Question

The dissociation constant of $n$-butyric acid is $1.6\times 10^{-5}$ and the molar conductivity at infinite dilution is $380\times 10^{-4}S{m}^{2}mo{l}^{-1}$. The specific conductance of the $0.01M$ acid solution is :-

• A. $1.52\times 10^{-5}S{m}^{-1}$
• B. $1.52\times 10^{-2}S{m}^{-1}$
• C. $1.52\times 10^{-3}S{m}^{-1}$
• D. $6.08\times 10^{-4}S{m}^{-1}$

Answer 1

Answer: (B)

$K_a=C{\alpha }^2$
$\alpha =\sqrt{K_a/C}$
$\alpha =\sqrt{\frac{1.6\times 10^{-5}}{0.01}}$
$\alpha =0.04$
$\alpha =\frac{\lambda m}{{\lambda }_m\infty }$
${\lambda }_m=0.04\times 380\times 10^{-4}$
$=15.2\times 10^{-4}S{m}^{2}mo{l}^{-1}$
or
$=15.2Sc{m}^{2}mo{l}^{-1}$
${\lambda }_m=\frac{1000\times K}{M}$
$K=\frac{15.2\times 0.01}{1000}$
$K=152\times 10^{-6}$
$K=1.52\times 10^{-4}Sc{m}^{-1}$
or $1.52\times 10^{-2}S{m}^{-1}$