Question

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

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

Answer 1

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

Degree of dissociation,

$\alpha =\frac{{\lambda }^c}{{\lambda }^{infinite}}$

$\alpha =\frac{8}{400}=2\times {10}^{{}^{-2}}$

$Ka=\frac{C{\alpha }^2}{1-\alpha }$

= $C{\alpha }^2$ ( 1>>${\alpha }^2$)

= $1.25\times {10}^{-5}$