Λ_m° for NaCl, HCl, and NaA is 126.4, 425.9, and 100.5 S cm²mol^-1, respectively. If the conductivity of 0.001 M HA is 5 × 10^-5 S cm^-1, the degree of dissociation of HA is

Explanation:

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

Calculation:

According to Kohlrausch’s law, the molar conductivity of HA at infinite dilution is given as,

\({\Lambda_m}^\circ \left( {{\rm{HA}}} \right) = \left[ {{\Lambda _m}^\circ \left( {{{\rm{H}}^ + }} \right) + {\Lambda _m}^\circ \left( {{\rm{C}}{{\rm{l}}^ - }} \right)} \right] + \left[ {{\Lambda _m}^\circ \left( {{\rm{N}}{{\rm{a}}^ + }} \right) + {\Lambda _m}^\circ \left( {{A^ - }} \right)} \right] - \left[ {{\Lambda _m}^\circ \left( {{\rm{N}}{{\rm{a}}^ + }} \right) + {\Lambda _m}^\circ \left( {{\rm{C}}{{\rm{l}}^ - }} \right)} \right.\) ]

= 425.9 + 100.5 – 126.4

= 400 S cm² mol^-1

Also, molar conductivity (Λm°) at given concentration is given as,

\(\Lambda_m = \frac{{1000 \times k}}{M}\)

Given, k = conductivity ⟹ 5 × 10^-5 S cm^-1

M = Molarity ⟹ 0.001 M

\({\therefore \Lambda_m} = \frac{{1000 \times 5 \times {{10}^{ - 5}}{\rm{\;Sc}}{{\rm{m}}^{ - 1}}}}{{{{10}^{ - 3}}{\rm{M}}}}\)

= 50 S cm² mol^-1

Therefore, degree of dissociation (α), of HA is,

\(\alpha = \frac{{{\Lambda_m}}}{{{\Lambda_m}^\circ {\rm{\;}}}} = \frac{{50\;S\;c{m^2}mo{l^{ - 1}}}}{{400\;S\;c{m^2}\;mo{l^{ - 1}}}} = 0.125\)