Question

The equivalent conductance of an infinitely dilute solution $N{H}_{4}Cl$ is $150$ and that of $C{l}^{-}$ and $O{H}^{-}$ is 76 and 198 respectively. The equivalent conductance of a $0.01N$ solution $N{H}_{4}OH$ is $9.6$. If its degree of dissociation of $N{H}_{4}OH$ is $353\times {10}^{-x}$, then what is the value of $x$?

Answer 1

${\Delta }_{N{H}_{4}Cl}^{\circ }={\lambda }_{N{H}_4^{+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }$

$\therefore {\lambda }_{N{H}_4^{+}}^{\circ }={\Delta }_{N{H}_{4}Cl}^{\circ }-{\lambda }_{C{l}^{-}}^{\circ }=150-76=74$

$\therefore {\Delta }_{N{H}_{4}OH}^{\circ }={\lambda }_{N{H}_4^{+}}^{\circ }+{\lambda }_{O{H}^{-}}^{\circ }=74+198=272$

Further,

Degree of dissociation, $\alpha =\frac{{\Delta }_{N{H}_{4}OH}}{{\Delta }_{N{H}_{4}OH}^{\infty }}=\frac{9.6}{272}=0.0353=353\times {10}^{-4}$

$\therefore x=4$