Question

The molar conductances of Ba(OH)$_2$, BaCl$_2$ and NH$_4$Cl at infinite dilution are 523.28, 280.0 and 129.8 S cm$^2$ mol$^{-1}$ respectively. The molar conductance of NH$_4$OH at infinite dilution will be

• A. 125.72 S cm$^2$ mol$^{-1}$
• B. 251.44 S cm$^2$ mol$^{-1}$
• C. 502.88 S cm$^2$ mol$^{-1}$
• D. 754.32 S cm$^2$ mol$^{-1}$

Hint:

Using a simple frame or just bolding for the box Key Points: Kohlrausch's Law: $\Lambda_m^0(\text{Electrolyte) = \nu_+ \lambda^0_+ + \nu_- \lambda^0_-$, where $\nu$ are stoichiometric coefficients. Limiting molar conductivity ($\Lambda_m^0$) of weak electrolytes can be found by algebraic manipulation of $\Lambda_m^0$ values of strong electrolytes containing the constituent ions. Ensure the ions you don't need (Ba$^{2+$, Cl$^-$) cancel out correctly. Pay attention to stoichiometry (e.g., using $\frac{1{2$ for Ba(OH)$_2$ and BaCl$_2$).

Answer 1

The Correct Option is B

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).