Question

A weak monobasic acid is $5$% dissociated in $0.01mold{m}^{-3}$ solution. The limiting molar conductivity at infinite dilution is $4.00\times {10}^{-2}oh{m}^{-1}{m}^{2}mo{l}^{-1}$. Calculate the conductivity of a $0.05mold{m}^{-3}$ solution of the acid.

Answer 1

Dissociation constant of acid $K_a=C{\alpha }^2$
$=0.01\times (0.05{)}^2$
$=2.5\times {10}^{-5}$
$K_a=C{\alpha }^2$
$2.5\times {10}^{-5}=0.05\times {\alpha }^2$
$\alpha =0.0223$
We know that, $\alpha =\frac{{\wedge }_m^c}{{\wedge }_m^{\infty }}$
$0.0223=\frac{{\wedge }_m^c}{4\times {10}^{-2}}$
${\wedge }_m^c=8.92\times {10}^{-4}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$.