Question

Limiting molar conductivity of $NH_4OH $ $[i.e. \,\wedge^\circ {m (NH_4OH)}]$ is equal to

• A. ${ \wedge^{\circ}_{m(NH_4OH)} + \wedge^{\circ}_{m(Nacl)} - \wedge^{\circ}_{m(NaOH)} } $
• B. ${ \wedge^{\circ}_{m(NH4Cl)} + \wedge^{\circ}_{m(NHOH)} + \wedge^{\circ}_{m(NaCl)} } $
• C. ${ \wedge^{\circ}_{m(NH4Cl)} + \wedge^{\circ}_{m(NaCl)} - \wedge^{\circ}_{m(NaOH)} } $
• D. $\Lambda^\circ_{m}\left(N H_{4} C l\right)+\Lambda_{m}^{\circ}(N a O H)-\Lambda_{m}^{\circ}(N a C l)$

Answer 1

The Correct Option is D

The problem asks for the expression for the limiting molar conductivity of ammonium hydroxide (NH₄OH), denoted as \wedge^\circ_{m}({NH_4OH}). To solve this, we consider the concept of the limiting molar conductivity, which can be explained using Kohlrausch's Law. According to this law, the limiting molar conductivity of an electrolyte can be expressed as the sum of the limiting ionic conductivities of its constituent ions.

According to Kohlrausch's Law of Independent Migration of Ions:

\Lambda^\circ_{m} = \lambda^\circ_+ + \lambda^\circ_-

where:

• \lambda^\circ_+ is the limiting ionic conductivity of the cation.
• \lambda^\circ_- is the limiting ionic conductivity of the anion.

For $NH_4OH$, it dissociates into $NH_4^+$ and $OH^-$ ions. However, the compound indirectly involves a calculation process through a cycle involving other known electrolytes:

Given the options, we are required to determine which expression correctly represents \wedge^\circ_{m}({NH_4OH}).

Let's look at each one's components:

1. \wedge^\circ_{m(NH_4Cl)} + \wedge^\circ_{m(NaOH)} - \wedge^\circ_{m(NaCl)}:
   • From this expression, \wedge^\circ_{m(NH_4Cl)} = \lambda^\circ_{NH_4^+} + \lambda^\circ_{Cl^-}
   • \wedge^\circ_{m(NaOH)} = \lambda^\circ_{Na^+} + \lambda^\circ_{OH^-}
   • \wedge^\circ_{m(NaCl)} = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-}
   • Taking the sum and subtracting appropriately, we get: (\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^-}, which indeed represents the ions of NH₄OH upon dissociation.

Hence, the correct expression for the limiting molar conductivity of $NH_4OH$ is:

Correct Answer: \Lambda^\circ_{m}(NH_4Cl)+\Lambda_{m}^\circ(NaOH)-\Lambda_{m}^\circ(NaCl)