Categories > electrochemistry > Conductors and Conductance > Aniline hydrochloride $({{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl)$ is a salt of a weak base and strong acid. The hydrolysis equilibria ${{K}_{h}}=\frac{{{h}^{2}}C}{(1-h)}$. On the basis of conductivity, h is given by (where $\Lambda$, ${{\Lambda }_{1}},\,{{\Lambda }_{2}}$ are equivalent conductance of hydrolysed salt, unhydrolysed salt and HCl respectively, Kh = hydrolysis constant, C = concentration of ${{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl$ and h = degree of hydrolysis).

electrochemistry

Question: Aniline hydrochloride $({{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl)$ is a salt of a weak base and strong acid. The hydrolysis equilibria ${{K}_{h}}=\frac{{{h}^{2}}C}{(1-h)}$. On the basis of conductivity, h is given by (where $\Lambda$, ${{\Lambda }_{1}},\,{{\Lambda }_{2}}$ are equivalent conductance of hydrolysed salt, unhydrolysed salt and HCl respectively, Kh = hydrolysis constant, C = concentration of ${{C}_{6}}{{H}_{5}}N{{H}_{3}}Cl$ and h = degree of hydrolysis).


1) $h=\frac{\Lambda -{{\Lambda }_{1}}}{{{\Lambda }_{2}}-\Lambda }$
2) $h=\frac{\Lambda -{{\Lambda }_{1}}}{{{\Lambda }_{2}}-{{\Lambda }_{1}}}$
3) $h=\frac{{{\Lambda }_{2}}+{{\Lambda }_{1}}}{\Lambda }$
4) $h=\frac{\Lambda +{{\Lambda }_{1}}}{{{\Lambda }_{2}}-{{\Lambda }_{1}}}$
Solution: Explanation: No Explanation
Conductors and Conductance

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)

Previous Question Next Question

Related Questions