Question

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

• A. $11.4\%$, $3.67\times 10^{-4}mol{L}^{-1}$
• B. $22.8\%$, $1.83\times 10^{-4}mol{L}^{-1}$
• C. $52.2\%$, $4.25\times 10^{-4}mol{L}^{-1}$
• D. $1.14\%$, $3.67\times 10^{-6}mol{L}^{-1}$

Answer 1

Answer: (A)

${\lambda }_{HCOOH}^{\circ }={\lambda }_{H^{+}}^{\circ }+{\lambda }_{HCO{O}^{-}}^{\circ }$
$=349.6+54.6$
$=404.2Sc{m}^{2}mo{l}^{-1}$
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{46.1}{404.2}=11.4\%$
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
$=\frac{0.025\times {\left(0.114\right)}^2}{1-0.114}$
$=\frac{0.025\times 0.114\times 0.114}{0.886}$
$=3.67\times 10^{-4}mol{L}^{-1}$