Question

The ionization constant of a weak acid is $1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $400\times {10}^{-4}S{m}^{2}mo{l}^{-1}$. If the cell constant is $0.01m^{-1}$ then conductance of 0.01 M acid solution is (in S):

Answer 1

$K_a=\frac{c{\alpha }^2}{1-\alpha }\Rightarrow 1.6\times {10}^{-5}=\frac{0.01\times {\alpha }^2}{1-\alpha }$
$\alpha =\sqrt{\frac{1.6\times {10}^{-5}}{0.01}}\Rightarrow 0.04$
$\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\infty }};{\wedge }_m=0.04\times 400\times {10}^{-4}$
$=16\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
$\kappa ={\wedge }_m\times C=16\times {10}^{-4}\times {10}^3\times {10}^{-2}$
$(\because 1m^3=1000litre)$
$\kappa =G.G^{\ast }and{G}^{\ast }=0.01m^{-1},$
where $G^{\ast }=\frac{l}{A}$
$\therefore G=\frac{1.6\times {10}^{-2}}{0.01}\Rightarrow 1.6S$