Question

The molar conductance of 0.1 M $C{H}_{3}COOH$ at pH=4 is $Sc{m}^{2}mo{l}^{-1}$.
(Given the limiting molar conductance of acetic acid is 390 $Sc{m}^{2}mo{l}^{-1}$).

Answer 1

$pH=-log\left[H^{+}\right]log\left[H^{+}\right]=-4\Rightarrow \left[H^{+}\right]={10}^{-4}$
$\left[H^{+}\right]=c\alpha$, for acetic acid where, 'c' is the concentration and $\alpha$ is the degree of dissociation.
$\Rightarrow c\alpha ={10}^{-4}$
Given $c=0.1M$
$\Rightarrow \alpha ={10}^{-3}$
Also, $\alpha =\frac{{\lambda }_m}{{\lambda }_m^0}={10}^{-3}$
Given, ${\lambda }_m^0=390Sc{m}^{2}mo{l}^{-1}$
Thus ${\lambda }_m=390\times {10}^{-3}=0.390Sc{m}^{2}mo{l}^{-1}$