# Λm° for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol-1, respectively. If the conductivity of 0.001 M HA is 5 × 10-5 S cm-1, degree of dissoc

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Λm° for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol-1, respectively. If the conductivity of 0.001 M HA is 5 × 10-5 S cm-1, degree of dissociation of HA is
1. 0.025
2. 0.50
3. 0.75
4. 0.125

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Correct Answer - Option 4 : 0.125

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 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,

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