Question

$\Lambda^\circ_m (NH_4OH)$ is equal to ___________

• A. $\Lambda^\circ_m (NH_4OH) + \Lambda^\circ_m (NH_4Cl) - \Lambda^\circ_m (HCl)$
• B. $\Lambda^\circ_m (NH_4Cl) + \Lambda^\circ_m (NaOH) - \Lambda^\circ_m (NaCl)$
• C. $\Lambda^\circ_m (NH_4Cl) + \Lambda^\circ_m (NaCl) - \Lambda^\circ_m (NaOH)$
• D. $\Lambda^\circ_m (NaOH) + \Lambda^\circ_m (NaCl) - \Lambda^\circ_m (NH_4Cl)$

Hint:

To construct the $\Lambda^\circ_m$ for a weak electrolyte (e.g., $AB$), you generally need $\Lambda^\circ_m$ of its strong salt ($AX$), a strong base containing its cation ($BX$), and a strong acid containing its anion ($BY$), then $\Lambda^\circ_m (AB) = \Lambda^\circ_m (AX) + \Lambda^\circ_m (BY) - \Lambda^\circ_m (XY)$. In this case, we have $NH_4OH$, so we use $NH_4Cl$ (strong salt), $NaOH$ (strong base), and $NaCl$ (strong salt from remaining ions).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to express the limiting molar conductivity of a weak base, ammonium hydroxide ($\Lambda^\circ_m (NH_4OH)$), using Kohlrausch's Law of Independent Migration of Ions. This law allows calculating the limiting molar conductivity of a weak electrolyte from those of strong electrolytes.
Step 2: Key Formula or Approach:
Kohlrausch's Law states that $\Lambda^\circ_m$ of an electrolyte can be written as the sum of the limiting molar conductivities of its constituent ions:
\[ \Lambda^\circ_m (A_xB_y) = x\lambda^\circ_{A^{y+}} + y\lambda^\circ_{B^{x-}} \]
For weak electrolytes like $NH_4OH$, which do not dissociate completely, $\Lambda^\circ_m$ cannot be directly measured by extrapolation. Instead, it is calculated by combining $\Lambda^\circ_m$ values of strong electrolytes that share the relevant ions.
We need to find a combination that yields $\lambda^\circ_{NH_4^+} + \lambda^\circ_{OH^-}$.
Step 3: Detailed Explanation:
Consider the following strong electrolytes:
1. Ammonium chloride ($NH_4Cl$, a strong salt):
\[ \Lambda^\circ_m (NH_4Cl) = \lambda^\circ_{NH_4^+} + \lambda^\circ_{Cl^-} \]
2. Sodium hydroxide ($NaOH$, a strong base):
\[ \Lambda^\circ_m (NaOH) = \lambda^\circ_{Na^+} + \lambda^\circ_{OH^-} \]
3. Sodium chloride ($NaCl$, a strong salt):
\[ \Lambda^\circ_m (NaCl) = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} \]
To obtain $\Lambda^\circ_m (NH_4OH) = \lambda^\circ_{NH_4^+} + \lambda^\circ_{OH^-}$, we can sum the conductivities of $NH_4Cl$ and $NaOH$ and then subtract the conductivity of $NaCl$:
\[ (\lambda^\circ_{NH_4^+} + \lambda^\circ_{Cl^-}) + (\lambda^\circ_{Na^+} + \lambda^\circ_{OH^-}) - (\lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-}) \]
\[ = \lambda^\circ_{NH_4^+} + \lambda^\circ_{OH^-} \]
Thus,
\[ \Lambda^\circ_m (NH_4OH) = \Lambda^\circ_m (NH_4Cl) + \Lambda^\circ_m (NaOH) - \Lambda^\circ_m (NaCl) \]
Step 4: Final Answer:
The correct expression for $\Lambda^\circ_m (NH_4OH)$ is given by option (2).