The measured equivalent conductance of a aqueous solution of is exactly . The equivalent conductance — Electrochemistry Chemistry Question
Question
The measured equivalent conductance of a $0.01\text{ N}$ aqueous solution of $NH_4OH$ is exactly $13.6\text{ S cm}^2\text{ eq}^{-1}$. The equivalent conductance of an infinitely dilute solution of $NH_4Cl$ is $150$, while the ionic conductances of individual $OH^-$ and $Cl^-$ ions are $198$ and $76\text{ S cm}^2\text{ eq}^{-1}$ respectively. What is the precise percentage degree of dissociation ($\% \alpha$) of the $0.01\text{ N}$ $NH_4OH$ solution?
💡 Solution & Explanation
First, find the limiting equivalent conductance of $NH_4OH$ ($\Lambda_{eq}^\circ$) using Kohlrausch's Law: $\Lambda_{eq}^\circ(NH_4OH) = \Lambda_{eq}^\circ(NH_4Cl) - \lambda_{eq}^\circ(Cl^-) + \lambda_{eq}^\circ(OH^-) = 150 - 76 + 198 = 272\text{ S cm}^2\text{ eq}^{-1}$. The degree of dissociation $\alpha$ is the ratio of conductance at a given concentration to that at infinite dilution: $\alpha = \Lambda_{eq} / \Lambda_{eq}^\circ = 13.6 / 272 = 0.05$. In percentage, this is $0.05 \times 100 = 5\%$.