Question

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

• A. \(\Lambda^o_m\)(NH\(_4\)OH) + \(\Lambda^o_m\)(NH\(_4\)Cl) - \(\Lambda^o_m\)(HCl)
• B. \(\Lambda^o_m\)(NH\(_4\)Cl) + \(\Lambda^o_m\)(NaOH) - \(\Lambda^o_m\)(NaCl)
• C. \(\Lambda^o_m\)(NH\(_4\)Cl) + \(\Lambda^o_m\)(NaCl) - \(\Lambda^o_m\)(NaOH)
• D. \(\Lambda^o_m\)(NaOH) + \(\Lambda^o_m\)(NaCl) - \(\Lambda^o_m\)(NH\(_4\)Cl)

Hint:

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).

Answer 1

The Correct Option is B

Step 1: Understanding Kohlrausch's Law and its Application.
Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte (\(\Lambda^0_m\)) is the sum of the limiting ionic conductivities of its constituent ions. This law is particularly useful for determining the \(\Lambda^0_m\) of weak electrolytes (like NH\(_4\)OH), which cannot be found by extrapolating a graph of \(\Lambda_m\) vs. \(\sqrt{C}\). We can calculate it using the \(\Lambda^0_m\) values of strong electrolytes.
Step 2: Expressing the target \(\Lambda^0_m\) in terms of ionic conductivities.
We want to find \(\Lambda^0_m\)(NH\(_4\)OH). According to Kohlrausch's law: \[ \Lambda^0_m(\text{NH}_4\text{OH}) = \lambda^0(\text{NH}_4^+) + \lambda^0(\text{OH}^-) \] Our goal is to combine the \(\Lambda^0_m\) values of strong electrolytes in a way that results in this expression.
Step 3: Analyzing the options.
Let's express the \(\Lambda^0_m\) of the strong electrolytes in each option in terms of their ionic conductivities. We need to combine them to get \(\lambda^0(\text{NH}_4^+) + \lambda^0(\text{OH}^-)\). Consider option (B): \(\Lambda^o_m\)(NH\(_4\)Cl) + \(\Lambda^o_m\)(NaOH) - \(\Lambda^o_m\)(NaCl)

• \(\Lambda^o_m\)(NH\(_4\)Cl) = \(\lambda^0(\text{NH}_4^+) + \lambda^0(\text{Cl}^-)\)
• \(\Lambda^o_m\)(NaOH) = \(\lambda^0(\text{Na}^+) + \lambda^0(\text{OH}^-)\)
• \(\Lambda^o_m\)(NaCl) = \(\lambda^0(\text{Na}^+) + \lambda^0(\text{Cl}^-)\)

Now, let's perform the combination: \[ [\lambda^0(\text{NH}_4^+) + \lambda^0(\text{Cl}^-)] + [\lambda^0(\text{Na}^+) + \lambda^0(\text{OH}^-)] - [\lambda^0(\text{Na}^+) + \lambda^0(\text{Cl}^-)] \] \[ = \lambda^0(\text{NH}_4^+) + \lambda^0(\text{Cl}^-) + \lambda^0(\text{Na}^+) + \lambda^0(\text{OH}^-) - \lambda^0(\text{Na}^+) - \lambda^0(\text{Cl}^-) \] The \(\lambda^0(\text{Cl}^-)\) and \(\lambda^0(\text{Na}^+)\) terms cancel out, leaving: \[ = \lambda^0(\text{NH}_4^+) + \lambda^0(\text{OH}^-) \] This is exactly equal to \(\Lambda^0_m\)(NH\(_4\)OH).
Step 4: Final Answer.
The correct combination is given in option (B).