Exhaustive Guide: Conductance & Kohlrausch's Law | Class 12 Chemistry

Conductance & Kohlrausch's Law

Module 3 | CBSE Class 12 Chemistry | Electrochemistry Chapter

Table of Contents

• 1. Metallic vs. Electrolytic Conductance
• 2. Basic Terminology (Resistance, Conductivity)
• 3. Measurement of Conductivity & Cell Constant
• 4. Molar Conductivity (Λ_m)
• 5. Variation of Conductivity and Molar Conductivity with Concentration
• 6. Kohlrausch's Law of Independent Migration of Ions
• 7. Applications of Kohlrausch's Law
• 8. NCERT Solved Examples
• 9. Previous Year Questions (PYQs)

1. Metallic vs. Electrolytic Conductance

Substances that allow the passage of electricity are called conductors. There are two primary mechanisms of conduction:

Electronic (Metallic) Conductance	Electrolytic (Ionic) Conductance
Current is carried by the flow of electrons.	Current is carried by the flow of positive and negative ions.
No chemical decomposition occurs.	Involves chemical decomposition at the electrodes.
Conductivity decreases with an increase in temperature (due to kernel vibrations).	Conductivity increases with an increase in temperature (due to increased ion mobility).
Depends on the nature and structure of the metal, and the number of valence electrons.	Depends on the nature of the electrolyte, size of ions, nature of solvent (viscosity), and concentration.

2. Basic Terminology

A. Resistance (R) and Resistivity (ρ)

The electrical resistance (R) of any object is directly proportional to its length (l) and inversely proportional to its area of cross-section (A).

R = ρ × (l / A)

The constant of proportionality, ρ (rho), is called resistivity (or specific resistance).
Units: Resistance (R) is measured in Ohms (Ω). Resistivity (ρ) is measured in Ω m or Ω cm.

B. Conductance (G) and Conductivity (κ)

The inverse of resistance is called conductance (G). It is a measure of the ease with which current flows.

G = 1 / R = (A / ρ l) = κ × (A / l)

Units of G: Siemens (S) or Ω^-1 or mho.

Conductivity (κ - kappa): The inverse of resistivity is called conductivity (or specific conductance). It can be defined as the conductance of a material of 1 cm length and 1 cm² area of cross-section. (i.e., conductance of 1 cm³ volume of the solution).

Units of κ: S m^-1 or S cm^-1.

3. Measurement of Conductivity & Cell Constant

To measure the resistance of an ionic solution, we face two problems:
1. Passing Direct Current (DC) changes the composition of the solution (electrolysis). Solution: We use Alternating Current (AC).
2. A solution cannot be connected to a Wheatstone bridge like a metallic wire. Solution: We use a specially designed vessel called a conductivity cell.

Cell Constant (G*): For a given conductivity cell, the ratio of the distance between the electrodes (l) to the area of cross-section (A) remains constant.

G* = l / A

Since κ = G × (l / A), we get:
κ = G × G*   &implies;   G* = R × κ

Units of G*: m^-1 or cm^-1.

4. Molar Conductivity (Λ_m)

Conductivity (κ) depends on the concentration of ions. To compare the conducting power of different electrolytes, we must normalize for concentration. We use a quantity called Molar Conductivity.

Molar Conductivity (Λ_m): The conducting power of all the ions produced by dissolving one mole of an electrolyte in a given volume of solution.

Λ_m = κ / c

For calculations in chemistry (where κ is in S cm^-1 and M is Molarity in mol L^-1):
Λ_m = (κ × 1000) / M

Units of Λ_m: S cm² mol^-1.

5. Variation of Conductivity and Molar Conductivity with Concentration

This is one of the most frequently asked concepts in CBSE Board exams.

A. Variation of Conductivity (κ) with Dilution

Conductivity ALWAYS DECREASES with a decrease in concentration (increase in dilution) for both strong and weak electrolytes.

Reason: Conductivity is the conductance of 1 cm³ of solution. Upon dilution, the total volume increases, but the number of current-carrying ions per unit volume (per cm³) decreases. Hence, conductivity drops.

B. Variation of Molar Conductivity (Λ_m) with Dilution

Molar conductivity ALWAYS INCREASES with a decrease in concentration (increase in dilution).
This happens because the total volume (V) containing one mole of electrolyte increases significantly upon dilution, outweighing the decrease in κ.

For Strong Electrolytes (e.g., KCl, NaCl):

Strong electrolytes are completely dissociated in solution. At high concentrations, interionic attractions (solute-solute interactions) are strong, which retards the motion of ions. Upon dilution, ions move further apart, decreasing these attractive forces, thus increasing mobility and Λ_m.

The variation follows the Debye-Hückel-Onsager equation:

Λ_m = Λ^°_m - A √c

Where:
Λ^°_m = Limiting Molar Conductivity (molar conductivity as concentration approaches zero or infinite dilution).
A = Constant depending on the nature of solvent and temperature.
c = Concentration.

For Weak Electrolytes (e.g., CH₃COOH):

Weak electrolytes dissociate only partially. Upon dilution, their degree of dissociation (α) increases (Ostwald's dilution law). This drastically increases the total number of ions in the solution, leading to a steep, exponential increase in Λ_m at high dilutions.

Graphical Note: The plot of Λ_m vs √c for a strong electrolyte is a straight line, allowing extrapolation to zero concentration to find Λ^°_m. For a weak electrolyte, the curve becomes parallel to the y-axis, making extrapolation impossible. We need Kohlrausch's law to find Λ^°_m for weak electrolytes.

6. Kohlrausch's Law of Independent Migration of Ions

Friedrich Kohlrausch analyzed the Λ^°_m values of various strong electrolytes and observed a pattern. He deduced that at infinite dilution, where dissociation is complete and interionic attractions are zero, each ion makes a definite contribution towards molar conductivity, regardless of the other ion present.

Statement of Kohlrausch's Law: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Λ^°_m = ν₊ λ^°₊ + ν_- λ^°_-

Where λ^°₊ and λ^°_- are limiting molar conductivities of the cation and anion respectively. ν₊ and ν_- are the number of cations and anions generated per formula unit.

Example:
Λ^°_m (BaCl₂) = λ^°(Ba²⁺) + 2 λ^°(Cl^-)
Λ^°_m (CH₃COOH) = λ^°(CH₃COO^-) + λ^°(H⁺)

7. Applications of Kohlrausch's Law

A. Calculation of Λ^°_m for Weak Electrolytes

Since we cannot extrapolate the graph for weak electrolytes, we use Kohlrausch's law by algebraically adding and subtracting the Λ^°_m of strong electrolytes.
Example: To find Λ^°_m for Acetic Acid (CH₃COOH), we use strong electrolytes HCl, CH₃COONa, and NaCl:
Λ^°_m(CH₃COOH) = Λ^°_m(CH₃COONa) + Λ^°_m(HCl) - Λ^°_m(NaCl)

B. Calculation of Degree of Dissociation (α)

At any concentration 'c', if the molar conductivity is Λ^c_m, and the limiting molar conductivity is Λ^°_m, then the degree of dissociation is the ratio of the two:

α = Λ^c_m / Λ^°_m

C. Calculation of Dissociation Constant (K_a) for Weak Acids

Using Ostwald's dilution law for a weak acid (HA → H⁺ + A^-):

K_a = (c α²) / (1 - α)

Substituting α = Λ^c_m / Λ^°_m :
K_a = ( c × (Λ^c_m)² ) / [ Λ^°_m × (Λ^°_m - Λ^c_m) ]

8. NCERT Solved Examples (Step-by-Step)

NCERT Example 3.4: Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L^-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 S m^-1.

Solution:
Step 1: Find the Cell Constant (G*) using the 0.1 M solution
R = 100 Ω; κ = 1.29 S m^-1 = 0.0129 S cm^-1 (Converting to cm is easier for Λ_m later).
G* = R × κ = 100 Ω × 0.0129 S cm^-1 = 1.29 cm^-1

Step 2: Find Conductivity (κ) of the 0.02 M solution
The cell constant remains the same because it is the same cell.
R = 520 Ω; G* = 1.29 cm^-1
κ = G* / R = 1.29 / 520 = 2.48 × 10^-3 S cm^-1

Step 3: Find Molar Conductivity (Λ_m) of 0.02 M solution
C = 0.02 mol L^-1
Λ_m = (κ × 1000) / C
Λ_m = (2.48 × 10^-3 × 1000) / 0.02 = 2.48 / 0.02 = 124 S cm² mol^-1

NCERT Example 3.8: The molar conductivity of 0.001028 mol L^-1 acetic acid is 48.15 S cm² mol^-1. Calculate its dissociation constant if Λ^°_m for acetic acid is 390.5 S cm² mol^-1.

Solution:
Given:
C = 0.001028 mol L^-1
Λ^c_m = 48.15 S cm² mol^-1
Λ^°_m = 390.5 S cm² mol^-1

Step 1: Calculate degree of dissociation (α)
α = Λ^c_m / Λ^°_m
α = 48.15 / 390.5 = 0.1233

Step 2: Calculate Dissociation Constant (K_a)
K_a = (c α²) / (1 - α)
K_a = (0.001028 × (0.1233)²) / (1 - 0.1233)
K_a = (0.001028 × 0.0152) / 0.8767 = 1.56 × 10^-5 / 0.8767 = 1.78 × 10^-5 mol L^-1

9. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual & Give Reasons (1-2 Marks)

[CBSE 2018, 2020]

Q1. Why does the conductivity of a solution decrease with dilution?

Answer: Conductivity (κ) is the conductance of 1 cm³ of a solution. When a solution is diluted, the total volume increases, leading to a decrease in the number of ions per unit volume. Since fewer ions are available to carry the charge per cm³, the overall conductivity decreases.

[CBSE 2017, 2019]

Q2. State Kohlrausch's law of independent migration of ions. Why does the molar conductivity of a weak electrolyte increase steeply on dilution?

Answer: Statement: Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion.
Reason for steep increase: Weak electrolytes dissociate partially at normal concentrations. Upon dilution, the degree of dissociation (α) increases significantly. This large increase in the total number of ions in the solution causes a steep, exponential increase in molar conductivity.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): Limiting molar conductivity for weak electrolytes cannot be determined directly by extrapolation of the Λ_m vs √c graph.
Reason (R): The plot of Λ_m against √c for weak electrolytes becomes parallel to the y-axis at very low concentrations.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Due to the sudden increase in the degree of dissociation as concentration approaches zero, the curve becomes asymptotic to the y-axis, making linear extrapolation impossible. Hence, we use Kohlrausch's law.

Part C: Numerical Problems (3 Marks)

[CBSE 2016, 2021]

Q4. Calculate Λ^°_m for acetic acid, given that:
Λ^°_m (HCl) = 426 S cm² mol^-1
Λ^°_m (NaCl) = 126 S cm² mol^-1
Λ^°_m (CH₃COONa) = 91 S cm² mol^-1

Answer:
According to Kohlrausch's Law, we can rearrange the given strong electrolytes to form the weak electrolyte (CH₃COOH).
We need CH₃COO^- and H⁺ ions. We get them from CH₃COONa and HCl. We must subtract NaCl to remove the unwanted Na⁺ and Cl^- ions.

Λ^°_m (CH₃COOH) = Λ^°_m (CH₃COONa) + Λ^°_m (HCl) - Λ^°_m (NaCl)
Λ^°_m (CH₃COOH) = 91 + 426 - 126
Λ^°_m (CH₃COOH) = 517 - 126 = 391 S cm² mol^-1.

[CBSE 2015]

Q5. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.

Answer:
Given:
Concentration (M) = 0.20 mol L^-1
Conductivity (κ) = 0.0248 S cm^-1

Formula: Λ_m = (κ × 1000) / M
Λ_m = (0.0248 × 1000) / 0.20
Λ_m = 24.8 / 0.20 = 124 S cm² mol^-1.

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