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NEET 2021
Exam
NEET
Year
2021
Subject
Chemistry
Chapter
Electrochemistry

NEET 2021 · Chemistry · Electrochemistry

The molar conductivity of 0 . 007M acetic acid is 20S cm ^ 2 mol ^...

The molar conductivity of 0.007M0 . 007M acetic acid is 20Scm2mol−1.20S{cm}^{2}{mol}^{- 1} . What is the dissociation constant of acetic acid? Choose the correct option.

λH+∘=350Scm2mol−1λCH3COO−∘=50Scm2mol−1{\lambda}_{{H}^{+}}^{∘} = 350S{cm}^{2}{mol}^{- 1}{\lambda}_{{CH}_{3}{COO}^{-}}^{∘} = 50S{cm}^{2}{mol}^{- 1}

Options

  1. A1.75×10−4molL−11 . 75 \times {10}^{- 4}mol{L}^{- 1}
  2. B2.50×10−4molL−12 . 50 \times {10}^{- 4}mol{L}^{- 1}
  3. C1.75×10−5molL−11 . 75 \times {10}^{- 5}mol{L}^{- 1}
  4. D2.50×10−5molL−12 . 50 \times {10}^{- 5}mol{L}^{- 1}

Correct answer

C. 1.75×10−5molL−11 . 75 \times {10}^{- 5}mol{L}^{- 1}

Step-by-step solution

λCH3COOH∘=λCH3COO−∘+λH+∘{\lambda}_{{CH}_{3}COOH}^{∘} = {\lambda}_{{CH}_{3}{COO}^{-}}^{∘} + {\lambda}_{{H}^{+}}^{∘}

λCH3COO−∘=350+50=400Scm2mol−1{\lambda}_{{CH}_{3}{COO}^{-}}^{∘} = 350 + 50 = 400S{cm}^{2}{mol}^{- 1}

λCH3COOH=20Scm2mol−1{\lambda}_{{CH}_{3}COOH}^{} = 20{Scm}^{2}{mol}^{- 1} given

α=λCH3COOHλCH3COOH∘=20400=5×10−2\alpha = \lambda{CH}_{3}COOH{\lambda}_{{CH}_{3}COOH}^{∘} = \dfrac{20}{400} = 5 \times {10}^{- 2}

Ka=cα2=0.007×(5×10−2)2=1.75×10−5molL−1{K}_{a} = {c \alpha}^{2} = 0 . 007 \times {\left(5 \times {10}^{- 2}\right)}^{2} = 1 . 75 \times {10}^{- 5}{molL}^{- 1}

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