Question

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

• A. $11.4\%,3.67\times {10}^{-4}$ mol $L^{-1}$
• B. $22.8\%,1.83\times 1^{-4}$ 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

The correct option is A $11.4\%,3.67\times {10}^{-4}$ mol $L^{-1}$

We know,

$\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\infty }}$

where, ${\wedge }_m=$ Molar conductivity

and, ${\wedge }_m^{\infty }=$ Molar conductivity at $\infty$ dilution

${\wedge }_m^{\infty }={\lambda }_{H^{+}}^o+{\lambda }_{HCO{O}^{⊝}}^o=349.6+54+6$

${\wedge }_m^{\infty }=404.2$ $Sc{m}^{2}mo{l}^{-1}$

$\alpha =\frac{46.1}{404.2}=0.114=11.4\%$

Dissociation constant, $K=C{\alpha }^2$

$K=0.025\times (0.114{)}^2$

$⟹K=3.6\times {10}^{-4}$ $mol$ $l^{-1}$