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 applies Kohlrausch's Law of independent migration of ions. This law states that the limiting molar conductivity of an electrolyte (\(\Lambda_m^0\)) equals the sum of the limiting ionic conductivities (\(\lambda^0\)) of its ions. We'll use the provided values for strong electrolytes to find the limiting molar conductivity of the weak electrolyte NH₄OH. We aim to find \(\Lambda_m^0(NH_4OH)\). According to Kohlrausch's Law:
\(\Lambda_m^0(NH_4OH) = \lambda^0(NH_4^+) + \lambda^0(OH^-)\)
Given:
\(\Lambda_m^0(Ba(OH)_2) = \lambda^0(Ba^{2+}) + 2\lambda^0(OH^-) = 523.28\)
\(\Lambda_m^0(BaCl_2) = \lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-) = 280.0\)
\(\Lambda_m^0(NH_4Cl) = \lambda^0(NH_4^+) + \lambda^0(Cl^-) = 129.8\)
Our goal is to combine these equations to get \(\Lambda_m^0(NH_4OH)\). We need one \(\lambda^0(NH_4^+)\) and one \(\lambda^0(OH^-)\).

• We can get \(\lambda^0(NH_4^+)\) from the third equation.
• We can get \(\lambda^0(OH^-)\) from the first equation, but it includes \(\lambda^0(Ba^{2+})\).
• We must cancel out \(\lambda^0(Cl^-)\) and \(\lambda^0(Ba^{2+})\).

Consider the combination: \(\Lambda_m^0(NH_4Cl) + (1/2)\Lambda_m^0(Ba(OH)_2) - (1/2)\Lambda_m^0(BaCl_2)\)
= \([\lambda^0(NH_4^+) + \lambda^0(Cl^-)] + (1/2)[\lambda^0(Ba^{2+}) + 2\lambda^0(OH^-)] - (1/2)[\lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-)]\)
= \(\lambda^0(NH_4^+) + \lambda^0(Cl^-) + (1/2)\lambda^0(Ba^{2+}) + \lambda^0(OH^-) - (1/2)\lambda^0(Ba^{2+}) - \lambda^0(Cl^-)\)
= \(\lambda^0(NH_4^+) + \lambda^0(OH^-)\)
= \(\Lambda_m^0(NH_4OH)\)
Therefore, we calculate \(\Lambda_m^0(NH_4OH)\) using the provided values:
\(\Lambda_m^0(NH_4OH) = \Lambda_m^0(NH_4Cl) + (1/2)\Lambda_m^0(Ba(OH)_2) - (1/2)\Lambda_m^0(BaCl_2)\)
= \(129.8 + (1/2)(523.28) - (1/2)(280.0)\)
= \(129.8 + 261.64 - 140.0\)
= \(251.44 S cm^2 mol^{-1}\)
This matches option (B).