含有水分,标定时消耗AgNO3体积偏小,标定结果则偏高。

H2C2O42H2O长期保存于干燥器中,标定NaOH浓度时,标定结果会偏低。因H2C2O4

2H2O

试剂较稳定,一般温度下不会风化,只需室温下干燥即可。若将H2C2O42H2O长期保存于干燥器中,则会失去结晶水,标定时消耗NaOH体积偏大,标定结果则偏低。 2-2 计算下列各题:

(1) H2C2O42H2O和KHC2O4 H2C2O42H2O两种物质分别和NaOH作用时,

△n(H2C2O42H2O):-△n(NaOH)=1:2 ;△n(NaOH): △n(KHC2O4 H2C2O42H2O)=3:1 。 (2) 测定明矾中的钾时,先将钾沉淀为KB(C6H5)4,滤出的沉淀溶解于标准EDTA—Hg(溶液中,在以已知浓度的Zn标准溶液滴定释放出来的 EDTA: KB(C6H5)4+4HgY+3H2O+5H=4Hg(C6H5)+4H2Y+H3BO3+K H2Y+Zn=ZnY+2H

K与Zn的物质的量之比为1:4 。 2-3解:

+

2+

2-2+

2-+2-+

+

2-+

2+

)

c(H2C2O4)?m(H2C2O4?2H2O)1.6484g?M(H2C2???2H2O)V(H2C2O4?2H2O)126.1g?mol-1?0.2500L

?0.05229 mol?L-12-4解:NaOH + KHC8H4O4 = NaKC8H4O4 + H2O

-△n(NaOH)=-△n(KHC8H4O4)

m(KHC8H4O4) = c(NaOH)Vv(NaOH)M(KHC8H4O4)= L-10.020L204.2gmol-1=0.4g

△n(H2C2O42H2O) = (1/2)△n(NaOH)

-1

-1

m(H2C2O42H2O)=(1/2)L0.020L126gmol=0.13g

RE?E?0.0002g???0.2% T0.13g2-5解:滴定反应:Na2B4O710H2O+2HCl=4H3BO3+2NaCl+5H2O

△ n(Na2B4O710H2O)= (1/2)△n(HCl) △ n(B) = 2△n(HCl)

1c(HCl)V(HCl)M(Na2B4O7?10H2O)2w(Na2B4O7?10H2O)?mS1?0.2000mol?L?1?0.02500L?381.4g?mol-1?2?0.9536?95.36%1.000gM(Na2B4O7)201.2g?mol-1w(Na2B4O7)??w(Na2B2O7?10H2O)??95.36%?50.30%-1M(Na2B2O7?10H2O)381.4g?mol4M(B)4?10.81g?mol-1w(B)??w(Na2B4O7?10H2O)??95.36%?10.81%-1M(Na2B4O7?10H2O)381.4g?mol或:w(B)?3+

2c(HCl)V(HCl)M(B)mS2--+

2-6解:Al+H2Y=AlY+2H

△n(Al) = △n(EDTA) △n(Al2O3) = (1/2)△n(EDTA) Zn+ H2Y=ZnY+2H △n(Zn) = △n(EDTA)

2+

2+

2-2-+

3+

1[c(EDTA)V(EDTA)?c(Zn)V(Zn)]M(Al2O3)w(Al2O3)?2mS1(0.05010mol?L?1?0.02500L?0.05005mol?L?1?0.00550L)?102.0g?mol?1?2?24.9%0.2000g2-7解:ClO3+6Fe+6H=Cl+6Fe+3H2O

-2+

+

-3+