分析化学教材习题参考答案doc 下载本文

2-解:CH3OH?Cr2O7?8H??2Cr3??CO2?6H2O n?CH3OH??n1?K2Cr2O7? 2?Cr2O7?6Fe2??14H??2Cr3??6Fe3??7H2O nFe2??6n?K2Cr2O7?

??1?n?K2Cr2O7??n1?K2Cr2O7??n2?K2Cr2O7??n?CH3OH??nFe2?

61n?CH3OH??n?K2Cr2O7??nFe2?

6????1?2?2?????cKCrOVKCrO?cFeVFe227227?6??CH3OH???ms???????M?CHOH??3?100%

1?1?3?3??0.1000?25.00?10??0.1000?10.00?10?32.04??66????100%?8.01%0.1000也可以用等物质规则:n?CH3OH??nFe?1?6???2? 1??n??KCrO6?227?? ?

6-8 测定柠檬果汁中的维生素C(抗坏血酸,Mr=176.1)。取100.0mL柠檬果汁,用H2SO4

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酸化,加入20.00mL 0.02500 mol·LI2标准液与之反应(维生素为还原剂,半反应为,过量的I2用Na2S2O3标准溶液返滴,消耗Na2S2O3 C6H10O6?C6H8O6?2H??2e)

10.00mL,计算柠檬果汁中维生素C的质量浓度ρ(以mg·mL表示)

解:C6H10O6?I2?C6H8O6?2H??2I- I2?2Na2S2O3?2NaI?Na2S4O6

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?1??1?n?C6H10O6??n?Na2S2O3??n?I2? ?2??2???1????nI?nNaSO??223??M?Vc??22c1I2V?I2??c?Na2S2O3?V?Na2S2O3??M?Vc?????2??Vc???V?Vc?V?Vc??2?0.02500?20.000?0.02000?10.00??176.1?0.198?100.0????

6-9 称取苯酚试样0.5005g,用NaOH溶解后,准确配制成250mL试液,移取25.00mL试液与碘量瓶中,加入KBrO3-KBr标准溶液25.00mL及HCl,使苯酚溴化为三溴苯酚。加入KI

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溶液,使未反应的Br2还原并析出定量的I2,然后用0.1008 mol·L Na2S2O3标准溶液滴定,用去15.05mL。另取25.00mL KBrO3-KBr标准溶液,加入HCl和KI溶液,析出I2,用上述Na2S2O3标准溶液滴定,用去40.20mL,计算苯酚的质量分数[苯酚Mr(C6H5OH)=94.11]。 解:BrO3?5Br?6H?3Br2?3H2O

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2I-?Br2?2Br-?I2

I2?2Na2S2O3?2NaI?Na2S4O6

由第二个实验可知25.00mL的KBrO3-KBr溶液可知生成的Br2的量。

n?Br2??n'?Br2??11n?Na2S2O3??c?Na2S2O3??V?Na2S2O3? V?Na2S2O3??40.20mL 22由上述反应关系可知,在第一个实验中生成的Br2中过量的部分为:

1'1n?Na2S2O3??c?Na2S2O3??V'?Na2S2O3? V'?Na2S2O3??15.05mL 2211?n?苯酚??n?Br2??n'?Br2??c?Na2S2O3??V?Na2S2O3??V'?Na2S2O3?

36????c?Na2S2O3?V?Na2S2O3??V'?Na2S2O3??M?苯酚????苯酚???100%6ms???0.1008??40.20?15.05?10?3?94.11??100%?79.45%6?0.5005?250

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6-10 今有I2标准溶液对As2O3的滴定度为0.9892g·mL及Na2S2O3标准溶液对KBrO3的滴定

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度为0.5567 g·mL,求:

⑴ 此两种溶液的浓度各为多少? ⑵ Na2S2O3对Cu的滴定度为多少?

2-⑶ I2对硫化物中S的滴定度为多少?

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⑷ 用0.002000 mol·LKMnO4标准液30.00mL与过量KI的作用,析出的I2用上述Na2S2O3标准溶液滴定,需消耗Na2S2O3多少毫升?

?⑸ 称取含有Na2S和Sb2S5的试样0.2000g,用酸溶解后,使Sb全部变为SbO33后,在

NaHCO3介质中,用0.01000 mol·LI2标准溶液滴定,消耗20.00mL。另取同样质量的试样,溶于酸后,将产生的H2S完全吸收在70.00mL上述I2溶液中,过量的I2用上述的标准Na2S2O3溶液滴定,消耗10.00mL,计算试样中Na2S和Sb2S5的质量分数。 解:⑴ 2I2?As2O3?5H2O?2H3AsO4?4HI

-2-2- BrO3?6I??6H??Br-?3I2?3H2O I2?2S2O3?2I-?S4O6-1

TI2?As2O31c?I2??10?3?M?As2O3? 2As2O32?0.9892?10-3c?I2???3??0.01000mol?L?1 -310?M?As2O3?10?197.842TI2n?KBrO3??1n?Na2S2O3? 6TNa2S2O3?KBrO31c?Na2S2O3??10?3?M?KBrO3? 6KBrO36?0.5557?10-3?1 c?Na2S2O3???3??0.02000mol?L-310?M?As2O3?10?167.002-2-⑵ 2Cu2??4I??2CuI?I2 I2?2S2O3 ?2I-?S4O66TNa2S2O3n?Cu??n?Na2S2O3? TNa2S2O3Cu?c?Na2S2O3??10?3?M?Cu??0.02000?10?3?63.55?1.271?10?3g?mL?1

⑶ I2?S2-?2I-?S?

TI2?c?I2??10?3?M?S??0.01000?10?3?32.06?3.206?10?4g?mL?1

S-?2?⑷ 2MnO-?5I2?8H2O 4?10I?16H?2MnI2?2Na2S2O3?2NaI?Na2S4O6 V?Na2S2O3??⑸ 第一步测定 Sb

? I2?H3SbO3?H2O?2I??H2SbO-4?3H

5c?KMnO4?V?KMnO4?5?0.002000?30.00??15.00mL

??cNa2S2O30.02000 n?H3SbO3??n?I2??2n?Sb2O5?

?n?Sb2O5??11n?I2???0.01000?20.00?10?3?1.00?10?4mmol22n?S2O5??M?Sb2O5?1.000?10?4?403.85???Sb2O5???100%??100%?20.19%

ms0.2000第二步测定 Na2S

2-2- H2S?I2?S??2H??2I? I2?2S2O3?2I-?S4O6?1??1?n'?I2??n?Na2S2O3??n?H2S? ?2??2?即 2n?I2??n?Na2S2O3??2n?H2S?

'?n?Sb2O5???1n2n'?I2??n?Na2S2O3??10n?Sb2O5?2???1?2?0.01000?70.00?10?3?0.02000?10?3?10.00?10?1.000?10?4 2?1.000?10?4mol?n?Na2S??M?Na2S?1.000?10?4?78.04???Na2S???100%??100%?3.90%

ms0.2000

6-11 称取含Na2S和Sb2S3试样0.2000g溶于浓HCl中,反应生成的H2S用50.00mL 0.01000

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mol·L I2标液吸收(使H2S?S),然后用0.02000 mol·L Na2S2O3标液滴定剩余的I2,用去10.00mL;此后将试液(已除去H2S)调制微碱性.再用上述I2标液滴定Sb(III),耗用10.00mL,试计算试样中Sb2S3与Na2S的质量分数。已知:M(Na2S)=78.04,M(Sb2S3)=339.7 解:H2S?I2?S??2HI I2?2Na2S2O3?2NaI?Na2S4O6 第一步测定 n??1??1?H2S??n?Na2S2O3??n?I2? ?2??2?即 2n?H2S??n?Na2S2O3??2n?I2? n?H2S??n?Na2S??3n?Sb2S3?

2n?Na2S??6n?Sb2S3??2n?I2??n?Na2S2O3??2?0.01000?50.00?10?0.02000?10.00?10?8.000?10mol第二步测定

?3?3?4

I2?H3SbO3?H2O?2I??H2SbO-4?3H? n'?I2??n?H3SbO3??2n?Sb2O5?

?n?Sb2O5??n?Na2S??1‘1n?I2???0.01000?10.00?10?3?5.000?10?5mmol 221'n8.000?10-4?6?5.000?10?5?2.5000?10?4mol 2??n?S2O5??M?Sb2O5?5.000?10?5?339.7???Sb2O5???100%??100%?8.49%

ms0.2000n?Na2S??M?Na2S?2.5000?10?4?78.04???Na2S???100%??100%?3.90%

ms0.2000

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6-12 取同量KIO3·HIO3溶液,其一直接用0.1000 mol·L NaOH滴定,耗用NaOH V mL;其

二溶

液酸化后,加入过量KI,以Na2S2O3标液滴定,用去4V mL,试求Na2S2O3的浓度。 解:HIO3?NaOH?NaIO3?H2O