Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
244 views
in Chemistry by (96.4k points)
closed by
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

1 Answer

0 votes
by (85.7k points)
selected by
 
Best answer
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,

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

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...