Question

${\Lambda }_m^o$ for $NaCl,HCl$ and $NaA$ are $126.4,425.9$ and $100.5Sc{m}^{2}mo{l}^{-1}$ respectively. If the conductivity of $0.001MHA$ is $5\times 10^{-5}Sc{m}^{-1}$, degree of dissociation of $HA$ is :

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

Answer 1

Answer: (B)

${\Lambda }_m^o(HA)={\Lambda }_m^o(HCl)+{\Lambda }_m^o(NaA)-{\Lambda }_m^o(NaCl)$
= $425.9+100.5-126.4$
= $400Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_m=\frac{1000k}{M}=\frac{1000\times 5\times 10^{-5}}{10^{-3}}=50Sc{m}^{2}mo{l}^{-1}$
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^o}=\frac{50}{400}=0.125$