A 25.5 mL volume of a 0.01140 M KMnO4 solution was used to reach the equivalence

Question

A 25.5 mL volume of a 0.01140 M KMnO4 solution was used to reach the equivalence point in the titration of 0.1055 g of an unknown sample containing the oxalate ion. Determine the mass (in grams) of C2O42- that would be present in a 100 g sample.

2. (6 pts) The strong halide acids (HCl, HBr, HI) cannot be used in place of sulfuric acid. Briefly explain why. Acidity is not the issue here.

3. (8 pts) Assume that there are 58.00 g of C2O42- and 12.27 g of iron in a 100 g sample of your unknown. What is the mass of water in 100 g of your unknown sample? Remember that iron is present in the +3 oxidation state and that oxalate is a bidentate ligand.

Explanation / Answer

1)

moles of KMnO4 = 25.5 x 0.01140 / 1000 = 2.907 x 10^-4

Thus, moles of MnO4- ions produced = 2.907 x 10^-4

reaction between oxalate ion and permanganate ion:

2 MnO4- + 5 H2C2O4 + 6 H+ =>10 CO2 + 2 Mn2+ + 8 H2O

2 mol MnO4- -------------> 5 mol H2C2O4

2.907 x 10^-4 mol MnO4-   -----------------> ??

moels of C2O42- = 2.907 x 10^-4 x 5 / 2 = 7.268 x 10^-4 mol

moels of C2O42- = 7.268 x 10^-4 mol

mass of C2O42- = 7.268 x 10^-4 x 88 = 0.06395 g

Now, 0.1055 g sample contains = 7.268 x 10^-4 moles oxalate ions

So, 100 g sample will contain = 7.268 x 10^-4)/0.1055 x 100

                                              = 0.6889 mol

mass of oxalate ions in 100 g sample = 0.6889 x 88

mass of C2O42- that would be present = 60.6 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.