Question

When potassium permanganate is titrated against ferrous ammonium sulphate in acidic.. medium, the equivalent weight of potassium permanganate is: (a) \(\frac{\text { molecular weight }}{3}\) (b) \(\frac{\text { molecular weight }}{5}\) \(\therefore \therefore\) (c) \(\frac{\text { molecular weight }}{2}\) (d) \(\frac{\text { molecular weight }}{10}\)

Short answer

\( \frac{\text{Molecular weight}}{5} \)

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation for the redox reaction of potassium permanganate (KMnO_4) with ferrous ammonium sulphate (FeSO_4.(NH_4)_2SO_4.6H_2O) in an acidic medium is as follows:MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2OAnd for ferrous ion,Fe^2+ → Fe^3+ + e^-

Determine the n-factor for potassium permanganate

The n-factor for a substance in a redox reaction is the number of electrons that each molecule can accept or donate. From the equation, we see that each molecule of KMnO_4 (potassium permanganate) accepts 5 electrons. Therefore, the n-factor of potassium permanganate is 5.

Calculate the equivalent weight of potassium permanganate

The equivalent weight of a compound is calculated by dividing its molecular weight (M) by its n-factor.Equivalent weight of KMnO_4 = \( \frac{Molecular weight of KMnO_4}{n-factor of KMnO_4} \)Where the n-factor of KMnO_4 is 5. As a result, the equivalent weight of potassium permanganate will be:Equivalent weight of KMnO_4 = \( \frac{Molecular weight of KMnO_4}{5} \)

Key concepts

Understanding Redox Reactions

Redox reactions, or oxidation-reduction reactions, are processes where electrons are transferred between two substances. In the case of potassium permanganate titrated against ferrous ammonium sulphate, the change in oxidation states of the atoms involved in the reaction is essential. Potassium permanganate (KMnO₄) acts as an oxidizing agent and reduces while the ferrous ion (Fe²⁺) is the reducing agent and gets oxidized. This transfer of electrons is crucial in determining the n-factor of the substances involved in a redox reaction, which in turn is key to understanding the stoichiometry of the reaction. The balanced chemical equation showcases that the MnO₄^- ion accepts five electrons to convert to Mn²⁺, while Fe²⁺ loses an electron to become Fe³⁺. Understanding the balanced equation is fundamental because it guides us toward quantitative aspects of the reaction, such as the number of moles of reactants needed and the calculation of equivalent weights.

Determining n-factor in Chemistry

In chemistry, the n-factor, also known as the valence factor, is a term used to indicate the change in oxidation states of atoms in a redox reaction or the number of protons or hydroxide ions a substance can donate or accept in an acid-base reaction. For redox reactions, the n-factor is determined by the number of electrons gained or lost by one mole of a substance.

For potassium permanganate, which serves as an oxidizing agent, the n-factor is 5 because one mole of KMnO₄ can accept 5 moles of electrons to undergo a reduction. This n-factor is vital when we perform the calculations for equivalent weight, especially in titration problems. It's a conversion factor that helps relate the molecular weight of a substance to its reacting capacity, giving us a grasp of its chemical equivalent in reaction stoichiometry. Students often conceptualize n-factor as the 'bridge' between actual weight and reactive capacity.

Titration Calculations

Titration is a quantitative analytical technique used to determine the concentration of an unknown solution using a solution of known concentration (titrant). In the context of the potassium permanganate and ferrous ammonium sulphate titration, calculations involve using the n-factor to find the equivalent weight of the titrant.

In such a titration, the stoichiometry of the reaction implies how much of one reactant will react with a given amount of another. To find the equivalent weight of potassium permanganate, divide its molecular weight by the n-factor (which, in this case, is 5). This calculation provides the amount of the substance that will react with one equivalent of the titratee. It is crucial in titration to know the equivalent weight, as it enables us to calculate how much of the titrant is needed to fully react with a certain amount of the substance being titrated. For students, mastering titration calculations is significant for laboratory work and understanding concepts of molarity, normality, and stoichiometry in reactions.