Kohlrausch's Law

Kohlrausch's Law

Shivani PooniaUpdated on 07 Aug 2025, 11:48 PM IST

Have you ever wondered how the conductivity of an dilute electrolyte behaves? why some electrolytes dissociate more efficiently than others? You will find these answer by reading Kohlrausch's Law. Kohlrausch's Law helps us understand the relationship between the conductivity of a solution and its concentration. Kohlrausch's Law states that the limiting molar conductivity $\left(\wedge_0\right)$ of an electrolyte is the sum of the limiting molar conductivities of its constituent ions, each ion's contribution being proportional to its concentration.

This Story also Contains

  1. Kohlraush's Law
  2. Application Of Kohlrausch's Law
  3. Molar Conductance At Infinite Dilution
  4. Some Solved Examples
  5. Summary
Kohlrausch's Law
Kohlrausch's Law

Kohlrausch's Law of Independent Migration of Ions was derived from his work on the conductivity of electrolytes. The law states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the contributions from the individual ions, each ion contributing independently to the total conductivity.

Kohlraush's Law

Kohlrausch law States that 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 that is, at infinite dilution, the contribution of any ion towards equivalent conductance is constant; it does not depend upon the presence of any ion.

$\Lambda_0=\lambda_0^{+}+\lambda_0^{-}$

Where:

  • $\Lambda_0$ is the limiting molar conductivity of the electrolyte at infinite dilution.
  • $\lambda_0^{+}$is the limiting molar conductivity of the cation.
  • $\lambda_0^{-}$is the limiting molar conductivity of the anion.
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For any electrolyte:
$
\begin{aligned}
& \mathrm{Px}_{\mathrm{XY}} \rightarrow \mathrm{XP}^{+\mathrm{Y}}+\mathrm{YQ}^{-\mathrm{X}} \\
& \Lambda^{\circ}\left(P_X Q_Y\right)=\mathrm{X} \lambda_{P^{+}}^{\circ}+\mathrm{Y}_{Q^{-X}}^{\circ} \\
& \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}
\end{aligned}
$

$\begin{aligned} & \Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=\left(\lambda_{\mathrm{H}^{+}}^{\infty}+\lambda_{\mathrm{Cl}^{-}}^{\infty}\right)+\left(\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}+\lambda_{\mathrm{Na}^{+}}^{\infty}\right)-\left(\lambda_{\mathrm{Na}^{+}}^{\infty}-\lambda_{\mathrm{Cl}^{-}}^{\infty}\right) \\ & =\Lambda_{\mathrm{HCl}}^{\infty}+\Lambda_{\mathrm{CH}, \mathrm{COONa}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty}\end{aligned}$

Application Of Kohlrausch's Law

  • Determination of $\Lambda_M^o$of a weak electrolyte:
    In the case of weak electrolytes, the degree of ionization increases which increases the value of Λm. However, it cannot be obtained by extrapolating the graph. The limiting value, Λm, for weak electrolytes can be obtained by Kohlrausch law.
  • To determine the degree of dissociation and equilibrium constant of weak electrolyte:
    $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
    C 0 0
    C-C$\alpha$ C$\alpha$ C$\alpha$

    Here $\mathrm{C}=$ Initial concentration
    $\alpha=$ Degree of dissociation
    $
    \alpha=\frac{\Lambda_{\mathrm{M}}}{\Lambda_{\mathrm{M}}}
    $

    Here $\Lambda^{\circ}$ or $\Lambda^{\infty}=$ Molar conductance at infinite dilution or zero concentration.

    $\begin{aligned} & \Lambda_{\mathrm{M}}=\text { Molar conductance at given conc. } \mathrm{C} \\ & \mathrm{K}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & \mathrm{K}=\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}\end{aligned}$

    $
    \mathrm{K}=\frac{\mathrm{C} \alpha^2}{1-\alpha}=\frac{\mathrm{C} \cdot\left(\Lambda / \Lambda_{\mathrm{M}}^o\right)^2}{\left(1-\Lambda / \Lambda_{\mathrm{M}}^o\right)^2}=\frac{\mathrm{C} \Lambda_M^2}{\Lambda^{\circ}\left(\Lambda^{\circ}-\Lambda_{\mathrm{m}}\right)}
    $

    These are Ostwald's relations.

  • To determine the solubility of salt and Ksp:

$\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$

If the solubility of AgCl is M and K and has values in $S \mathrm{~cm}^{-1}$ and $S \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, then

  • $
    \begin{aligned}
    & \Lambda^{\circ}=\frac{1000 \mathrm{~K}}{\mathrm{M}} \\
    & \Lambda^{\circ}=\lambda^{\circ} \mathrm{Ag}^{+}+\lambda^{\circ} \mathrm{Cl}^{-} \\
    & \mathrm{M}=\frac{1000 \mathrm{~K}}{\Lambda^{\circ}}
    \end{aligned}
    $
    . Here $\mathrm{M}=$ Solubility of AgCl
    Solubility product:
    $
    \begin{aligned}
    & \mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-1}\right] \\
    & \mathrm{As}\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{Cl}^{-}\right] \\
    & \mathrm{Ksp}=\frac{1000 \mathrm{~K}}{\Lambda^{\circ}} \times \frac{1000 \mathrm{~K}}{\Lambda^{\circ}} \\
    & \mathrm{Ksp}=\left(1000 \mathrm{~K} / \Lambda^{\circ}\right)^2
    \end{aligned}
    $

Molar Conductance At Infinite Dilution

When the addition of water doesn’t bring about any further change in the conductance of a solution, this situation is referred to as Infinite Dilution.

  • Strong Electrolytes: When infinite dilution is approached, the conductance of a solution of strong electrolyte approaches a limiting value and can be obtained by extrapolating the curve between Λm and c1/2
    The molar conductivity of strong electrolytes is found to vary with concentration as
    $\wedge_{\mathrm{m}}=\lambda_{\mathrm{m}}^0-\mathrm{B} \sqrt{\mathrm{c}}$
    where B is a constant depending upon the type of electrolyte, the nature of the solvent, and the temperature. This equation is known as the Debye Huckel-Onsage equation and is found to hold good at low concentrations.

    All electrolytes having the same formula type have the same value of B e.g. (KCl, NaCl) and (CaCl2, MgCl2)
  • Weak Electrolytes: When infinite dilution is approached, the conductance of a solution of the weak electrolyte increases very rapidly and thus, cannot be obtained through extrapolation. Also, the variation between Λm and c1/2 is not linear at low concentrations.

Recommended topic video on (Kohlrausch's Law)

Some Solved Examples

Example.1

1. The molar conductivities $\Lambda_{\mathrm{NaOAc}}$ and $\Lambda_{\mathrm{HCl}}^{\circ}$ at infinite dilution in water at $25^{\circ} \mathrm{C}$ are 91.0 and 426.2 S cm2/mol respectively. To calculate $\Lambda_{H O A}^\gamma$, the additional value required is

1)$\Lambda^{\circ} \mathrm{H}_2 \mathrm{O}$

2)$\Lambda_{\mathrm{KCl}}$

3)$\Lambda^{\circ} \mathrm{NaOH}^2$

4) (correct)$\Lambda^{\circ} \mathrm{NaCl}$

Solution

$\mathrm{CH}_3 \mathrm{COONa}+\mathrm{HCl} \rightarrow \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaCl}$

From the reaction,

$\Lambda_{\mathrm{CH} \mathrm{H}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}=\Lambda_{\mathrm{CH} \mathrm{H}_3 \mathrm{COOH}}^{\circ}+\Lambda_{\mathrm{NaCl}}^{\circ}$

or $\Lambda_{\mathrm{CH} \mathrm{H}_3 \mathrm{COOH}}^{\circ}=\Lambda_{\mathrm{CH} \mathrm{H}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}$

Thus to calculate the value of $\Lambda_{\mathrm{CH}}^3 \mathrm{COOH}$one should know the value of $\Lambda_{\mathrm{NaCl}}^{\circ}$ along with$\Lambda_{\mathrm{CH}}^3 \mathrm{COONa}$ and$\Lambda_{\mathrm{HCl}}^{\circ}$.

Hence, the answer is the option (4).

Example.2

2. The equivalent conductances of two strong electrolytes at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ (where ions move freely through a solution ) at 25°C are given below:

$\begin{aligned} & \Lambda_{\mathrm{CH}_3 \mathrm{COONa}}=91.0 \mathrm{Scm}^2 \text { /equiv. } \\ & \Lambda_{\mathrm{HCl}}^{\circ}=426.2 \mathrm{Scm}^2 / \text { equiv. }\end{aligned}$

What additional information/quantity one needs to calculate $\Lambda^{\circ}$ of an aqueous solution of acetic acid?

1)$\Lambda^{\circ}$of chloroacetic acid $\left(\mathrm{ClCH}_2 \mathrm{COOH}\right)$

2) (correct)$\Lambda^{\circ}$ of NaCl

3)$\Lambda^{\circ}$ of $\mathrm{CH}_3 \mathrm{COOK}$

4)The limiting equivalent conductance of $H^{+}\left(\lambda^{\circ}{ }_{H^{+}}\right)$

Solution

According to Kohlrausch’s law, the molar conductivity at infinite dilution $\left(\Lambda^{\circ}\right)$ for weak electrolyte, $\mathrm{CH}_3 \mathrm{COOH}$

$\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\circ}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}$

So, for calculating the value of $\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\circ}$ , value of $\Lambda_{\mathrm{NaCl}}^{\circ}$ should also be known.

Hence, the answer is the option (2).

Example.3

3. Match the column I with column II

a) Kohlrausch Law p) $\frac{\Lambda_m}{\Lambda_m^o}$

b) $\Lambda_m$ q) $\frac{1}{R} \times \frac{l}{A}$

c) K r) $\Lambda_{m \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2}^o=3 \lambda_{\mathrm{Ca}^{2+}}^o+2 \lambda_{P O_4^3}^o$

d) $\alpha$ s) $K \times \frac{1000}{M}$

1) (correct)a -r, b- s, c- q, d -p

2)a -s, b- r, c- q, d -p

3)a -r, b- s, c- p, d -q

4)a -s, b- r, c- p, d -q

Solution

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

According to Kohlrausch's law, $\Lambda_{\mathrm{eq}}^0=\Lambda_{\mathrm{c}}^0+\Lambda_{\mathrm{a}}^0$

Molar conductivity is given by, $\Lambda_{\mathrm{m}}=\frac{\kappa}{\mathrm{C}}$

The degree of dissociation is given by, $\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{n}}^0}$

So, Correct Match => a -r, b- s, c- q, d -p

Hence, the answer is the option (1).

Example.4

4. A = At infinite Dilution, the equivalent conductance is the sum contribution of its constituent ions.

R = At infinite dilution, each ion makes a definite contribution towards equivalent conductance of electrolyte irrespective of the nature of the ion it is associated with

1) (correct)A & R are correct and R explains A

2)A & R are correct and R doesn't explain A

3)A is correct but R is not

4)A & R are incorrect

Solution

Kohlrausch's law of independent migration of ions - The law states that limiting the molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. R explains A correctly
Hence, the answer is the option (1).

Example.5

5. The conductivity of 0.02 M Acetic acid is $7.8 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^2$$17.8 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^2$. Calculate its molar conductivity and if $\Lambda_{\mathrm{CH} \mathrm{H}_3 \mathrm{COOH}}^o$ is 390 S cm2 mol-1. Calculate its dissociation constant based on the given data.

1)$2.54 \times 10^{-7}$

2)$2.68 \times 10^{-7}$

3) (correct)$2 \times 10^{-6}$

4)$2.23 \times 10^{-7}$

Solution

Application of Kohlrausch's law - Calculation of molar conductivities of weak electrolytes at infinite dilution.

$\begin{aligned} & \Lambda=K \times \frac{1000}{M}=\frac{7.8 \times 10^{-5} \times 100}{0.02}=3.9 \\ & \alpha=\frac{\Lambda_m}{\Lambda_m^o}=\frac{3.9}{390.5}=0.01 \\ & K_\alpha=\frac{c \alpha^2}{1-\alpha}=\frac{0.02 \times(0.01)^2}{1}=2 \times 10^{-6}\end{aligned}$

Hence, the answer is the option (3).

EXAMPLE.6

6. The incorrect equation is:

1) (correct)$\left(\Lambda_m^{\circ}\right)_{N a B r}-\left(\Lambda_m^{\circ}\right)_{N a I}=\left(\Lambda_m^{\circ}\right)_{K B r}-\left(\Lambda_m^{\circ}\right)_{N a B r}$

2)$\left(\Lambda_m^{\circ}\right)_{\mathrm{NaBr}}-\left(\Lambda_m^{\circ}\right)_{\mathrm{NaCl}}=\left(\Lambda_m^{\circ}\right)_{\mathrm{KBr}}-\left(\Lambda_m^{\circ}\right)_{\mathrm{KCl}}$

3)$\left(\Lambda_m^{\circ}\right)_{K C l}-\left(\Lambda_m^{\circ}\right)_{N a C l}=\left(\Lambda_m^{\circ}\right)_{K B r}-\left(\Lambda_m^{\circ}\right)_{N a B r}$

4)$\left(\Lambda_m^{\circ}\right)_{\mathrm{H}_2 \mathrm{O}}=\left(\Lambda_m^{\circ}\right)_{\mathrm{HCl}}+\left(\Lambda_m^{\circ}\right)_{\mathrm{NaOH}}-\left(\Lambda_m^{\circ}\right)_{\mathrm{NaCl}}$

Solution

Kohlrausch's law follows here

$\begin{aligned} & \left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{NaBr}}-\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{NaI}}=\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{KBr}}-\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{NaBr}} \\ & \left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Na}}+\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Br}}-\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Na}}-\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{I}}=\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{K}}+\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Br}}-\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Na}}+\left(\Lambda_{\mathrm{m}}^0\right)_{\mathrm{Br}}\end{aligned}$

Both sides are not equal.

Hence, the answer is the option(1).

Practice More Questions With The Link Given Below

Effect of Dilution on Conductance, Λm, Λeq and Conductivity Practice Question and MCQs
Kohlrausch's Law Practice Question and MCQs

Summary

Kohlrausch’s Law provided a fundamental understanding of ionic mobility and conductivity. Kohlrausch’s Law allows for the calculation of the individual mobilities of ions in an electrolyte solution. By measuring the conductivity of the solution, one can determine the mobility of each ion, which is crucial for understanding how ions move in various conditions.

Frequently Asked Questions (FAQs)

Q: What is the relationship between Kohlrausch's Law and the Beer-Lambert law in the context of solution chemistry?
A:

Both Kohlrausch's Law

Q: How does the hydration of ions affect their contribution to conductivity in the context of Kohlrausch's Law?
A:

Ion hydration affects the mobility of ions in solution, which in turn influences their limiting molar ionic conductivity. Generally, smaller ions with higher charge density are more heavily hydrated, which can reduce their mobility and thus their contribution to conductivity as described by Kohlrausch's Law.

Q: What is the relationship between Kohlrausch's Law and the concept of ionic atmospheres?
A:

Kohlrausch's Law applies at infinite dilution where ionic atmospheres are negligible. As concentration increases, ionic atmospheres form around each ion, leading to deviations from the law. Understanding these deviations helps in developing more comprehensive theories of electrolyte conductivity.

Q: How can Kohlrausch's Law be used to estimate the limiting molar conductivity of ions that cannot exist independently in solution?
A:

For ions that cannot exist independently in solution (like OH- or H+), Kohlrausch's Law can be used indirectly. By measuring the conductivity of compounds containing these ions and subtracting the known contributions of other ions, the limiting molar conductivity of these special ions can be estimated.

Q: What is the significance of Kohlrausch's Law in the development of super-ionic conductors?
A:

While Kohlrausch's Law itself doesn't directly apply to solid-state ionic conductors, the principles of additive ionic conductivities and the importance of ion mobility that it embodies are crucial in understanding and developing super-ionic conductors for applications like solid-state batteries.

Q: How does Kohlrausch's Law help in understanding the concept of conductivity bridges used in analytical chemistry?
A:

Conductivity bridges measure the resistance of electrolyte solutions. Kohlrausch's Law helps in interpreting these measurements by providing a theoretical framework for how different ions contribute to overall conductivity, especially in dilute solutions where the law is most applicable.

Q: Can Kohlrausch's Law be applied to non-electrolyte solutions?
A:

Kohlrausch's Law specifically applies to electrolyte solutions and cannot be directly applied to non-electrolyte solutions. Non-electrolytes do not dissociate into ions and thus do not contribute significantly to electrical conductivity in the way that Kohlrausch's Law describes.

Q: How does Kohlrausch's Law relate to the concept of ion conductance in ion-selective electrodes?
A:

While ion-selective electrodes operate on different principles, Kohlrausch's Law provides insight into how individual ions contribute to overall solution conductivity. This understanding is valuable in interpreting the behavior of reference solutions and in understanding the limitations of conductivity-based measurements in complex solutions.

Q: How does Kohlrausch's Law relate to the concept of equivalent conductivity?
A:

Equivalent conductivity is similar to molar conductivity but is based on the equivalent concentration rather than the molar concentration. Kohlrausch's Law can be expressed in terms of equivalent conductivity by adjusting for the number of equivalents per mole of electrolyte.

Q: Can Kohlrausch's Law be applied to polyelectrolytes?
A:

Applying Kohlrausch's Law to polyelectrolytes is challenging due to their complex structure and behavior. While the principle of additive ionic conductivities still holds, factors like conformational changes, counterion condensation, and intramolecular interactions complicate the analysis and limit the direct application of the law.