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Electrochemistry question

2021 · Q123
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Electrochemistry question

2021 · Q123

NEETChemistryElectrochemistryMCQ+4 / −1
The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol−-−1. What is the dissociation constant of acetic acid? Choose the correct option.

[ΛH+o\Lambda _{{H^ + }}^oΛH+o​ = 350 S cm2 mol−-−1

ΛCH3COO−o\Lambda _{C{H_3}CO{O^ - }}^oΛCH3​COO−o​ = 50 S cm2 mol−-−1]
  1. A
    2.50 ×\times× 10−-−5 mol L−-−1
  2. B
    1.75 ×\times× 10−-−4 mol L−-−1
  3. C
    2.50 ×\times× 10−-−4 mol L−-−1
  4. D
    1.75 ×\times× 10−-−5 mol L−-−1
View written solutionFree

Correct answer: D

Λm\Lambda _m^{}Λm​ = 20 S cm2 mol−-−1

According to Kohlrausch’s law,

Λm  CH3COOHo=ΛCH3COO−o+Λm  H+o\Lambda _{m\,\,C{H_3}COOH}^o = \Lambda _{C{H_3}CO{O^ - }}^o + \Lambda _{m\,\,{H^ + }}^oΛmCH3​COOHo​=ΛCH3​COO−o​+ΛmH+o​

= 50 + 350 = 400 S cm2 mol−-−1

Degree of dissociation,

α=ΛmΛmo=20400=120\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{20} \over {400}} = {1 \over {20}}α=Λmo​Λm​​=40020​=201​



NEET 2021 Chemistry - Electrochemistry Question 15 English Explanation

Ka=Cα21−α≃Cα2{K_a} = {{C{\alpha ^2}} \over {1 - \alpha }} \simeq C{\alpha ^2} Ka​=1−αCα2​≃Cα2

As α\alpha α << 1 so 1 - α\alpha α ≈\approx≈ 1

=7×10−3×(120)2= 7 \times {10^{ - 3}} \times {\left( {{1 \over {20}}} \right)^2}=7×10−3×(201​)2

=7×10−3×14×10−2 = 7 \times {10^{ - 3}} \times {1 \over 4} \times {10^{ - 2}}=7×10−3×41​×10−2

=1.75×10−5 = 1.75 \times {10^{ - 5}}=1.75×10−5 mol L-1

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