化学分析(武汉大学编 - 第五版)课后习题答案

内容发布更新时间 : 2024/12/23 7:39:13星期一 下面是文章的全部内容请认真阅读。

?为1.05?10?5?100?233.4=0.245mg

?1HSO24洗涤时 100 mL0.010 mol?L0.010mol?L?1H2SO4的[H?]?1.41?10?2mol?L?1

4Ksp?1.1?10?10?[Ba2?][SO42?]?s?(s?0.01)??SO2??s?0.01?

Ka21.41?10?2?Ka2

?s=2.65?10-8mol?L?1,BaSO4损失mg数为:2.65?10-8?100?233.4?6.2?10?4mg16.解:

NH4HF2???NH4F?HFHF(1)

0.005?[H?][H]H???0.005?[H?]F?

[H?][F?]Ka?[HF]?[H?]?5.84?10?4mol?L?10.001Ka?[Ca2?][F?]2??(2?0.005??F?)2?0.0005?(0.01??)22[H]?Ka?0.0005?(0.01?0.56)2?1.57?10?8?KspAgCl?有沉淀生成

?Ag(NH)= (2)

3cAg[Ag?]?1?103.24?0.5?107.0(0.5)2?2.8?106

0.05?0.5?8.9?10?9?KspAgCl62.8?10?有沉淀生成

0.05pH?9.26?lg=8.260.5(3) [Ag?][Cl?]?pOH?5.74,[OH]?1.82?10?6mol?L?1?[Mg2?][OH?]2?0.005?(1.82?10?6)2?1.66?10?14?KspMg(OH)2?无沉淀生成19.解:

s?[Zn2?]?[ZnOH?]?[Zn(OH)2]+[Zn(OH)-+[Zn(OH)2-3]4] ?[Zn2?]{1??1[OH?]??2[OH?]2??3[OH?]3??4[OH?]4} ?Ksp??2?3?4?{1??[OH]??[OH]??[OH]??[OH]}1234?2[OH]

?2.5?10-7mol?L?1 主要状态可由数值得 22.解:

F?(1)

M(Cr2O3)?0.23512M(PbCrO4)

F?(2)

2M(MgSO4?7H2O)?2.215M(Mg2P2O7)

F?(3)

M[Ca3(PO4)2]?0.082662M[(NH4)3PO4?12MoO3]

F?(4)

M(P2O5)?0.037832M2M[(NH4)3PO4?12MoO3]

25.解: 设CaC2O4为x,MgC2O4

y=0.6240-x

?x?

M(CaCO3)M(MgCO3)?(0.6240?x)??0.4830M(CaC2O4)M(MgC2O4)

x?0.4773g,CaC2O4%?76.49(.解:

y?0.1467g,MgC2O4%?23.51%M(AgCl)107.868?35.453?0.5805??1.4236M(NaCl)Na?35.453

解得Na?22.988865 0.5805?30.有关电对反应为

Cr2O7?14H??6e?2Cr3??7H2OO2?4H??4e?2H2O

2?故

22?molCr2O7?4mole312?1molFe?molCr2O76 1molO2?31[(cV)K2Cr2O7?(cV)Fe2?]?32.00?1036COD?2V水样31?(0.01667?25.00??0.1000?15.00)?32.00?1036?2100.0?80.02(mg/L)31.解: 设为

FexOy

?x?55.85?y?16?0.5434?x?55.85?0.3801?0.3801则x??0.00680655.85 y0.010203???x0.0068062

???

?为Fe2O350-1

34.解: AgCl:10?143.3=0.035(mol·L)

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