Question

$\Lambda_m^o $ 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 $5 \times 10^{-5}\,S\,cm^{-1}$, degree of dissociation of $HA$ is :

• A. 0.75
• B. 0.125
• C. 0.25
• D. 0.50

Answer 1

Answer: (B)

$\Lambda_m^o (HA) \, = \, \Lambda_{m}^{o} (HCl) \, + \, \Lambda_{m}^{o} \, (NaA) \, - \, \Lambda_{m}^{o} (NaCl) $
= $425.9 + 100.5 - 126.4$
= $400 \,S \,cm^2 mol^{-1}$
$\Lambda_m \, \, = \, \frac{1000k}{M} \, = \, \frac{1000 \times 5 \times \, 10^{-5}}{10^{-3}} \, \, =50 \, S \, cm^2 \, mol^{-1}$
$\alpha \, \, = \, \frac{\Lambda_m}{\Lambda_m^o} \, = \, \frac{50}{400} \, = 0.125$