Electrochemistry includes the concept of equivalent conductance, which is the conductance of a volume of solution containing one equivalent of an electrolyte. Let’s study the idea of the equivalent conductance formula.

### Equivalent Conductance

The term “equivalent conductance” refers to the conductance (or “conducting power”) of all the ions (of a solution) created by dissolving one gram equivalent of an electrolyte in a specific solution.

We can state that an electrolytic solution’s conductance is influenced by the ion concentration present in the solution. Having comparable outcomes for various electrolytes is beneficial. It’s represented by the symbol âˆ§_{e}. From specific conductance, equivalent conductance is computed.

**Unit of Equivalent Conductance is ohm**^{-1}**cm**^{2}**eq**^{-1}**.**

** Equivalent Conductance Formula** is as follows:

âˆ§_{e}= K Ã— VWhere,

- âˆ§
_{e}= Equivalent Conductance,- K = Specific Conductance (Reciprocal of Specific Resistance),
- V = Volume (in ml) of 1 gm-equivalent electrolyte.

Also,

âˆ§_{e}= (K Ã— 1000) / NWhere,

N = Normality

### Derivation of Equivalent Conductance

Conductance of V cm

^{3}= âˆ§_{e}Conductance of 1 cm

^{3}= KTherefore, âˆ§

_{e}= K Ã— V â€¦(Equation 1)We are aware that the equation below provides information about a solution’s normality (N).

N = (n/V) Ã— 1000

âˆ´ V = (1000 Ã— n) / N

Number of equivalents, n = 1, for the electrolytic solution mentioned above.

V = 1000 / N

Substitute value of V in Equation 1,

âˆ´ âˆ§

_{e}= (K Ã— 1000) / N

### Conductors and Insulators

Conductors are substances that make it simple and unhindered for electrons to go from one end to the other. Conductors contain electric charges in the form of electrons, which facilitates the electrons’ free movement.

On the other hand, insulators are the kind of substances that obstruct the easy movement of electrons from one end to other. Any charge that is transferred through an insulator only settles at the point where the two materials first come together; it does not expand outward.

### Equivalent conductance at infinite dilution

The value of equivalent conductance rises as the solution becomes more diluted as ionization, or the amount of ions in a solution, rises. However, there comes a point where further dilution of the solutions is impossible, meaning it has no impact whatsoever on the solution’s concentration. The infinite dilution is the name given to the entire idea when diluting ceases.

Since a solution already has the maximum amount of solvent that may be added, infinite dilution is the condition in which no further concentration can be achieved with any amount of dilution. In this infinite dilution condition, all ions are fully dissociated.

### Kohlrauschâ€™s Law

According to Kohlrausch’s law, each ion contributes significantly to the equivalent conductance of the electrolyte at infinite dilution when dissociation is complete, regardless of the type of ion it is associated with. The value of equivalent conductance at infinite dilution for any electrolyte is the sum of the contributions of its constituent ions (cations and anions). We can therefore interpret it to mean that “conductivity of an electrolyte’s ions at infinite dilution is constant and does not depend on the nature of co-ions.”

Î»^{âˆž}_{eq}= Î»^{âˆž}_{c}+ Î»^{âˆž}_{a}Where,

- Î»
^{âˆž}_{eq}= Molar conductivity at infinity of dilution,- Î»
^{âˆž}_{c}= Conductivity of cation at infinity of dilution,- Î»
^{âˆž}_{a}= Conductivity of anion at infinity of dilution.

The term “limiting molar conductivity” refers to the molar conductivity that exists when the electrolyte concentration is almost zero.

When held between two electrodes with a unit area of cross-section and a unit distance, the volume of the solution that conducts and also contains one mole of electrolyte is known as the molar conductivity. Molar conductivity rises as concentration falls. The volume that makes up one mole of electrolytes increases, which results in an increase in the molar conductivity. When the electrolyte concentration becomes close to zero, the molar conductivity is known as the limiting molar conductivity.

**Uses of Kohlrauschâ€™s Law**

**Uses of Kohlrauschâ€™s Law**

- The molar conductivity at infinite dilution for the weak electrolytes is determined using Kohlrausch’s law. Calculating the molar conductivity of weak electrolytes at infinite dilution is highly challenging or impossible. due to the extremely low conductance of these sorts of solutions and the fact that the dissociation of these electrolytes is incomplete even at high dilutions.
- The solubility of a moderately soluble salt is determined using Kohlrausch’s law. Some salts are referred to as weakly or sparingly soluble salts because they only slightly dissolve in water. For instance, silver chloride, lead sulfate, barium sulfate, etc.
- The term “limiting molar conductivity” refers to the molar conductivity that exists when the electrolyte concentration is almost zero. We can calculate the limiting molar conductivity of an electrolyte using Kohlrausch’s law.

### Molar Conductivity

Molar conductivity, which can be determined by a solution’s ionic strength or salt concentration, is the conductance of a solution containing one mole of electrolyte. It is therefore not a constant.

Molar conductivity, then, is the sum of the conductivities of all the ions produced when a mole of an electrolyte is dissolved in a solution. The ability of an electrolyte to transmit electricity in a solution is generally assessed using the property of an electrolyte solution called molar conductivity. It is therefore not a constant. The unit of Molar conductivity is Sm^{2}mol^{-1}.

The following expression is used to numerically represent molar conductivity:

Î¼ = K/CWhere,

- K = Specific Conductivity,
- C = the concentration of moles per liter.

### Factors Affecting Equivalent Conductivity

- Temperature: As temperature rises, more ions are produced, which results in an increase in an electrolyte solution’s conductance.
- Strong electrolytes completely ionize, producing more ions as a result of which they have higher conductivities.
- On the other hand, weak electrolytes only experience partial ionization, which results in low conductivity in their solutions.
- Ionic size and mobility: As an ion’s size grows, so does its mobility, and its conductivity also decreases.
- Because of the solvent’s composition and viscosity, ionic mobility is reduced in more viscous solvents. The conductivity consequently declines.

### Sample Questions

**Question 1: What Does Equivalent Conductance Mean in Chemistry?**

** Answer**:

Equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes spaced 1 cm apart and big enough to hold the entire solution between them.

**Question 2: Write Two Factors Affecting Equivalent Conductivity.**

** Answer**:

Factors Affecting Equivalent Conductivity:

- Strong electrolytes fully ionize, resulting in the production of more ions and higher conductivities.
- Ionic size and mobility: An ion’s mobility increases with its size while its conductivity decreases.

**Question 3: List the situations in which the independent ion migration law of Kohlrausch is applicable.**

**Answer:**

For weak electrolytes, it is used to determine the limiting molar conductivity, level of dissociation, and dissociation constant. It is also employed in the computation of the salt’s solubility.

### Sample Problems

**Problems 1: A 0.7 N salt solution put between two platinum electrodes separated by 2 cm and covering an area of 6 cm**^{2}** has a 25-ohm resistance. calculate equivalent conductivity.**

**Solution:**

Since,

âˆ§

_{e}= (K Ã— 1000) / Nâˆ´ âˆ§

_{e}= 1/25 Ã— 2/6 Ã— 1000/0.7âˆ´ âˆ§

_{e}= 0.04 Ã— 0.33 Ã— 1428.5

âˆ´ âˆ§_{e}= 18.8562 ohm^{-1}cm^{2}eq^{-1}

**Problems 2: A salt solution in N/10 is found to have a resistance of 1.2 Ã— 10**^{3}** ohms. Calculate the solution’s equivalent conductance. 1.5 cm**^{-1}** is the cell constant.**

**Solution:**

Since,

âˆ§

_{e}= (K Ã— 1000) / NK = cell constant Ã— conductance

âˆ´ K = 1.5 Ã— (1/1.2 Ã— 10

^{3})âˆ´ K = 1.25 Ã— 10-3

âˆ§

_{e}= K Ã— 1000 / Nâˆ´ âˆ§

_{e}= 1.25 Ã— 10^{-3}/ (1/10)âˆ´ âˆ§

_{e}= 1.25 Ã— 10^{-3}/ 0.1

âˆ´ âˆ§_{e}= 12.5 Ã— 10^{-3}ohm^{-1}cm^{2}eq^{-1}

**Problems 3: Calculate the volume of the solution if the equivalent conductance is 10.255 ohm**^{-1}**cm**^{2}**eq**^{-1}** and the specific conductance is 2.17 cm**^{-1}**ohms**^{-1}**.**

**Solution:**

Since,

âˆ§

_{e}= K Ã— Vâˆ´ V = âˆ§

_{e}/ Kâˆ´ V = 10.255 / 2.17

âˆ´ V = 4.7258 cm^{3}

**Problems 4: Calculate the Equivalent conductance if the specific conductance is 1.83 cm**^{-1}**ohms**^{-1}** and the volume of the solution is 3.91 ml.**

**Solution:**

Since,

âˆ§

_{e}= K Ã— Vâˆ´ âˆ§

_{e}= 1.83 Ã— 3.91

âˆ´ âˆ§_{e}= 7.1553 ohm^{-1}cm^{2}eq^{-1}

**Problems 5: It is discovered that a salt solution in N/100 has a resistance of 5.21 Ã— 10**^{3}** ohms. Calculate the equivalent conductance of the solution. The cell constant is 3.71 cm**^{-1}**.**

**Solution:**

Since,

âˆ§

_{e}= (K Ã— 1000) / NK = cell constant Ã— conductance

âˆ´ K = 3.71 Ã— (1/5.21 Ã— 10

^{3})âˆ´ K = 0.7120 Ã— 10

^{-3}âˆ§

_{e}= K Ã— 1000 / Nâˆ´ âˆ§

_{e}= 0.7120 Ã— 10^{-3}/ (1/100)âˆ´ âˆ§

_{e}= 0.7120 Ã— 10^{-3}/ 0.01

âˆ´ âˆ§_{e}= 71.2 Ã— 10^{-3}ohm^{-1}cm^{2}eq^{-1}

**Problems 6: Calculate the Equivalent conductance if the specific conductance is 3.11 cm**^{-1}**ohms**^{-1}** and the volume of the solution is 2.95 ml.**

**Solution:**

Since,

âˆ§

_{e}= K Ã— Vâˆ´ âˆ§

_{e}= 3.11 Ã— 2.95

âˆ´ âˆ§_{e}= 9.1745 ohm^{-1}cm^{2}eq^{-1}

**Problems 7: If the equivalent conductance is 2.916 Ã— 10**^{3}** ohm**^{-1}**cm**^{2}**eq**^{-1}** and the specific conductance is 2.87 cm**^{-1}**ohms**^{-1}** then calculate the volume of the solution.**

**Solution:**

Since,

âˆ§

_{e}= K Ã— Vâˆ´ V = âˆ§

_{e}/ Kâˆ´ V = 2.916 Ã— 10

^{3}/ 2.87

âˆ´ V = 1.0160 Ã— 10^{3 }cm^{3}

**Problems 8: Calculate Î»**_{m}** for Cacl**_{2}** if Î»Ca**^{2+}** = 137.0 Scm**^{2}**mol**^{-1}** and Î»cl**^{–}** = 65.9 Scm**^{2}**mol**^{-1}**.**

**Solution:**

Since,

Î»

^{o}_{m}Cacl_{2}= Î»^{o }Ca^{2+}+ 2Î»^{o}cl^{–}âˆ´ Î»

^{o}_{m}Cacl_{2}= 137 + (2 Ã— 65.9)âˆ´ Î»

^{o}_{m}Cacl_{2}= 137 + 131.8

âˆ´ Î»^{o}_{m}Cacl_{2}= 268.8 Scm^{2}mol^{-1}

**Problems 9: Calculate Î»**^{o}_{m}** for MgSO**_{4}** if Î»**^{o}**Mg**^{2+}** = 119 Scm**^{2}**mol**^{-1}** and Î»**^{o}**SO**_{4}^{2-}** = 241 Scm**^{2}**mol**^{-1}**.**

**Solution:**

Since,

Î»

^{o}_{m}MgSO_{4}= Î»^{o}Mg^{2+}+ Î»^{o }SO_{4}^{2-}âˆ´ Î»

^{o}_{m}MgSO_{4}= 119 + 241

âˆ´ Î»^{o}_{m}MgSO_{4}= 360 Scm^{2}mol^{-1}