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.
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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.
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:
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}
$
$\begin{aligned} & \Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=\left(\lambda_{\mathrm{H}^{+}}^{\infty}+\lambda_{\mathrm{Cl}^{-}}^{\infty}\right)+\left(\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}+\lambda_{\mathrm{Na}^{+}}^{\infty}\right)-\left(\lambda_{\mathrm{Na}^{+}}^{\infty}-\lambda_{\mathrm{Cl}^{-}}^{\infty}\right) \\ & =\Lambda_{\mathrm{HCl}}^{\infty}+\Lambda_{\mathrm{CH}, \mathrm{COONa}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty}\end{aligned}$
$
\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.
$\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
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
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.
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.
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).
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)
Both Kohlrausch's Law
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.
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.
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.
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.
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.
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.
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.
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.
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.