(完整版)分析化学各章节习题(含答案) 下载本文

∵21.42%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 = 0.6% s2 = 0.9%

?s2大0.92Fs2?2?2?F表?9.12 ∴s1和s2间无显著性差异 小0.6t?96.4?93.94?50.64?5?6?t表?2.37

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:30.12%为可疑值 (1)用Q值检验法:

Q?30.12?30.4930.60?30.12?0.77?Q表?0.76

∴30.12%应舍弃。 (2)用4d检验法:

x?30.60%?30.52%?30.49%3?30.54%d?0.06%?0.02%?0.05%3?0.04%

4d?0.16%xi?x?30.12%?30.54%?0.42%?0.16%∴30.12%应舍弃。

x?30.54%0.062?0.022?0.052s?3?1%?0.06%

??x?tsn?30.54?2.92?0.063?(30.54?0.10)%1-10 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

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计算结果应以三位有效数字报出。

(2) 已知pH=4.75,c(H+) = 1.8?10-5 (pH=4.75为两位有效数字)

(3) pH = 2.658

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500~600?C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。

H2C2O4?2H2O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H2C2O4?2H2O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) H2C2O4?2H2O和KHC2O4? H2C2O4?2H2O两种物质分别和NaOH作用时,

△n(H2C2O4?2H2O):-△n(NaOH)=1:2 ;△n(NaOH): △n(KHC2O4? H2C2O4?2H2O)=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Zn2+标准溶液滴定释放出来的 EDTA: KB(C6H5)4+4HgY2-+3H2O+5H+=4Hg(C6H5)++4H2Y2-+H3BO3+K+ H2Y2-+Zn2+=ZnY2-+2H+ K+与Zn2+的物质的量之比为1:4 。 2-3解:

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)= 0.1mol?L-1?0.020L?204.2g?mol-1=0.4g △n(H2C2O4?2H2O) = (1/2)△n(NaOH)

m(H2C2O4?2H2O)=(1/2)?0.1mol?L-1?0.020L?126g?mol-1=0.13g

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RE?E?0.0002g???0.2% T0.13g2-5解:滴定反应:Na2B4O7?10H2O+2HCl=4H3BO3+2NaCl+5H2O

△ n(Na2B4O7?10H2O)= (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%M(Na2B4O7?10H2O)381.4g?mol-1或:w(B)?2c(HCl)V(HCl)M(B)mS2-6解:Al3++H2Y2-=AlY-+2H+

△n(Al3+) = △n(EDTA) △n(Al2O3) = (1/2)△n(EDTA) Zn2++ H2Y2-=ZnY2-+2H+ △n(Zn2+) = △n(EDTA)

1[c(EDTA)V(EDTA)?c(Zn)V(Zn)]M(Al2O3)2w(Al2O3)?mS1(0.05010mol?L?1?0.02500L?0.05005mol?L?1?0.00550L)?102.0g?mol?1?2?24.9%0.2000g2-7解:ClO3-+6Fe2++6H+=Cl-+6Fe3++3H2O

△n(ClO3-)=(1/6)△n(Fe2+) △n[Ca(ClO3)2] = (1/12) △n(Fe2+) Cr2O72-+6Fe2++14H+=2Cr3++6Fe3++7H2O △n(Fe2+) = 6△n(Cr2O72-)

1(0.1000mol?L?1?0.02600L?6?0.02000mol?L?1?0.01000L)?207.0g?mol?1w[Ca(ClO3)2]?12?12.08%0.2000g2-8 解:Ca2++C2O42-=CaC2O4

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CaC2O4+2H+= H2C2O4+ Ca2+

5 H2C2O4+2MnO4-+6H+ = 2Mn2++10CO2+8H2O △n(CaO) = △n(Ca) = (5/2)△n(MnO4-)

55c(KMnO4)V(KMnO4)M(CaO)?0.02mol?L?1?0.030L?56.08g?mol?1mS?2?2?0.2g

w(CaO)40%

第三章 酸碱滴定法

3-1 EDTA在水溶液中是六元弱酸(H6Y2+),其pKa1~pKa6分别为0.9、1.6、2.07、2.75、6.24、10.34、则Y4-的pKb3为: pKb3 = pKw-pKa4=14-2.75 =11.25 3-2解:

Ka1.8?10?5 x(Ac)???7?0.99 ???5c(H)/c?Ka10?1.8?10- x(HAc) = 1-0.99 = 0.01 c(Ac) = 0.99?0.1mol·L

-1

= 0.099 mol·L

-1

c(HAc) = 0.01?0.1mol·L

-1

= 0.001 mol·L1

3-3

(1) H3PO4 的PBE:c(H+) = c(H2PO4)+2c(HPO42)+3c(PO43)+c(OH) (2) Na2HPO4的PBE:c(H+)+c(H2PO4)+2c(H3PO4) = c(PO43)+c(OH) (3) Na2S的PBE:c(OH-) = c(HS)+2c(H2S)+c(H+)

(4) NH4H2PO4的PBE:c(H+) = c(NH3)+2c(PO43)+c(HPO42) +c(OH)-c(H3PO4) (5) Na2C2O4的PBE:c(OH) = c(HC2O4)+2c(H2C2O4)+c(H+) (6) NH4Ac的PBE:c(H+)+c(HAc)=c( NH3) +c(OH) (7) HCl+HAc的PBE:c(H+)=c(OH)+c(HCl)+ c(Ac) (8) NaOH+NH3的PBE:c(OH)=c(NH4+)+c(H+)+c(NaOH) 3-4计算下列溶液的pH值: (1)c(H3PO4)= 0.20mol?L1

因为Ka1/Ka2>10,(c/c?)/Ka2>102.44,?只考虑H3PO4的第一步解离 又因为(c/c?)/Ka1=29<400,?用近似式计算:

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