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Chemistry Question on Electrochemistry

Limiting molar conductivity of NH4OHNH_4OH [i.e. ∧∘m(NH4OH)][i.e. \,\wedge^\circ {m (NH_4OH)}] is equal to

A

∧m(NH4OH)∘+∧m(Nacl)∘−∧m(NaOH)∘{ \wedge^{\circ}_{m(NH_4OH)} + \wedge^{\circ}_{m(Nacl)} - \wedge^{\circ}_{m(NaOH)} }

B

∧m(NH4Cl)∘+∧m(NHOH)∘+∧m(NaCl)∘{ \wedge^{\circ}_{m(NH4Cl)} + \wedge^{\circ}_{m(NHOH)} + \wedge^{\circ}_{m(NaCl)} }

C

∧m(NH4Cl)∘+∧m(NaCl)∘−∧m(NaOH)∘{ \wedge^{\circ}_{m(NH4Cl)} + \wedge^{\circ}_{m(NaCl)} - \wedge^{\circ}_{m(NaOH)} }

D

Λm∘(NH4Cl)+Λm∘(NaOH)−Λm∘(NaCl)\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

Λm∘(NH4Cl)+Λm∘(NaOH)−Λm∘(NaCl)\Lambda^\circ_{m}\left(N H_{4} C l\right)+\Lambda_{m}^{\circ}(N a O H)-\Lambda_{m}^{\circ}(N a C l)

Explanation

Solution

The limiting molar conductivity that is denoted by Λ°. It is the measure of the conductivity of the solution that contains one mole of an electrolyte that is dissolved in the solution.

For the question, the limiting molar conductivity of NH4OH is given by:
Λ°(NH4Cl) +Λ°( NaOH) - Λ° (NaCl)

According to Kohlrausch law of independent migration of ions:

Λm(NH4OH)=Λm(NH4+)+Λm∗(OH−)\Lambda_{m}\left(N H_{4} O H\right)=\Lambda_{m}\left(N H_{4}^{+}\right)+\Lambda_{m}^{*}\left(O H^{-}\right)
=Λm(NH4+)+Λm(Cl−)+Λm∘(OH−)=\Lambda_{m}\left(N H_{4}^{+}\right)+\Lambda_{m}\left(C l^{-}\right)+\Lambda_{m}^{\circ}\left(O H^{-}\right)
=Λm∗(NH4+)−Λm−(Cl−)−Λmn(Na+)=\Lambda_{m}^{*}\left(N H_{4}^{+}\right)-\Lambda_{m}^{-}\left(C l^{-}\right)-\Lambda_{m}^{n}\left(N a^{+}\right)
=Λm∗(NH4Cl)+Λm−(NaOH)−Λm∘(NaCl)=\Lambda_{m}^{*}\left(N H_{4} C l\right)+\Lambda_{m}^{-}(N a O H)-\Lambda_{m}^{\circ}(N a C l)