The conductivity of 0.001 M acetic acid is 5 × 10–5 S cm–1 and is 390.5 S cm2 mol–1 then the calculated value of dissociation constant of acetic acid would be

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81.78 × 10–4

81.78 × 10–5

18.78 × 10–6

18.78 × 10–7

answer is C.

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Detailed Solution

α=λcλo=1000κMλoα=5×10-5×100010-3×390.5=50390.5Ka=Cα21-α=10-350390.521-50390.5=10-3[0.128]20.872=18.78×10-6

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