Electrochemistry

This problem uses Kohlrausch's Law of independent migration of ions. The law states that the limiting molar conductivity of an electrolyte (Λ_m⁰) is the sum of the limiting ionic conductivities (λ⁰) of its constituent ions. We can use the given values for strong electrolytes to find the limiting molar conductivity of the weak electrolyte NH₄OH. We want to find Λ_m⁰(NH₄OH). According to Kohlrausch's Law:
Λ_m⁰(NH₄OH) = λ⁰(NH₄⁺) + λ⁰(OH^-)
We are given:
Λ_m⁰(Ba(OH)₂) = λ⁰(Ba²⁺) + 2λ⁰(OH^-) = 523.28
Λ_m⁰(BaCl₂) = λ⁰(Ba²⁺) + 2λ⁰(Cl^-) = 280.0
Λ_m⁰(NH₄Cl) = λ⁰(NH₄⁺) + λ⁰(Cl^-) = 129.8
Our goal is to combine these equations to get the expression for Λ_m⁰(NH₄OH). We need one λ⁰(NH₄⁺) and one λ⁰(OH^-).

• We can get λ⁰(NH₄⁺) from the third equation.
• We can get λ⁰(OH^-) from the first equation, but it comes with λ⁰(Ba²⁺).
• We need to cancel out λ⁰(Cl^-) and λ⁰(Ba²⁺).

Consider the combination: Λ_m⁰(NH₄Cl) + (1/2)Λ_m⁰(Ba(OH)₂) - (1/2)Λ_m⁰(BaCl₂)
= [λ⁰(NH₄⁺) + λ⁰(Cl^-)] + (1/2)[λ⁰(Ba²⁺) + 2λ⁰(OH^-)] - (1/2)[λ⁰(Ba²⁺) + 2λ⁰(Cl^-)]
= λ⁰(NH₄⁺) + λ⁰(Cl^-) + (1/2)λ⁰(Ba²⁺) + λ⁰(OH^-) - (1/2)λ⁰(Ba²⁺) - λ⁰(Cl^-)
= λ⁰(NH₄⁺) + λ⁰(OH^-)
= Λ_m⁰(NH₄OH)
So, we can calculate Λ_m⁰(NH₄OH) using the given values:
Λ_m⁰(NH₄OH) = Λ_m⁰(NH₄Cl) + (1/2)Λ_m⁰(Ba(OH)₂) - (1/2)Λ_m⁰(BaCl₂)
= 129.8 + (1/2)(523.28) - (1/2)(280.0)
= 129.8 + 261.64 - 140.0
= 251.44 S cm² mol^-1
This corresponds to option (B).