Question

Equivalent conductance at infinite dilution, ${\lambda }^0$ of NH${}_4$Cl, NaOH and NaCl are 128.0, 217.8 and 109.9 $oh{m}^{-1}c{m}^{2}e{q}^{-1}$ respectively. The equivalent conductance of 0.01 N $N{H}_{4}OH$ is 9.30 $oh{m}^{-1}c{m}^{2}e{q}^{-1}$, then the degree of ionization of $N{H}_{4}OH$ at this temperature would be:

• A. 0.04
• B. 0.1
• C. 0.39
• D. 0.62

Answer 1

The correct option is A 0.04

According to Kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivities of its ions.
Hence,

${\Lambda }_{eq}^{\infty }\left({NH}_{4}OH\right)={\Lambda }_{eq}^{\infty }\left({NH}_{4}Cl\right)+{\Lambda }_{eq}^{\infty }\left(NaOH\right)-{\Lambda }_{eq}^{\infty }\left(NaCl\right)$

Substitute values in the above equation

${\Lambda }_{eq}^{\infty }\left({NH}_{4}OH\right)=128.0+217.8-109.9=235.9{ohm}^{-1}{cm}^2{eq}^{-1}$

The degree of dissociation is the ratio of the equivalent conductance at given concentration to the equivalent conductance at zero concentration. Hence,

$\alpha =\frac{{\Lambda }_{eq}}{{\Lambda }_{eq}^{\infty }}=\frac{9.30}{235.9}=\mathrm{0.04.}$

Option A is correct.