Question

At 291 K, the molar conductivities at infinite dilution of $N{H}_{4}Cl,NaOH$ and $NaCl$ are 129.8, 217.4 and $108.9Sc{m}^{2}mo{l}^{-1}$ respectively . If the molar conductivity of a centinormal solution of $N{H}_{4}OH$ is $9.532Sc{m}^{2}mo{l}^{-1}$, then calculate the dissociation constant of $N{H}_{4}OH$?

• A. $1.6\times {10}^{-5}$
• B. $5.2\times {10}^{-5}$
• C. $9.6\times {10}^{-15}$
• D. $1.6\times {10}^{-8}$

Answer 1

The correct option is A $1.6\times {10}^{-5}$

Here, we are given
${\Lambda }_m^{\circ }$ for $N{H}_{4}Cl=129.8Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_m^{\circ }$ for $NaOH=217.4Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_m^{\circ }$ for $NaCl=108.9Sc{m}^{2}mo{l}^{-1}$
By Kohlrausch's law,
${\Lambda }_m^{\circ }$ for $N{H}_{4}OH={\Lambda }_{N{H}_4^{+}}^{\circ }+{\Lambda }_{O{H}^{-}}^{\circ }$
${\Lambda }_m^{\circ }\left(N{H}_{4}Cl\right)+{\Lambda }_m^{\circ }\left(NaOH\right)$
$-{\Lambda }_m^{\circ }\left(NaCl\right)$
$=129.8+217.4-108.9=238.3Sc{m}^{2}mo{l}^{-1}$
Given ${\Lambda }_m^c=9.532Sc{m}^{2}mo{l}^{-1}$
$\therefore$ Degree of dissociation $\left(\alpha \right)=\frac{{\Lambda }_m^c}{{\Lambda }_m^{\circ }}$
$=\frac{9.532}{238.3}=0.04$
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
Initial conc. C - -
Equilibrium conc. $C-C\alpha$ $C\alpha$ $C\alpha$
$\therefore K=\frac{C{\alpha }^2}{1-\alpha }$
Now, putting concentration = $0.01N=0.01M$ and $\alpha =0.04,$
We get $K=1.6\times {10}^{-5}$