Question

Which of the following expressions correctly relates the limiting molar conductivity (\(\Lambda_m^\circ\)) of aluminum sulfate, \(\text{Al}_2(\text{SO}_4)_3\), to its individual ionic components according to Kohlrausch's Law?

• A. \(\Lambda_m^\circ = \lambda^\circ(\text{Al}^{3+}) + \lambda^\circ(\text{SO}_4^{2-}) \)
• B. \(\Lambda_m^\circ = 2\lambda^\circ(\text{Al}^{3+}) + 3\lambda^\circ(\text{SO}_4^{2-}) \)
• C. \(\Lambda_m^\circ = 3\lambda^\circ(\text{Al}^{3+}) + 2\lambda^\circ(\text{SO}_4^{2-}) \)
• D. \(\Lambda_m^\circ = \frac{1}{2}\lambda^\circ(\text{Al}^{3+}) + \frac{1}{3}\lambda^\circ(\text{SO}_4^{2-}) \)

Hint:

Always scale the individual ionic values by the balance coefficients for molar conductivity. For equivalent conductivity, these stoichiometric coefficients are normalized.

Answer 1

The Correct Option is B

Step 1 : Understanding the Question:
The topic of this question is Electrolytic Conductance, specifically Kohlrausch's Law of Independent Migration of Ions. This law states that at infinite dilution (limiting molar conductivity), each ion makes a definite contribution towards the total molar conductivity of an electrolyte, regardless of the nature of the other ion with which it is associated. The question asks us to identify the correct mathematical summation for Aluminum Sulfate based on its stoichiometric dissociation.
Step 2 : Key Formulas and approach:
Kohlrausch’s law is mathematically expressed as:
\[ \Lambda_m^\circ = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_- \]
Where:
- $\Lambda_m^\circ$ is the limiting molar conductivity of the electrolyte.
- $\lambda^\circ_+$ and $\lambda^\circ_-$ are the limiting molar conductivities of the individual cation and anion.
- $\nu_+$ and $\nu_-$ are the stoichiometric coefficients (number of ions) produced per formula unit of the electrolyte.
The approach is to correctly write the dissociation equation for Aluminum Sulfate and identify the coefficients.
Step 3 : Detailed Explanation:

First, we must write the balanced dissociation equation for Aluminum Sulfate, $\text{Al}_2(\text{SO}_4)_3$, in an aqueous solution.

When one mole of $\text{Al}_2(\text{SO}_4)_3$ dissolves, it dissociates completely into its constituent ions: $\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$.

From this equation, we can see that:
- Number of Aluminum cations ($\nu_+$) = 2.
- Number of Sulfate anions ($\nu_-$) = 3.

According to Kohlrausch's Law, the total molar conductivity is the sum of the conductivities of all individual ions produced.

Therefore, we multiply the limiting molar conductivity of $\text{Al}^{3+}$ by its coefficient (2) and the limiting molar conductivity of $\text{SO}_4^{2-}$ by its coefficient (3).

Summing these together gives: $\Lambda_m^\circ = 2\lambda^\circ(\text{Al}^{3+}) + 3\lambda^\circ(\text{SO}_4^{2-})$.

This matches exactly with option (B). Option (A) is wrong because it ignores stoichiometry, while Option (C) swaps the coefficients incorrectly.

Step 4 : Final Answer:
Based on the dissociation stoichiometry of Aluminum Sulfate, the correct expression is option (B).