Question

At ${18}^{\circ }C$, the conductivities at infinite dilution of $N{H}_{4}Cl,NaOH$ and $NaCl$ are $129.8,217.4$ and $108.9$ mho respectively. If the equivalent conductivity of $N/100$ solution of $N{H}_{4}OH$ is $9.93mho$, calculate the degree of dissociation of $N{H}_{4}OH$ at this dilution.

Answer 1

${\wedge }_{\infty N{H}_{4}Cl}={\lambda }_{N{H}_4^{+}}+{\lambda }_{C{l}^{-}}=129.8....\left(i\right)$
${\wedge }_{\infty NaOH}={\lambda }_{N{a}^{+}}+{\lambda }_{O{H}^{-}}=217.4....\left(ii\right)$
${\wedge }_{\infty NaCl}={\lambda }_{N{a}^{+}}+{\lambda }_{C{l}^{-}}=108.9.....\left(iii\right)$
Adding eqs. (i) and (ii) and subtracting eq. (iii)
${\lambda }_{N{H}_4^{+}}+{\lambda }_{C{l}^{-}}+{\lambda }_{N{a}^{+}}+{\lambda }_{O{H}^{-}}-{\lambda }_{N{a}^{+}}-{\lambda }_{C{l}^{-}}$
$={\lambda }_{N{H}_4^{+}}+{\lambda }_{O{H}^{-}}=129.8+217.4-108.9$
${\wedge }_{\infty N{H}_{4}OH}=238.3mho$
Degree of dissociation, $\alpha =\frac{{\wedge }_v}{{\wedge }_{\infty }}=\frac{9,93}{238.3}=0.04167$
or $4.17$% dissociated.