Question

From the following molar conductivities at infinite dilution, \( \Lambda_m^0 \), for \( \mathrm{NH_4OH} \) is
\( \Lambda_m^0 \) for \( \mathrm{Ba(OH)_2} = 446.8 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{BaCl_2} = 241.6 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{NH_4Cl} = 130 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)

Hint:

Use subtraction to eliminate common ions quickly in Kohlrausch law problems.

Answer 1

Step 1: Understanding the Concept:
According to Kohlrausch's Law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the molar ionic conductivities of its constituent ions. This allows us to calculate the \(\Lambda_{m}^{\circ}\) of weak electrolytes using values from strong electrolytes.
Step 2: Key Formula or Approach:
To find \(\Lambda_{m}^{\circ}(NH_{4}OH)\), we need the values for \(NH_{4}^{+}\) and \(OH^{-}\).
The relation is:
\[ \Lambda_{m}^{\circ}(NH_{4}OH) = \Lambda_{m}^{\circ}(NH_{4}Cl) + \frac{1}{2}\Lambda_{m}^{\circ}(Ba(OH)_{2}) - \frac{1}{2}\Lambda_{m}^{\circ}(BaCl_{2}) \]
Step 3: Detailed Explanation:
Given:
\(\Lambda_{m}^{\circ}(Ba(OH)_{2}) = 446.8\)
\(\Lambda_{m}^{\circ}(BaCl_{2}) = 241.6\)
\(\Lambda_{m}^{\circ}(NH_{4}Cl) = 130.0\)

The formula expands to:
\[ \Lambda_{m}^{\circ}(NH_{4}OH) = 130.0 + \frac{1}{2}(446.8) - \frac{1}{2}(241.6) \]
\[ \Lambda_{m}^{\circ}(NH_{4}OH) = 130.0 + 223.4 - 120.8 \]
\[ \Lambda_{m}^{\circ}(NH_{4}OH) = 353.4 - 120.8 = 232.6 \]
Step 4: Final Answer:
The molar conductivity at infinite dilution for \(NH_{4}OH\) is 232.6 \(\Omega^{-1} \text{ cm}^{2} \text{ mol}^{-1}\).