1. 1.75×10⁻⁴ mol L ⁻¹ 2. 2.50×10⁻⁴ mol L ⁻¹ 3. 1.75×10⁻⁵ mol L ⁻¹ 4. 2.50×10⁻⁵ mol L ⁻¹

Answer: 3. 1.75×10⁻⁵ mol L ⁻¹ Key idea: α=Lambda_m/Lambda^°_m , and K_a=Cα²/1-α≈ Cα² . Lambda^°=350+50=400 ; α=20/400=0.05 . K_a≈0.007×(0.05)²=1.75×10⁻⁵\ mathrmmol L⁻¹ (using the usual approximation 1-α≈1 ).

Topic: Conductance, molar conductivity and its variation with concentration· Redox Reactions and Electrochemistry

The molar conductivity of 0.007 M acetic acid is 20 S cm^2 mol^-1. What is the dissociation constant of acetic acid ? Choose the correct option.^°_H^+ = 350 S cm^2 mol^-1, ^°_CH_3COO^- = 50 S cm^2 mol^-1

1. \(1.75\times10^{-4}\) mol L\(^{-1}\)

2. \(2.50\times10^{-4}\) mol L\(^{-1}\)

3. \(1.75\times10^{-5}\) mol L\(^{-1}\)

4. \(2.50\times10^{-5}\) mol L\(^{-1}\)

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Answer: 3. \(1.75\times10^{-5}\) mol L\(^{-1}\)

Key idea: \(\alpha=\Lambda_m/\Lambda^\circ_m\), and \(K_a=\dfrac{C\alpha^2}{1-\alpha}\approx C\alpha^2\).

\(\Lambda^\circ=350+50=400\); \(\alpha=20/400=0.05\).

\(K_a\approx0.007\times(0.05)^2=1.75\times10^{-5}\ \mathrm{mol\,L^{-1}}\) (using the usual approximation \(1-\alpha\approx1\)).

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The molar conductivity of 0.007 M acetic acid is 20 S cm^2 mol^-1. What is the dissociation constant of acetic acid ? Choose the correct option.^°_H^+ = 350 S cm^2 mol^-1, ^°_CH_3COO^- = 50 S cm^2 mol^-1

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