0.0982mol/L?0.480L?0.5000mol/L?V2?0.1000mol/L?(0.480L?V2)
V2?2.16mL 2． 解：①
nNaOH:nKHC8H4O4?1:1
m1?n1M?cV1M?0.2mol/L?0.025L?204.22g/mol?1.0g m2?n2M?cV2M?0.2mol/L?0.030L?204.22g/mol?1.2g 应称取邻苯二甲酸氢钾1.0~1.2g ②
nNaOH:nH2C2O4?2H2O?2:1m1?n1M?
1cV1M21??0.2mol/L?0.025L?126.07g/mol?0.3g2 1m2?n2M?cV2M21??0.2mol/L?0.030L?126.07g/mol?0.4g2
应称取
3． 解：
H2C2O4?2H2O0.3~0.4g
S:SO2:H2SO4:2KOH
w?nM?100%m01?0.108mol/L?0.0282L?32.066g/mol2??100%0.471g?10.3%
4． 解：
CaCO3:2HCl NaOH:HCl
1(cV?cV)MnM2w??100%??100%m0m01(0.2600mol/L?0.025L?0.2450mol/L?0.0065L)?100.09g/mol2??100%0.2500g?98.24%
5．
Sb2S3:2Sb:3SO2:6Fe2?:6KMnO45
解：
55nSb2S3?cV??0.0200mol/L?0.03180L?0.00053mol66nSb?2nSb2S3?2?0.00053mol?0.00106mol0.00053mol?339.68g/mol?100%?71.64#0.2513g0.00106mol?121.76g/molwSb??100%?51.36%0.2513gwSbS? 6.
3?Q5As2O3:10AsO3:4MnO4?解：
4mcVKMnO4???10005M 4?0.2112?1000cKMnO4?5?0.02345(mol/L)36.42?197.8
7．解：
mH2C2O4MH2C2O4500?1000?5.553?10?3(mol)90.035nH2C2O4?
QH2C2O4:2NaOH
nNaOH?2nH2C2O4?2?5.553?10?3?11.106?10?3(mol)VNaOHnNaOH11.106?10?3???0.111(L)?111(mL)cNaOH0.100
QH2C2O4:nKMnO42KMnO45 22?nH2C2O4??5.553?10?3?2.221?10?3(mol)55
VKMnO4?
8. 解：
nKMnO4cKMnO42.221?10?3??0.0222(L)?22.2(mL)0.100
cK2Cr2O7?nK2Cr2O7VK2Cr2O75.442?294.18?0.01850(mol/L)1
TFe3O4/K2Cr2O7?cK2Cr2O7?MFe3O4?2?0.01850?231.54?2?8.567(mg/mL) 9.解：
?Q5Fe2?:MnO4
4?nFeSO4?7H2O?5nMnO??5?35.70?0.02034?3.631?10?3(mol)1000
?FeSO?7HO?42mFeSO4?7H2Om?nFeSO4?7H2O?MFeSO4?7H2Om
3.631?10?3?278.04??99.76%1.012
故为一级化学试剂。
10. 解：第一步以酚酞作指示剂，被滴定的是Na2CO3
滴定反应为：Na2CO3 + HCl = NaHCO3 + NaCl
?NaCO?23CHClVHClMNa2CO3ms?100%0.1060?20.10?10?3?105.99??100% -------（4分）
0.3010?75.02%第二步以甲基橙作指示剂，被滴定的是原有的NaHCO3及第一步滴定生成的NaHCO3
滴定反应为：NaHCO3 + HCl = NaCl + H2O + CO2
第二步消耗盐酸的体积为：V' = 47.70 - 20.10 = 27.60 mL
?NaHCO?3CHCl(VHCl'?VHCl)MNaHCO3ms?100%0.1060?(27.60?20.10)?10?3?84.01??100%
0.3010?22.19%
11. 解：根据分布分数计算公式计算：
[H?][HAc]=?HAc?cHAc= ?cHAc ?[H]?Ka=
10?410?4?0.10 ?5?1.8?10 = 0.085 (mol?L-1 )
[Ac－ ] =?Ac-?cHAc =
[Ka]?cHAc ?[H]?Ka1.8?10?5= ?4?0.10 ?510?1.8?10 = 0.015 (mol?L-1) 或：
∵ [HAc]+[Ac－ ]=0.10mol?L-1 ∴ [Ac－ ]= 0.1－0.085 = 0.015 (mol?L-1)
12.解：（1）加入8.24 mLNaOH溶液后构成缓冲体系，
pH?pKa?A?，设此时溶液体积为VmL，
?lg??HA??A??C?NaA?HA??CHA8.24?0.09000?mol/L?
V(41.20?8.24)?0.09000?mol/L? ?V?