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  4. The conductivity of a weak acid HA of concentration 0.001 mol L-1 is 2.0 × 10-5 S cm -1 . If Lambdam °(H A)=190 S cm 2 mol -1 the ionization constant (Ka) of H A is equal to × 1-6 (Round off to the Nearest Integer)

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Q. The conductivity of a weak acid $HA$ of concentration $0.001\, mol L^{-1}$ is $2.0 \times 10^{-5} S\,cm ^{-1} .$ If $\Lambda_{m}{ }^{\circ}(H A)=190\, S\,cm ^{2} mol ^{-1}$ the ionization constant $\left(K_{a}\right)$ of $H A$ is equal to_____ $\times 1^{-6}$
(Round off to the Nearest Integer)

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Solution:

$\Lambda_{m}=1000 \times \frac{K}{M}$
$=1000 \times \frac{2 \times 10^{-5}}{0.001}=20\, S\,cm ^{2} mol ^{-1}$
$\Rightarrow \alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{\infty}}=\frac{20}{190}$
$=\left(\frac{2}{19}\right)$
$H A \rightleftharpoons H^{+}+A^{-}$
$0.001(1-\alpha) 0.001 \alpha 0.001 \alpha$
$\Rightarrow k_{a}=0.001\left(\frac{\alpha^{2}}{1-\alpha}\right)$
$=\frac{0.001 \times\left(\frac{2}{19}\right)^{2}}{1-\left(\frac{2}{19}\right)}$
$=12.3 \times 10^{-6}$