Solubility, Ideal solution, deviation from Raoult's law

Solubility, Ideal Solution, Vapour Pressure and Deviation from Raoult's Law

In chemistry, the study of solutions is extremely important because most chemical reactions occur in solution form. When two or more substances mix together uniformly, the mixture formed is called a solution. A solution contains two major components: the solute and the solvent. The solute is the substance that dissolves, while the solvent is the substance that dissolves the solute. For example, when salt dissolves in water, salt is the solute and water is the solvent.

Understanding how substances dissolve and how solutions behave helps chemists explain many natural and industrial processes such as drug preparation, chemical manufacturing, biological reactions, and environmental chemistry. In this article we will study important concepts related to solutions such as solubility, ideal solution, vapour pressure, partial pressure, and positive and negative deviations from Raoult’s law.

1. Solubility

Solubility is defined as the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature and pressure. When the maximum amount of solute has dissolved, the solution becomes saturated. If less solute is present than the maximum amount, the solution is called an unsaturated solution.

For example, common salt dissolves easily in water. At room temperature, about 36 grams of sodium chloride can dissolve in 100 grams of water. If more salt is added after this limit, it will remain undissolved at the bottom of the container.

Solubility depends on several factors:

• Temperature
• Pressure (important for gases)
• Nature of solute and solvent

Generally, the solubility of solid substances in liquids increases with increase in temperature. However, the solubility of gases usually decreases when temperature increases. Pressure has a strong effect on gases; increasing pressure increases the solubility of gases in liquids.

2. Ideal Solution

An ideal solution is a solution that obeys Raoult's Law perfectly over the entire concentration range. In such solutions, the interactions between unlike molecules are almost the same as the interactions between like molecules.

This means that the forces between molecules of component A and component B are nearly equal to the forces between A-A and B-B molecules. Because of this similarity in intermolecular forces, mixing the two liquids does not produce any heat change.

The important characteristics of an ideal solution are:

• The solution follows Raoult’s law at all concentrations.
• No heat is absorbed or released during mixing.
• No change in volume occurs when the components are mixed.

Examples of nearly ideal solutions include mixtures such as benzene and toluene or hexane and heptane. These liquids have very similar molecular structures and intermolecular forces.

3. Vapour Pressure

Vapour pressure is the pressure exerted by the vapour of a liquid when the liquid and vapour are in dynamic equilibrium at a given temperature. When a liquid is kept in a closed container, some molecules escape from the liquid surface and enter the vapour phase.

As time passes, more molecules evaporate and the vapour concentration increases. Eventually, a stage is reached when the rate of evaporation becomes equal to the rate of condensation. At this point equilibrium is established and the pressure exerted by the vapour is called vapour pressure.

Different liquids have different vapour pressures. Liquids with weak intermolecular forces evaporate more easily and therefore have higher vapour pressure. Temperature also plays an important role. As temperature increases, molecules gain more kinetic energy and evaporation increases, resulting in higher vapour pressure.

4. Partial Pressure

When two or more gases are present in a mixture, each gas exerts its own pressure independently. This pressure exerted by an individual gas in a mixture is called its partial pressure.

According to Dalton’s law of partial pressures, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture.

For example, if a container contains oxygen, nitrogen and carbon dioxide, each gas contributes a certain pressure. If the partial pressure of oxygen is 200 mmHg, nitrogen is 500 mmHg, and carbon dioxide is 60 mmHg, the total pressure of the gas mixture will be:

Total Pressure = 200 + 500 + 60 = 760 mmHg

This concept is very important in chemistry, especially when studying gas mixtures, chemical reactions involving gases, and the behaviour of solutions containing volatile components.

5. Positive Deviation from Raoult's Law

In some solutions, the observed vapour pressure is higher than the vapour pressure predicted by Raoult’s law. Such solutions are said to show positive deviation from Raoult’s law.

Positive deviation occurs when the intermolecular forces between unlike molecules are weaker than those between like molecules. Because the attractive forces between different molecules are weaker, the molecules escape more easily into the vapour phase.

As a result, the vapour pressure of the solution becomes greater than expected.

Common examples of solutions showing positive deviation include:

• Ethanol and acetone
• Ethanol and benzene

In these mixtures, the interaction between different molecules is weaker, so evaporation becomes easier.

6. Negative Deviation from Raoult's Law

Negative deviation occurs when the vapour pressure of a solution is lower than the vapour pressure predicted by Raoult’s law.

This happens when the intermolecular forces between unlike molecules are stronger than the forces between like molecules. Because the molecules attract each other strongly, they remain in the liquid phase and do not escape easily into the vapour phase.

As a result, the vapour pressure of the solution decreases.

Examples of solutions showing negative deviation include:

• Acetone and chloroform
• Nitric acid and water

In these solutions strong intermolecular attractions such as hydrogen bonding are formed between the molecules of the two components.

Conclusion

The study of solutions plays an essential role in chemistry and many real-life applications. Concepts such as solubility, vapour pressure, and partial pressure help scientists understand how substances behave when mixed together.

Ideal solutions follow Raoult’s law perfectly, but many real solutions show deviations because the intermolecular forces between molecules are different. Positive deviation occurs when the forces between unlike molecules are weaker, while negative deviation occurs when these forces are stronger.

Understanding these concepts is very important for students studying chemistry, especially those preparing for competitive examinations and higher education in science. These principles are also applied in industries such as pharmaceuticals, chemical manufacturing, environmental science, and food technology.

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Written for educational purposes | Suitable for Class 11 and Chemistry learners

Amorphous and Crystalline Solids | Class 11 Chemistry

Amorphous and Crystalline Solids

Introduction

In Solid State Chemistry, solids are classified based on the arrangement of their particles. Two important types of solids are Crystalline Solids and Amorphous Solids. These two categories are important for understanding the structure and properties of many materials used in daily life and industry.

Atoms, molecules, or ions are the basic building blocks of solids. The way these particles arrange themselves determines whether the solid will be crystalline or amorphous. This concept is very important for students studying chemistry in Class 11 and Class 12.

Crystalline Solids

A crystalline solid is a solid in which the particles are arranged in a regular and repeating pattern in three dimensions. This regular arrangement forms a structure known as a crystal lattice.

Characteristics of Crystalline Solids

• Particles are arranged in an ordered structure.
• They have a sharp and definite melting point.
• They possess a definite geometrical shape.
• They show anisotropic properties (physical properties change with direction).

Examples of Crystalline Solids

• Diamond
• Quartz
• Sodium Chloride (NaCl)
• Ice
• Sugar crystals

Amorphous Solids

An amorphous solid is a solid in which the particles are not arranged in a regular pattern. The arrangement of particles is random and does not show long-range order.

Because of this irregular arrangement, amorphous solids do not have a definite melting point. Instead, they soften over a range of temperatures.

Characteristics of Amorphous Solids

• Particles are arranged randomly.
• They do not have a sharp melting point.
• They have irregular shapes.
• They show isotropic properties (same properties in all directions).

Examples of Amorphous Solids

• Glass
• Rubber
• Plastic
• Wax

Difference Between Crystalline and Amorphous Solids

Property	Crystalline Solid	Amorphous Solid
Arrangement of Particles	Regular and ordered	Random and disordered
Melting Point	Sharp and definite	Not sharp
Shape	Definite geometrical shape	Irregular shape
Physical Properties	Anisotropic	Isotropic

Conclusion

Both crystalline and amorphous solids are important in chemistry and material science. Crystalline solids have an ordered structure and definite melting point, while amorphous solids have a random structure and soften over a range of temperatures. Understanding these differences helps scientists design materials for electronics, construction, and modern technology.

For students preparing for competitive exams such as engineering entrance exams, learning the structure and properties of these solids is essential because many conceptual questions are asked from this topic.

Ion electron method (Half reaction method)

Ion–Electron Method (Half Reaction Method) in Redox Reactions

The Ion–Electron Method, also known as the Half Reaction Method, is a systematic way to balance redox reactions. It is especially useful in aqueous solutions and is widely used in electrochemistry, titration calculations, and competitive examinations such as NEET and JEE.

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What is a Redox Reaction?

A redox reaction is a chemical reaction in which oxidation and reduction occur simultaneously.

• Oxidation → Loss of electrons
• Reduction → Gain of electrons

In such reactions, one species loses electrons while another gains electrons. The Ion–Electron Method helps us balance both mass and charge properly.

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Why Do We Need the Ion–Electron Method?

In many redox reactions, especially those occurring in solution, balancing by simple inspection becomes difficult. The Ion–Electron Method provides a step-by-step scientific approach to balance:

• All atoms
• All charges
• Electrons transferred

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Steps to Balance Redox Reaction in Acidic Medium

Let us understand the method with an example:

Example Reaction:

MnO₄^- + Fe²⁺ → Mn²⁺ + Fe³⁺

Step 1: Separate into Two Half Reactions

Oxidation Half:

Fe²⁺ → Fe³⁺

Reduction Half:

MnO₄^- → Mn²⁺

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Step 2: Balance Atoms Other Than Oxygen and Hydrogen

Check if atoms except O and H are balanced. In this case, manganese and iron are already balanced.

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Step 3: Balance Oxygen Using Water (H₂O)

MnO₄^- contains 4 oxygen atoms. Add 4H₂O to the right side:

MnO₄^- → Mn²⁺ + 4H₂O

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Step 4: Balance Hydrogen Using H⁺

There are 8 hydrogen atoms on the right side. Add 8H⁺ to the left:

8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

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Step 5: Balance Charge Using Electrons (e^-)

Calculate total charge on both sides:

• Left side charge = +8 -1 = +7
• Right side charge = +2

To balance charge, add 5 electrons to the left side:

8H⁺ + MnO₄^- + 5e^- → Mn²⁺ + 4H₂O

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Step 6: Balance Oxidation Half

Fe²⁺ → Fe³⁺ + e^-

To equalize electrons (since reduction uses 5 electrons), multiply the oxidation half by 5:

5Fe²⁺ → 5Fe³⁺ + 5e^-

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Step 7: Add Both Half Reactions

Now add the two half reactions and cancel electrons:

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

✔ All atoms balanced
✔ Charges balanced
✔ Electrons canceled

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Balancing in Basic Medium

For basic medium, follow these steps:

1. First balance the reaction in acidic medium.
2. Add equal number of OH^- ions to both sides to neutralize H⁺.
3. Combine H⁺ and OH^- to form H₂O.
4. Cancel extra water molecules if present on both sides.

This converts the equation into basic medium conditions.

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Advantages of Ion–Electron Method

• Scientifically accurate
• Works for complex reactions
• Useful in electrochemistry
• Important for competitive exams
• Helps in understanding electron transfer clearly

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Applications

• KMnO₄ titration
• K₂Cr₂O₇ reactions
• Electrochemical cells
• Galvanic and electrolytic cells
• Industrial redox processes

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Conclusion

The Ion–Electron Method is one of the most reliable and systematic techniques to balance redox reactions. By separating oxidation and reduction processes, balancing atoms, and finally balancing charges using electrons, we ensure that both mass and charge are conserved.

Mastering this method will strengthen your understanding of electrochemistry and help you perform well in board examinations as well as competitive exams like NEET and JEE.

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Practice Tip: Try solving 4–5 redox reactions daily using this method to gain confidence.

Ionic Product under Ionic Equilibrium

1. Introduction to Ionic Equilibrium

Ionic equilibrium is an important part of physical chemistry that deals with the equilibrium established in electrolyte solutions due to partial or complete ionization of substances. When acids, bases, or salts are dissolved in water, they produce ions. In many cases, this ionization is reversible, and a dynamic equilibrium is established between ions and undissociated molecules.

The concept of ionic equilibrium helps in understanding the behavior of weak electrolytes, strength of acids and bases, solubility of salts, and precipitation reactions. One of the most important ideas derived from ionic equilibrium is the ionic product.

2. Meaning of Ionic Product

The ionic product of a solution is defined as the product of the molar concentrations of the ions present in solution, each raised to the power of its stoichiometric coefficient at any given moment.

For a general salt:

AxBy ⇌ xA^y+ + yB^x−

The ionic product (IP) is given by:

IP = [A^y+]^x [B^x−]^y

Ionic product represents the current state of the solution and does not necessarily indicate equilibrium conditions.

3. Ionic Product and Solubility Product

Ionic product is closely related to solubility product (Ksp), but the two are not the same. Solubility product is a constant value for a sparingly soluble salt at a given temperature, whereas ionic product can have different values depending on the concentrations of ions present in solution.

Ionic product can be less than, equal to, or greater than the solubility product. Comparison of IP with Ksp helps in predicting whether a precipitate will form or not.

4. Relationship between IP and Ksp

Case 1: IP < Ksp

The solution is unsaturated. More solute can dissolve, and no precipitation occurs.

Case 2: IP = Ksp

The solution is saturated and the system is in equilibrium. No precipitation occurs.

Case 3: IP > Ksp

The solution is supersaturated. Excess ions combine to form a solid precipitate.

5. Ionic Product of Water

Water is a weak electrolyte and undergoes slight ionization as shown below:

2H₂O ⇌ H₃O⁺ + OH⁻

The ionic product of water is given by:

K_w = [H⁺][OH⁻]

At 25°C, the value of Kw is:

K_w = 1.0 × 10⁻¹⁴

This constant is useful in determining pH, acidity, and basicity of solutions.

6. Ionic Product in Weak Electrolytes

Weak acids and weak bases do not ionize completely in solution. For a weak acid HA:

HA ⇌ H⁺ + A⁻

The ionic product is:

IP = [H⁺][A⁻]

At equilibrium, this ionic product becomes the acid dissociation constant (Ka). Similarly, for weak bases, ionic product leads to the base dissociation constant (Kb).

7. Ionic Product and Precipitation

When two electrolyte solutions are mixed, precipitation may occur if the ionic product exceeds the solubility product of the resulting salt.

Example:

Ag⁺ + Cl⁻ → AgCl(s)

Ionic product:

IP = [Ag⁺][Cl⁻]

If IP is greater than Ksp of AgCl, precipitation takes place.

8. Selective Precipitation

Selective precipitation is the method of separating ions in a mixture based on differences in their solubility products.

For example, silver chloride precipitates before lead chloride when chloride ions are added slowly because AgCl has a much smaller Ksp value.

This method is widely used in qualitative analysis.

9. Common Ion Effect and Ionic Product

The addition of a common ion increases the ionic product of a solution. This shifts the equilibrium in accordance with Le Chatelier’s principle and reduces the solubility of the salt.

Example:

AgCl(s) ⇌ Ag⁺ + Cl⁻

Addition of NaCl increases chloride ion concentration, thereby increasing IP and decreasing solubility of AgCl.

10. Numerical Example

If the concentration of Ag⁺ is 2 × 10⁻⁴ M and Cl⁻ is 1 × 10⁻⁶ M, then:

IP = (2 × 10⁻⁴) × (1 × 10⁻⁶) = 2 × 10⁻¹⁰

Since this value is greater than Ksp of AgCl, precipitation occurs.

11. Importance of Ionic Product

The concept of ionic product is important in predicting precipitation, understanding solubility, explaining common ion effect, qualitative salt analysis, environmental chemistry, and industrial processes.

12. Common Errors by Students

Students often confuse ionic product with solubility product, ignore stoichiometric coefficients, or use incorrect concentrations. Careful application of the concept avoids these errors.

13. Conclusion

Ionic product is a fundamental concept of ionic equilibrium that helps in understanding the behavior of electrolyte solutions. By comparing ionic product with solubility product, it becomes possible to predict precipitation and solubility behavior accurately. This concept is essential for board examinations and competitive exams like JEE and NEET.

Mendius Reaction and Gattermann Reaction

Mendius reaction and Gattermann reaction are important named reactions of organic chemistry. These reactions are commonly studied in Class 12, NEET, and JEE syllabi. Both reactions are used for the preparation of important organic compounds.

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1. Mendius Reaction

Mendius reaction is the reduction of nitriles to primary amines using sodium metal in alcohol. It is an important method for the preparation of aliphatic primary amines.

General Reaction

R–C≡N + 4[H] → R–CH₂–NH₂

Reagents Used

• Sodium metal (Na)
• Alcohol (ethanol or methanol)

Example

CH₃–CN + Na / C₂H₅OH → CH₃–CH₂–NH₂

Important Points

• Nitrile group is reduced to primary amine
• Carbon chain length increases by one carbon atom
• Sodium acts as a reducing agent
• Reaction proceeds through nascent hydrogen

Uses

• Preparation of primary aliphatic amines
• Used in organic synthesis

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2. Gattermann Reaction

Gattermann reaction is used for the introduction of aldehyde group into an aromatic ring. It is also known as a formylation reaction of aromatic compounds.

General Reaction

Ar–H + HCN + HCl → Ar–CHO

Catalysts Used

• Anhydrous aluminium chloride (AlCl₃)
• Cuprous chloride (CuCl)

Example

C₆H₆ + HCN + HCl → C₆H₅–CHO

In this reaction, benzene is converted into benzaldehyde.

Important Points

• Aldehyde (–CHO) group is introduced into aromatic ring
• Reaction mainly occurs at ortho and para positions
• Used for preparation of aromatic aldehydes

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Difference Between Gattermann and Gattermann–Koch Reaction

Gattermann Reaction	Gattermann–Koch Reaction
Uses HCN and HCl	Uses CO and HCl
Uses AlCl3 and CuCl as catalyst	Uses AlCl3 and CuCl as catalyst
Less commonly used	More commonly used

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One-Line Examination Answers

Mendius Reaction: Reduction of nitriles to primary amines using sodium metal in alcohol.

Gattermann Reaction: Formylation of aromatic compounds using HCN and HCl in presence of AlCl₃ and CuCl.

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Prepared for academic and examination purposes.

Reactions in Which Sodium Mercury Amalgam is Used

Reactions in Which Sodium–Mercury Amalgam (Na–Hg) is Used

Sodium–mercury amalgam, written as Na–Hg, is an important reagent used in both organic and inorganic chemistry. It mainly acts as a mild and controlled reducing agent. Because of its gentle action, it is preferred over metallic sodium in many reactions.

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What is Sodium–Mercury Amalgam?

Sodium amalgam is a solution of sodium metal in mercury. Mercury controls the high reactivity of sodium and allows the gradual release of electrons or nascent hydrogen.

Due to this property, sodium amalgam is widely used in laboratory reductions.

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1. Reduction of Aldehydes

Sodium amalgam reduces aldehydes to primary alcohols in aqueous or alcoholic medium.

General Reaction:

R–CHO + [H] → R–CH₂OH

Here, sodium amalgam acts as a source of nascent hydrogen.

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2. Reduction of Ketones

Ketones are reduced to secondary alcohols using sodium amalgam.

General Reaction:

R–CO–R′ + [H] → R–CHOH–R′

This reaction is useful when mild reduction conditions are required.

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3. Reduction of Nitro Compounds

Nitro compounds are reduced to amines using sodium amalgam.

General Reaction:

R–NO₂ + Na–Hg + H₂O → R–NH₂

This reaction is preferred when strong reducing agents like tin and hydrochloric acid are not suitable.

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4. Reduction of Unsaturated Compounds

Alkenes and alkynes can be reduced by sodium amalgam under controlled conditions.

Example:

R–CH=CH–R → R–CH₂–CH₂–R

The hydrogen produced from sodium amalgam adds across the multiple bond.

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5. Reduction of Oximes and Imines

Oximes and imines are reduced to corresponding amines.

Example:

R–CH=NOH + Na–Hg → R–CH₂–NH₂

This reaction is important in organic synthesis.

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6. Reduction of Carbonyl Compounds in Aqueous Medium

Sodium amalgam reduces carbonyl compounds such as aldehydes and ketones in aqueous medium without affecting sensitive functional groups.

Hence, it is useful for selective reduction reactions.

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7. Reduction of Metal Ions (Inorganic Chemistry)

Sodium amalgam is also used to reduce metal ions in solution.

Example:

Fe³⁺ + Na–Hg → Fe²⁺

This reaction is commonly used in redox chemistry experiments.

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8. Preparation of Alcohols from Acyl Chlorides

Acyl chlorides can be reduced to primary alcohols using sodium amalgam.

General Reaction:

R–COCl + Na–Hg + H₂O → R–CH₂OH

This method avoids harsh reducing conditions.

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Why Sodium–Mercury Amalgam is Preferred

• It is safer than metallic sodium
• It provides controlled reduction
• It is a mild reducing agent
• It is suitable for selective reactions

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One-Line Examination Answer

Sodium–mercury amalgam is used as a mild reducing agent in reactions such as reduction of aldehydes, ketones, nitro compounds, oximes, imines, and metal ions.

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Educational content prepared for chemistry students.

Why Chemical Equation is Used to Find Molecularity and Strength of KMnO4

Why Chemical Equation is Used to Find Molecularity and Strength of KMnO₄ Using M/20 Mohr’s Salt Solution

In volumetric analysis, especially in Class 11 and Class 12 chemistry practicals, potassium permanganate (KMnO₄) is commonly standardized using M/20 Mohr’s salt solution. A very common question asked in exams is:

“Why is a chemical equation required to find the molecularity and strength of KMnO₄?”

The answer lies in the stoichiometry of the reaction and the nature of KMnO₄.

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1. Importance of Chemical Equation in Volumetric Analysis

In titration calculations, we do not depend only on the volume of solutions. All calculations are based on the balanced chemical equation.

KMnO₄ does not react with Mohr’s salt in a simple 1:1 ratio. Therefore, without a chemical equation:

• The reaction ratio cannot be known
• The number of electrons transferred cannot be determined
• The equivalent weight and n-factor cannot be calculated

Hence, a chemical equation is compulsory to calculate the molecularity and strength of KMnO₄.

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2. Why Mohr’s Salt is Used

Mohr’s salt (FeSO₄·(NH₄)₂SO₄·6H₂O) is used because:

• It is a primary standard
• It has high purity
• It has stable composition
• Its molarity (M/20) is accurately known

KMnO₄ is not a primary standard because it decomposes on standing and may contain impurities. Therefore, KMnO₄ solution must be standardized using Mohr’s salt.

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3. Chemical Equation of the Reaction

The titration is carried out in acidic medium using dilute sulphuric acid. The balanced ionic equation is:

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

From the above equation, we observe that:

• 1 mole of KMnO₄ reacts with 5 moles of Fe²⁺
• KMnO₄ acts as an oxidizing agent
• Mohr’s salt acts as a reducing agent

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4. Determination of Molecularity (Molarity / Normality)

From the chemical equation, it is clear that one mole of KMnO₄ gains five electrons. Therefore:

n-factor of KMnO₄ = 5

Using the titration formula:

N₁V₁ = N₂V₂

The normality of Mohr’s salt is known, and the volumes are measured experimentally. Using this equation, the normality of KMnO₄ is calculated.

Molarity of KMnO₄ is then obtained using:

M = N / n-factor

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5. Determination of Strength of KMnO₄

Once the molarity of KMnO₄ is known, its strength is calculated by:

Strength (g/L) = Molarity × Molar Mass

Molar mass of KMnO₄ = 158 g mol⁻¹

Thus, the strength of KMnO₄ solution can be accurately determined.

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6. Why Chemical Equation is Essential (Exam Point)

• To know the exact reaction ratio
• To calculate the n-factor of KMnO₄
• To determine molarity and strength correctly
• Because the reaction is not 1:1

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7. One-Line Board Answer

Chemical equation is used because KMnO₄ reacts with Mohr’s salt in a fixed stoichiometric ratio, which is necessary to calculate the molecularity and strength of KMnO₄.

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Prepared for educational and examination purposes.