Question

The equivalent conductance of M/32 solution of a weak monobasic acid is $8.0mhoc{m}^2$, and at infinite dilution is 400 $mhoc{m}^2$. The dissociation constant of this acid is:

• A. $1.25\times {10}^{-6}$
• B. $6.25\times {10}^{-4}$
• C. $1.25\times {10}^{-4}$
• D. $1.25\times {10}^{-5}$

Answer 1

The correct option is D $1.25\times {10}^{-5}$

We know that
$\alpha =\frac{{\lambda }_c}{{\lambda }_{\infty }^0}$

$\alpha =\frac{8}{400}=0.02$
For monobasic acid
$\underset{C-C\alpha }{HA}\leftrightharpoons \underset{C\alpha }{H^{+}}+\underset{C\alpha }{A^{-}}$
$K_a=\frac{(C\alpha {)}^2}{C-C\alpha }=\frac{C{\alpha }^2}{1-\alpha }$
$\alpha <<1$ thus
$K_a=C{\alpha }^2=1.25\times {10}^{-5}$