Question

The conductivity of 0.001 mol $L^{-1}$ acetic acid is 5 $\times {10}^{-5}sc{m}^{-1}$. Calculate it's dissociation constant if ${\lambda }_m^o$ m for acetic acid is $250sc{m}^{2}mo{l}^{-1}$?

Answer 1

The relationship between molar conductivity ${\Lambda }_m$ and conductivity $\kappa$
${\Lambda }_m=\frac{1000\kappa }{C}$
${\Lambda }_m=\frac{1000\times 5\times {10}^{-5}S/cm}{0.001mol/L}=50Sc{m}^2/mol$
The degree of dissociation $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{50Sc{m}^2/mol}{250Sc{m}^2/mol}=0.2$
The dissociation constant $K_a=\frac{C{\alpha }^2}{1-\alpha }$
$K_a=\frac{0.001\times (0.2{)}^2}{1-0.2}=5\times {10}^{-5}$