Equivalence conductance at infinite dilution of , and are 129.8, 217.4 and 108.9 Ω^-1 cm^2 mol^-1, r — Electrochemistry Chemistry Question
Question
Equivalence conductance at infinite dilution of $NH_4Cl$, $NaOH$ and $NaCl$ are 129.8, 217.4 and 108.9 Ω^-1 cm^2 mol^-1, respectively. If the equivalent conductance of 0.01 N - $NH_4OH$ at this temperature is 9.532 Ω^-1 cm^2 mol^-1, then the degree of dissociation of $NH_4OH$ at this temperature is
💡 Solution & Explanation
\noindent\textbf{Step 1 - Application of Kohlrausch's Law} According to Kohlrausch's law of independent migration of ions, the equivalent conductance of a weak electrolyte at infinite dilution ($\Lambda_{eq}^\infty$) can be determined by combining the equivalent conductances of strong electrolytes containing its constituent ions at infinite dilution. For ammonium hydroxide (\ce{NH4OH}), this relation is: $$\Lambda_{eq}^\infty(\ce{NH4OH}) = \Lambda_{eq}^\infty(\ce{NH4Cl}) + \Lambda_{eq}^\infty(\ce{NaOH}) - \Lambda_{eq}^\infty(\ce{NaCl})$$ \noindent\textbf{Step 2 - Substitution and Calculation of $\Lambda_{eq}^\infty(\ce{NH4OH})$} Substituting the given equivalent conductance values at infinite dilution of \ce{NH4Cl}, \ce{NaOH}, and \ce{NaCl}: $$\Lambda_{eq}^\infty(\ce{NH4OH}) = 129.8\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1} + 217.4\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1} - 108.9\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda_{eq}^\infty(\ce{NH4OH}) = 238.3\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$$ \noindent\textbf{Step 3 - Calculation of the Degree of Dissociation ($\alpha$)} The degree of dissociation ($\alpha$) is the ratio of equivalent conductance at a given concentration ($\Lambda_{eq}^c$) to the equivalent conductance at infinite dilution ($\Lambda_{eq}^\infty$): $$\alpha = \frac{\Lambda_{eq}^c}{\Lambda_{eq}^\infty}$$ Substituting the given equivalent conductance at $0.01\text{ N}$ concentration ($\Lambda_{eq}^c = 9.532\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$) and the calculated $\Lambda_{eq}^\infty$: $$\alpha = \frac{9.532}{238.3}$$ $$\alpha = 0.04$$ \noindent\textbf{Step 4 - Conversion to Percentage Degree of Dissociation} To express the degree of dissociation as a percentage: $$\alpha\% = \alpha \times 100\%$$ $$\alpha\% = 0.04 \times 100\%$$ $$\alpha\% = \boxed{4.0\%}$$ The degree of dissociation of \ce{NH4OH} at this temperature is $4.0\%$. Hence, the correct option is (C).