Question

The conductivity of 0.001028 mol ${\text{L}}^{\text{- 1}}$ acetic acid is $4.95\times {10}^{-5}\text{S}\text{c}{\text{m}}^{\text{- 1}}$. Calculate its dissociation constant if ${\Lambda }_{\text{m}}^{\text{o}}$ for $\text{390}\text{.5}\text{S}\text{c}{\text{m}}^{\text{2}}\text{mo}{\text{l}}^{-1}$.

Answer 1

$c=0.001028\frac{mol}{L}$

$k=4.25\times {10}^{-5}5c{m}^{2}mo{l}^{-1}$ conductivity

${\wedge }_m^0=390.55c{m}^{2}mo{l}^{-1}$

$K_a=?$ = dissociation constant

i) ${\wedge }_m=\frac{k\times 1000}{c}$

${\wedge }_m=\frac{4.95\times {10}^{-5}\times 10005c{m}^{-1}}{0.001028mol{l}^{-1}}$

${\wedge }_m=48.15175c{m}^{2}mo{l}^{-1}$

ii) $\alpha =\frac{{\wedge }_m}{{\wedge }_m^0}$ = degree of disso.

$\alpha =\frac{48.1517}{390.5}$

$\alpha =0.123307$

iii) Acid disso. constant $K_a=\frac{{\alpha }^{2}c}{1-\alpha }=\frac{(0.123307{)}^2\times 0.001028}{1-0.123307}$

$=\frac{1.56305\times {10}^{-5}}{0.876693}$

$K_a=1.7829\times {10}^{-5}\frac{mol}{L}$