Experiment details: Previously the coordination compound KwFex(C2O4)y(H2O)z was

Question

Experiment details:

Previously the coordination compound KwFex(C2O4)y(H2O)z was synthesized. A redox titration will be used to determine the moles of oxalate ion, and last week spectrophotometric analysis was used to determine the moles of iron(III) in KwFex(C2O4)y(H2O)z. The moles of potassium ion and water in the coordination compound will be determined from an interpretation of these two sets of data.

In this experiment the number of moles of oxalate in a known mass of the coordination compound will be determined using a standardized potassium permanganate solution as a titrant. The reaction of the permanganate ion with oxalate ions in a strongly acidic solution is described by the following equation:

2MnO4- (aq) + 5C2O42- (aq)+16H+ (aq) 2Mn2+ (aq) +10CO2 (g) +8H2O(l)

The oxalate ions become fully protonated in the acidic solution to form oxalic acid, H2C2O4. The permanganate ion, a strong oxidizing agent, oxidizes the oxalate ion to carbon dioxide and water. The deep purple permanganate ion serves as its own indicator. Its reaction with oxalate ion produces a colorless manganese(II) ion. The equivalence point in the titration occurs when excess purple permanganate ion persists after the complete oxidation of oxalate ions. Keeping the solution warm during the titration procedure removes the carbon dioxide as it is formed in the reaction.

The reaction of permanganate with oxalate ions is more complex than the balanced equation implies. The reaction is slow but is catalyzed by the presence of manganese(II) ion. The initial reaction rate is slow because of the absence of manganese(II) ion in the receiving flask. The reaction, however, becomes more rapid as the manganese(II) ion builds up. The reaction rate also increases with temperature and is a second reason for heating the reaction mixture.

Questions:

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

Answer for (2).

Reaction of KMnO4 with sulfuric acid follows as given below.

6 KMnO4 + 9 H2SO4 6 MnSO4 + 3 K2SO4 + 9 H2O + 5 O3

Reaction of KMnO4 with hydrochloric acid follows as given below.

2 KMnO4 + 16 HCl --> 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O

As can be seen chlorine is generated in reaction of KMnO4 with HCl and this generated chlorine also oxidizes oxalic acid formed in the reaction..

Similarly other acids HBr and HI also generate molecular halogens which react with K2MnO4.

Answer for (3).

Total weight of unknown sample = 100 g KwFex(C2O4)y(H2O)z

Oxalate weight in unknown sample = 58 g

Iron (3+) weight in unknown sample = 12.27 g

Find mass of hydrate (water) in the compound.

Oxalate being a bidentate ligand (with -2 charge on each ligand) and with iron being in Fe3+ oxidation state the compound should have the following composition with respect to oxalate and iron.

[Fe(C2O4)3]3- and this makes the compound in question to have 3 K+ ions which makes the composition of compound to be as follows:

K3[Fe(C2O4)].x H2O.

Thus in an unhydrated compound the ratio of K, Fe and oxalate can be calculated first as given below.

Molecular weight of unhydrated K3[Fe(C2O4)3] = (3xK) + Fe + (6xC) + (12 O)

                                                                                 = (3 x 39) + 56 + (6x12) + (12 x 16)

                                                                                 = 117 + 56 + 72 + 192

                                                                                  = 437

Percentage of K, Fe and oxalate in unhydrated compound = K: 26.77% (117/437 x 100)

                                                                                                       Fe: 12.81% (56/437 x 100)

                                                                                                       Oxalate = 60.42% (264/437 x 100)

Ratio of K, Fe and oxalate = 2.677:1.281:6.042

In the unknown sample oxalate = 58.00 g

                                       Iron = 12.27 g

Accordingly for 6.042 g of oxalate one should have 2.677 g potassium, For 58.0 g of oxalate how much potassium one should have

        58 x 2.677 / 6.042 = 25.697 g (approximated to two decimal point = 25.70 g)

Potassium = 25.70 g

Total weight of K, Fe and oxalate in unknown sample = 25.70 + 12.27 + 58 = 95.97 g

Therefore water in unknown sample should be = total weight - combinedweight of K, Fe and oxalate

                                                                           = 100 g - 95.97 g = 4.03 g

Weight of water (hydrate) in unknown sample = 4.03 g

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