Question

${\Lambda }_m^{\circ }$ 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 $5\times 10^{-5}Sc{m}^{-1}$, percent degree of dissociation of $HA$ is_______.

Answer 1

Answer: 12.5

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