Molar conductivity (Λm) is the conductivity of an electrolyte solution per unit concentration, and it provides a measure of how well the solution can conduct electricity. It depends on factors such as the concentration of ions and the mobility of the ions in the solution. As the concentration of the electrolyte decreases, the molar conductivity typically increases, reaching a limiting value at infinite dilution.

Equation for molar conductivity

Molar conductivity (Λm) can be defined as:

$\Lambda_m = \frac{\kappa}{c}$

Where:

• Λm: Molar conductivity (S·m²/mol)
• κ: Conductivity of the solution (S/m)
• c: Concentration of the electrolyte (mol/m³)

Limiting molar conductivity

The limiting molar conductivity (Λm,∞) is the molar conductivity of the electrolyte when the concentration approaches zero (infinite dilution). It represents the maximum conductivity achieved by the ions when they are not hindered by interactions with other ions.

$\Lambda_{m,\infty} = \lambda_+ + \lambda_-$

Where:

• Λm,∞: Limiting molar conductivity (S·m²/mol)
• λ+: Conductivity contribution of the cation (S·m²/mol)
• λ-: Conductivity contribution of the anion (S·m²/mol)

Degree of dissociation

The degree of dissociation (α) is the fraction of the total number of molecules that dissociate into ions in a solution. It is especially important for weak electrolytes, where only a portion of the electrolyte dissociates. The degree of dissociation can be calculated as:

$\alpha = \frac{\Lambda_m}{\Lambda_{m,\infty}}$

Where:

• α: Degree of dissociation (unitless, fraction between 0 and 1)
• Λm: Molar conductivity at a given concentration (S·m²/mol)
• Λm,∞: Limiting molar conductivity (S·m²/mol)

Example calculation

Consider the dissociation of acetic acid (CH₃COOH) in water. The limiting molar conductivity of acetic acid is 390 S·m²/mol, and its molar conductivity at a given concentration is 90 S·m²/mol. To find the degree of dissociation (α):

$\alpha = \frac{90}{390} = 0.231$

Therefore, the degree of dissociation of acetic acid at this concentration is 0.231, meaning 23.1% of the acetic acid molecules dissociate into ions.

Factors affecting degree of dissociation

1. Concentration: The degree of dissociation generally increases with decreasing concentration. At infinite dilution, all electrolytes dissociate completely, but at higher concentrations, some ions may pair up and form neutral molecules.
2. Temperature: Higher temperatures generally increase the degree of dissociation due to the increased kinetic energy of molecules.
3. Nature of the electrolyte: Strong electrolytes dissociate completely in solution, while weak electrolytes only partially dissociate.

Applications

1. Determining the ionization constant (Ka): The degree of dissociation is used to calculate the ionization constant of weak electrolytes.
2. Studying the conductivity of solutions: Molar conductivity helps in understanding how ions contribute to the conductivity of a solution.
3. Electrolyte solutions: It helps to design solutions with optimal conductivity for applications such as batteries and electrolytic cells.

Degree of ionization and the van 't Hoff factor

The degree of ionization (α) of a weak electrolyte can be determined from the ionization constant (Ka) and concentration (C) using the following relationship:

$K_a = \frac{\alpha \times C}{1 - \alpha}$

For small values of α (which is typical for weak electrolytes), we can approximate:

$\alpha \approx \sqrt{\frac{K_a}{C}}$

For example, for a 0.1 M acetic acid solution with a Ka of 1.8 × 10⁻⁵, the degree of ionization (α) is:

$\alpha \approx \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = \sqrt{1.8 \times 10^{-4}} = 0.0134$

Thus, only 1.34% of acetic acid dissociates into ions in this solution.

The van 't Hoff factor (i) describes the number of particles into which an electrolyte dissociates. For NaCl, i = 2 because it dissociates into two ions (Na⁺ and Cl⁻), while for Na₂SO₄, i = 3.

Calculations involving electrolytes

1. Molar conductivity calculation

For a solution of NaCl, if the conductivity (κ) is 1.25 S/m and the volume is 1 L, the molar conductivity (Λm) can be calculated as:

$\Lambda_m = \kappa \times V = 1.25 \times 1 = 1.25 \text{ S·m}^2/\text{mol}$

2. Osmotic pressure of electrolyte solutions

Osmotic pressure (π) for an electrolyte solution is calculated using the formula:

$\pi = i \times M \times R \times T$

Where:

• i is the van 't Hoff factor (e.g., 2 for NaCl)
• M is the molarity of the solution (mol/L)
• R is the gas constant (0.0821 L·atm/mol·K)
• T is the temperature in Kelvin

For a 1 M NaCl solution at 298 K, the osmotic pressure is:

$\pi = 2 \times 1 \times 0.0821 \times 298 = 48.9 \text{ atm}$

3. Freezing point depression and boiling point elevation

The depression in freezing point (ΔTf) and elevation of boiling point (ΔTb) for an electrolyte solution can be calculated using the formulas:

$\Delta T_f = i \times K_f \times m$

$\Delta T_b = i \times K_b \times m$

Where:

• i is the van 't Hoff factor (number of ions produced)
• Kf and Kb are the cryoscopic and ebullioscopic constants
• m is the molality of the solution (mol/kg of solvent)

Kohlrausch's law of independent ionic mobility states that the limiting molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the individual contributions from its constituent ions. Each ion contributes independently to the total conductivity, and these contributions are characteristic of the ion itself, irrespective of the other ions in the solution.

Equation for Kohlrausch's law

The limiting molar conductivity (Λm,∞) for an electrolyte dissociating into ions can be written as:

$\Lambda_{m,\infty} = \lambda_+ + \lambda_-$

Where:

• Λm,∞: Limiting molar conductivity of the electrolyte at infinite dilution (S·m²/mol)
• λ+: Molar conductivity of the cation at infinite dilution (S·m²/mol)
• λ-: Molar conductivity of the anion at infinite dilution (S·m²/mol)

Example calculation: sodium chloride (NaCl)

Let's consider the limiting molar conductivity of sodium chloride (NaCl) at infinite dilution. The individual ionic contributions at infinite dilution are:

$\Lambda_{m,\infty} = \lambda_{Na^+} + \lambda_{Cl^-}$

Given values at infinite dilution:

• λNa+ = 50.1 S·m²/mol
• λCl- = 76.3 S·m²/mol

The total limiting molar conductivity for NaCl is:

$\Lambda_{m,\infty} = 50.1 + 76.3 = 126.4 \text{ S·m}^2/\text{mol}$

Therefore, the limiting molar conductivity of NaCl at infinite dilution is 126.4 S·m²/mol.

Applications of Kohlrausch's law

• Estimating conductivity of electrolytes: Kohlrausch's law helps in estimating the molar conductivity of unknown electrolytes by using known ionic conductivities.
• Determining ionization constants: The law can be used to determine the degree of dissociation (α) of weak electrolytes.
• Analyzing ionic strength: It helps in studying how ionic strength affects conductivity in solutions.

Extension to weak electrolytes

For weak electrolytes, where dissociation is incomplete, Kohlrausch's law can still be applied. The degree of dissociation (α) can be estimated using the equation:

$\Lambda_m = \Lambda_{m,\infty} \times \alpha$

Where:

• Λm: Molar conductivity at a given concentration
• Λm,∞: Limiting molar conductivity at infinite dilution
• α: Degree of dissociation

Electrolytes are essential in maintaining biological functions, including nerve transmission, muscle contractions, and fluid balance. In industry, they are critical in processes like electroplating, batteries, and desalination.