Question:medium

\(\Lambda^o_m\)(NH\(_4\)OH) is equal to

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To solve these problems quickly, think like building with Lego blocks. You want the ions from your target weak electrolyte (NH\(_4\)\(^+\) and OH\(^-\)). 1. Pick a strong electrolyte that provides the cation you need: NH\(_4\)Cl provides NH\(_4\)\(^+\). 2. Pick a strong electrolyte that provides the anion you need: NaOH provides OH\(^-\). 3. Now you have extra ions you don't want (Cl\(^-\) and Na\(^+\)). To remove them, subtract the \(\Lambda^0_m\) of the strong electrolyte made from these unwanted ions: NaCl. So, you get: (NH\(_4\)Cl) + (NaOH) - (NaCl).
Updated On: Apr 28, 2026
  • \(\Lambda^o_m\)(NH\(_4\)OH) + \(\Lambda^o_m\)(NH\(_4\)Cl) - \(\Lambda^o_m\)(HCl)
  • \(\Lambda^o_m\)(NH\(_4\)Cl) + \(\Lambda^o_m\)(NaOH) - \(\Lambda^o_m\)(NaCl)
  • \(\Lambda^o_m\)(NH\(_4\)Cl) + \(\Lambda^o_m\)(NaCl) - \(\Lambda^o_m\)(NaOH)
  • \(\Lambda^o_m\)(NaOH) + \(\Lambda^o_m\)(NaCl) - \(\Lambda^o_m\)(NH\(_4\)Cl)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is the algebraic sum of the limiting molar conductivities of its constituent ions. This allows the calculation of $\Lambda_{m}^{o}$ for weak electrolytes using values from strong electrolytes.
Step 2: Key Formula or Approach:
We want to find $\Lambda_{m}^{o}(NH_{4}OH) = \lambda^{o}(NH_{4}^{+}) + \lambda^{o}(OH^{-})$. We must algebraically combine the conductivities of strong electrolytes to isolate these two specific ions.
Step 3: Detailed Explanation:
Let's evaluate the correct algebraic manipulation. We need $NH_{4}^{+}$ and $OH^{-}$.
Consider Option (B):
Expression: $\Lambda_{m}^{o}(NH_{4}Cl) + \Lambda_{m}^{o}(NaOH) - \Lambda_{m}^{o}(NaCl)$
Expand each term into its constituent ions based on Kohlrausch's law:
$= [\lambda^{o}(NH_{4}^{+}) + \lambda^{o}(Cl^{-})] + [\lambda^{o}(Na^{+}) + \lambda^{o}(OH^{-})] - [\lambda^{o}(Na^{+}) + \lambda^{o}(Cl^{-})]$
Group the terms:
$= \lambda^{o}(NH_{4}^{+}) + \lambda^{o}(OH^{-}) + \lambda^{o}(Cl^{-}) + \lambda^{o}(Na^{+}) - \lambda^{o}(Na^{+}) - \lambda^{o}(Cl^{-})$
The $Na^{+}$ and $Cl^{-}$ ion conductivities perfectly cancel out:
$= \lambda^{o}(NH_{4}^{+}) + \lambda^{o}(OH^{-})$
This exact sum represents $\Lambda_{m}^{o}(NH_{4}OH)$.
Step 4: Final Answer:
Option (B) correctly represents the relationship.
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