Question

$\Lambda_{ m }^{\circ}$ for $NaCl , HCl$ and $NaA$ are $126.4,425.9$ and $100.5\, S\, cm ^{2} mol ^{-1}$, respectively. If the conductivity of $0.001\, M\, HA$ is $5 \times 10^{-5} S\, cm ^{-1}$, percent degree of dissociation of $HA$ is_______.

Answer 1

Given, $\Lambda_{ m }^{\circ}( NaCl )=126.4\, S\, cm ^{2} mol ^{-1}$
$\Lambda_{ m }^{\circ}( HCl )=425.9\, S\, cm ^{2} mol ^{-1}$
$\Lambda_{ m }^{0}( NaA )=100.5\, S\, cm ^{2} mol ^{-1}$
$\Lambda_{ m }^{\circ}( HA )=\Lambda_{ m }^{\circ}( HCl )+\Lambda_{ m }^{\circ}( NaA )-\Lambda_{ m }^{\circ}( NaCl )$
$=425.9+100.5-126.4$
$=400\, S\, cm ^{2} mol ^{-1}$
$\Lambda_{ m }^{\circ}=\frac{1000 \kappa}{ C }=5 \times 10^{-5} \times \frac{1000}{0.001}=50$
$\alpha=\frac{\Lambda_{ m }}{\Lambda_{ m }^{\circ}}=\frac{50}{400}=0.125$
$\therefore \% \alpha=12.5$