!!!!!!!!!!!!!!!!calculate the concentration!!!!!!!!!!!

[匿名]

2008-10-01 8:20 am

25.0cm^3 of a solution containing ethanedioic acid and sodium ethanedioate required 14.75cm^3 of 0.1Msodium hydroxide solution for neutralization,and 30.5 cm^3 of 0.0205Mpotassium manganate(VII)solution for oxidation in acidic condition at about 60度 calculate the concentration of each constituent.

回答 (1)

Uncle Michael

2008-10-02 1:12 am

✔ 最佳答案

Only H2C2O4 can neutralize NaOH.
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
Mole ratio H2C2O4 : NaOH = 1 : 2
No. of moles of NaOH = MV = 0.1 x (14.75/1000) = 0.001475 mol
No. of moles of H2C2O4 = 0.001475 x (1/2) = 0.0007375 mol
Volume of the solution V = 25 cm3 = 0.025 dm3
Conc. of H2C2O4 = mol/volume = 0.0007375/0.025 = 0.0295 M

Both H2C2O4 and Na2C2O4 can be oxidized by acidified KMnO4.
2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 8H2O + 10CO2
Mole ratio MnO4- : C2O42- = KMnO4 : (H2C2O4 + Na2C2O4) = 2 : 5
No. of moles of KMnO4 = MV = 0.0205 x (30.5/1000) = 0.0006253 mol
No. of moles of H2C2O4 + Na2C2O4 = 0.0006253 x (5/2) = 0.001563 mol
No. of moles of Na2C2O4 = 0.001563 - 0.0007375 = 0.0008255 mol
Conc. of Na2C2O4 = mol/volume = 0.0008255/0.025 = 0.03302 M
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