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Standardize a Solution of Potassium

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1. To standardize a solution of potassium manganate(VII) by an iron(II) salt (ammonium iron(II) sulphate).

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In this experiment, it is a redox titration method to standardize a solution of potassium manganate(VII) by an iron(II) salt (ammonium iron(II) sulphate). So, the word of redox is related to the oxidation and reduction. Oxidation numbers describe the number of electrons the atom will gain or lose during a reaction. Each atom in an equation can be assigned an oxidation number according to certain rules.

Oxidation occurs when the oxidation number of an atom increases while reduction occurs when the oxidation number decreases.

Potassium manganate (VII) (KMnO4) solution is standardised by titration against the ammonium iron(II) sulphate, FeSO4 .(NH4)2SO4.6H2O. Potassium manganate is widely used as an oxidizing agent in volumetric analysis. While the ammonium iron(II) sulphate is used as a primary standard to standardize the KMnO4 solution. In this experiment, ammonium iron(II) sulphate crystals are the stable compound that remain as solid in room temperature.

Hence, the ammonium(II) sulphate solution is obtained by dissolving into the sulphuric acid, preferably oxygen.

During the titration process, the ammonium iron(II) sulphate ionises into iron(II), sulphate and ammonium ions. But, only the iron(II) ions (Fe2+) are oxidized by the managanate(VII) ions (MnO4-). The following equation represents the reaction:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

Throughout the experiment, there is no additional of indicator. Manganate(VII) is an intense dark purple colour. But, there is a colour change of manganate(VII) to Mn+2, which is from dark purple to light pink. At the end of the experiment, the colour of the potassium manganate disappears as it reacts with the iron(II) ions. This is because all the Fe2+ ions are fully reacted, whereas the extra drop of potassium manganate solution will make the titration mixture turn pink.


1. Calculate the number of moles of ammonium iron(II) sulphate crystals,

FeSO4 .(NH4)2SO4.6H2O in the weighed sample.

No. of moles of FeSO4 .(NH4)2SO4.6H2O

= mass of substance in gmolar mass in g

= 9.70 g56+32+416+214+4(1)+32+416+621+16

= 0.025 mol

2. Calculate the number of moles of Fe2+ ions in 10.0 or 25.0 cm3 of the solution pipetted.

No. of moles of Fe2+ = No. of moles of FeSO4 .(NH4)2SO4.6H2O

250.0 cm3 of FeSO4 .(NH4)2SO4.6H2O contain 0.025 mol of Fe2+ ions,

Thus, no. of moles of Fe2+ in 25.0cm3 = 0.025250 x 25

= 0.0025 mol

3. Calculate the number of moles of MnO4- which reacted during titration.

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

1 mole of MnO4- = 5 moles of Fe2+

Hence, no. of moles of MnO4-

= 0.00255

= 0.0005 mol

4. Calculate the concentration of the manganate(VII) ions, MnO4-, in mol dm-3 and in

g dm-3.

( a ) in mol dm-3

Concentration of MnO4-

= 0.0005 mol0.02663 dm-3

= 0.02 mol dm-3

( b ) in g dm-3

Mass = no. of moles × molar mass

= 0.0005 × 55+4(16)

= 0.06 g

Concentration of MnO4- = 0.06 0.02663

= 2.25 g dm-3

5. Calculate the mass of the potassium manganate(VII) ions KMnO4 in 1 dm3 of solution.

No. of moles of KMnO4 = no. of moles of MnO4

= 0.0005 mol

Mass of KMnO4 = no. of moles of KMnO4 × molar mass

= 0.0005 × 39+55+4(16)

= 0.08 g


Redox reactions are chemical reactions involving oxidation and reduction occurring simultaneously. Therefore, redox reaction is also known as oxidation-reduction reaction. The number of electrons lost and gained in the both half reactions must be equal.

The acidified ammonium iron(II) sulphate, ionises into iron(II), sulphate and ammonium ions. Manganate(VII) ions act as the oxidasing agent, accepting the electrons. Therefore, the iron(II) ions, Fe2+, are oxidized by the manganate (VII) ions in acid medium. As a result, it has undergoes reduction to become colourless manganese(II) ions, in which the oxidation number is decreased from +7 to +2.

Reduction half equation : MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Meanwhile, iron(II) ions act as the reducing agent, releasing the electrons to become iron(III) ions. So, each iron atom has undergone oxidation, in which the oxidation number is increased from +2 to +3. Thus, the colour changes from pale green to yellow.

Oxidation half equation: Fe2+ → Fe3+ + e-

Since the manganate(VII) ion is dark purple or violet in colour whereas the manganese (II) ion is nearly colorless, the endpoint in this experiment is determined by the appearing of the light pink in the solution. The redox reaction of this experiment is represented by the following overall ionic equation:

5Fe2+ +MnO4- + 8H+ → 5Fe3++ Mn2+ + 4H2O

In order to obtain the amount of potassium manganate(VII) which is used to fully react with acidified ammonium iron(II) sulphate, the initial and final readings of the burette is recorded and repeated 3 times, includes 1 for rough reading and another 2 for accurate readings. So, the average value of potassium manganate(VII) is calculated to get a more accurate result. In the summary of this experiment, 25.00 cm3 of the acidified ammonium iron(II) sulphate required 26.63 cm3 of the potassium manganate(VII) solution for reaction.

There are some precaution steps in this experiment in order to obtain the results more accurately and precisely. During the process of the experiment, the pipette is rinsed with a little sulphuric acid to remove water present inside the pipette. This is the way to prevent the water from diluting the acid that is poured inside the pipette. In the same way, burette is also rinsed with a little of potassium manganate(VII) solution. Besides, a piece of white tile that placed below the conical flask is used to enable us to detect the end point clearly, which the colourless solution changes to a light pink solution. Moreover, the eyes must be placed at the same level as the meniscus of the solution inside the burette in order to obtain an accurate of initial and final readings.

Reference Information

Molar Mass: This is the mass of one molecule of matter, expressed in atomic units of mass (this mass is 1 mole of substance).

Oxidation number: this is a conditional charge of the atoms of the chemical element in the compound, calculated from the assumption that all the bonds have an ionic type. The degrees of oxidation can have a positive, negative or zero value

Ammonium: it is a polyatomic cation. Its chemical form is displayed as NH4 +. Ammonium with anions forms ammonium salts, ammonium compounds. They are included in a large class of onium compounds.

FeSO4·(NH4)2SO4·6H2O: Mora salt is an inorganic compound, a double sulfuric acid salt of iron and ammonium. The molar mass is 392.1388 g/mol.
Iron (14,24 % mass), Hydrogen (5,14 % mass), Oxygen (57,12% mass), Sulfur (16,35% mass), Nitrogen (7,14 % mass).

FeSO4: ferrous sulfate, the inorganic compound. This is an iron salt of sulfuric acid with the formula FeSO4. Non-volatile, and has no smell. The anhydrous substance is colorless, opaque, very hygroscopic. Has a strong-astringent metallic taste. It is gradually weathered in the air (lose the crystallization water).

Potassium Permanganate: it is potassium permanganate, a potassium salt of the manganic acid. The chemical formula is KMnO4. On appearance, it is dark purple, even almost black crystals. In the process of dissolution in water, they form a brightly colored solution of fuchsia.

NH42SO4: Ammonium Sulphate synthetic (sulfuric-acid ammonium) (NH4) 2SO4-nitric-sulfur mineral fertilizer. Ammonium Sulphate contains 21% nitrogen and 24% sulfur. This is a chemically neutral crystalline salt of white color, readily soluble in water. Its hygroscopicity is weak, so for prolonged storage the flowability manganite: salts of unstable, non-existent in the free state of oxygenic manganese acids. They are in the following degrees of oxidation of V, VI, and VII. They also contain tetrahedral anions MnO43-, MnO42- and MnO4-, respectively.

Cite this Standardize a Solution of Potassium

Standardize a Solution of Potassium. (2017, Jun 29). Retrieved from https://graduateway.com/standardize-a-solution-of-potassium/

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