Question

The conductivity of a weak acid $HA$ of concentration $0.001mol{L}^{-1}$ is $2.0\times 10^{-5}Sc{m}^{-1}.$ If ${\Lambda }_m{}^{\circ }(HA)=190Sc{m}^{2}mo{l}^{-1}$ the ionization constant $\left(K_a\right)$ of $HA$ is equal to_____ $\times 1^{-6}$
(Round off to the Nearest Integer)

Answer 1

Answer: 12

${\Lambda }_m=1000\times \frac{K}{M}$
$=1000\times \frac{2\times 10^{-5}}{0.001}=20Sc{m}^{2}mo{l}^{-1}$
$\Rightarrow \alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}=\frac{20}{190}$
$=\left(\frac{2}{19}\right)$
$HA\rightleftharpoons H^{+}+A^{-}$
$0.001(1-\alpha )0.001\alpha 0.001\alpha$
$\Rightarrow k_a=0.001\left(\frac{{\alpha }^2}{1-\alpha }\right)$
$=\frac{0.001\times {\left(\frac{2}{19}\right)}^2}{1-\left(\frac{2}{19}\right)}$
$=12.3\times 10^{-6}$