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

Hint:

For weak electrolytes at infinite dilution: \begin{itemize} \item Use strong electrolytes to build ionic contributions \item Subtract equations to eliminate common ions \item Add required ionic combinations \end{itemize}

Answer 1

The Correct Option is B

Concept: Using Kohlrausch’s Law of Independent Migration of Ions: \[ \Lambda^\circ = \lambda^\circ_+ + \lambda^\circ_- \] We express unknown molar conductance in terms of known electrolytes. Step 1: Write ionic expressions \[ \Lambda^\circ (\text{Ba(OH)}_2) = \lambda^\circ_{\text{Ba}^{2+}} + 2\lambda^\circ_{\text{OH}^-} \] \[ \Lambda^\circ (\text{BaCl}_2) = \lambda^\circ_{\text{Ba}^{2+}} + 2\lambda^\circ_{\text{Cl}^-} \] \[ \Lambda^\circ (\text{NH}_4\text{Cl}) = \lambda^\circ_{\text{NH}_4^+} + \lambda^\circ_{\text{Cl}^-} \] Step 2: Eliminate common ions Subtract: \[ \text{Ba(OH)}_2 - \text{BaCl}_2 \] \[ 523.28 - 280 = 243.28 \] \[ 2(\lambda^\circ_{\text{OH}^-} - \lambda^\circ_{\text{Cl}^-}) = 243.28 \] \[ \lambda^\circ_{\text{OH}^-} - \lambda^\circ_{\text{Cl}^-} = 121.64 \] Step 3: Add NH$_4$Cl \[ \Lambda^\circ(\text{NH}_4\text{OH}) = \Lambda^\circ(\text{NH}_4\text{Cl}) + (\lambda^\circ_{\text{OH}^-} - \lambda^\circ_{\text{Cl}^-}) \] \[ = 129.8 + 121.64 \] \[ = 251.44 \text{ S cm}^2 \text{ mol}^{-1} \]