Question

The ionisation constant of a weak acid is $1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $380\times {10}^{-4}$ $S$ $m^2$ ${mol}^{-1}$. If the cell constant is ${0.01m}^{-1}$, then conductance of $0.01M$ acid solution is :

• A. $1.52\times$ ${10}^{-5}$ S
• B. $1.52$ S
• C. $1.52\times$ ${10}^{-3}$ S
• D. $1.52\times$ ${10}^{-4}$ S

Answer 1

The correct option is B $1.52$ S

$K_a=\frac{c{\alpha }^2}{1-\alpha }\Rightarrow 1.6\times {10}^{-15}=\frac{0.01\times {\alpha }^2}{1-\alpha }$
$\alpha =\sqrt{\frac{1.6\times {10}^{-5}}{0.01}}\Rightarrow 0.04$
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }};{\Lambda }_m=0.04\times 380\times {10}^{-4}$
$=15.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
$\kappa ={\Lambda }_m\times C=15.2\times {10}^{{}_4}\times {10}^3\times {10}^{-2}$
$(\because 1m^3=1000litre)$
$\kappa =G.G^{\ast }$ and ${}^{\ast }=0.01m^{-1}$,
where $G^{\ast }=\frac{l}{A}$

$\therefore G=\frac{1.52\times {10}^{-2}}{0.01}\Rightarrow 1.52S$

hence option (B) is correct.