Question

Given the following molar conductivities at ${25}^{\circ }C;HCl,426{\Omega }^{-1}c{m}^{2}mo{l}^{-1};NaCl,$ $126{\Omega }^{-1}c{m}^{2}mo{l}^{-1};NaC\left(sodiumcrotonate\right)$, $83{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$. What is the ionization constant of crotonic acid? If the conductivity of a $0.001M$ crotonic acid solution is $3.83\times {10}^{-5}{\Omega }^{-1}c{m}^{-1}$?

• A. ${10}^{-5}$
• B. $1.11\times {10}^{-5}$
• C. $1.11\times {10}^{-4}$
• D. $0.01$

Answer 1

Answer: (A)

${10}^{-5}$

From data:

${\lambda }_{HC}={\lambda }_{HCl}+{\lambda }_{NaC}-{\lambda }_{NaCl}$

${\lambda }^o=426+83-126=383Sc{m}^{2}mo{l}^{-1}$

${\lambda }_{eq}=3.83\times {10}^{-5}\times \frac{1000}{0.001}=38.3Sc{m}^{2}mo{l}^{-1}$

$\alpha =\frac{{\lambda }_{eq}}{{\lambda }^o}=38.3/383=0.1$

Dissociation constant, $K=C{\alpha }^2=0.001\times (0.1{)}^2=1\times {10}^{-5}$