Question

Molar conductivity of $0.025\, mol\, L^{-1}$ methanoic acid is $46.1\, S\, cm^2\, mol^{-1}$, the degree of dissociation and dissociation constant will be
(Given : $\lambda^{\circ}_{H^{+}}=349.6\,S\,cm^{2}\,mol^{-1}$ and $\lambda^{\circ}_{HCOO^{-}}=54.6\,S\,cm^{2}\,mol^{-1}$)

• A. $11.4\%$, $3.67\times10^{-4}\,mol\,L^{-1}$
• B. $22.8\%$, $1.83\times10^{-4}\,mol\,L^{-1}$
• C. $52.2\%$, $4.25\times10^{-4}\,mol\,L^{-1}$
• D. $1.14\%$, $3.67\times10^{-6}\,mol\,L^{-1}$

Answer 1

Answer: (A)

$\lambda^{\circ}_{HCOOH}=\lambda^{\circ}_{H^{+}}+\lambda^{\circ}_{HCOO^{-}}$
$=349.6+54.6$
$=404.2\,S\,cm^{2}\,mol^{-1}$
$\alpha=\frac{\Lambda_{m}}{\Lambda^{\circ}_{m}}=\frac{46.1}{404.2}=11.4\%$
$K_{a}=\frac{C\alpha^{2}}{1-\alpha}$
$=\frac{0.025\times\left(0.114\right)^{2}}{1-0.114}$
$=\frac{0.025\times0.114\times0.114}{0.886}$
$=3.67\times10^{-4}\,mol\,L^{-1}$