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Karnataka Board PUCPUC Science 2nd PUC Class 12

Conductivity of 0.00241 M acetic acid is 7.896 × 10^−5 S cm−1. Calculate its molar conductivity and if Λ0𝑚 for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

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Question

Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if `Lambda_m^0` for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Numerical
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Solution

Given: Conductivity (κ) = 7.896 × 10−5 S cm−1

Molar conductivity (M) = 0.00241

`Lambda_m^0` = 390.5 S cm2 mol−1

`∧_m^c = (kappa xx 1000)/"Molarity"`

= `(7.896 xx 10^-5 xx 1000)/0.00241`

= 32.763 S cm2 mol1

α = `(∧_m^c)/(∧_m^0)`

= `32.763/390.5`

= 0.084

Kα = `(alpha^2 c)/(1 - alpha)`

= `((0.084)^2 xx 0.00241)/(1 - 0.084)`

= `((0.084)^2 xx 0.00241)/(0.916)`

= 1.86 × 10−5

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Chapter 2: Electrochemistry - Exercises [Page 60]

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NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 2 Electrochemistry
Exercises | Q 2.11 | Page 60
Nootan Chemistry [English] Class 12 ISC
Chapter 2 Electrochemistry
'NCERT TEXT-BOOK' Exercises | Q 3.11 | Page 211

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