The conductivity of 1 × 1 0 - 3 M acetic acid is 5 × 1 0 - 5 S c m - 1 and Λ ° is 390.5 S c m 2 m o l - 1 . The dissociation constant of acetic acid is

Λ v = κ v × 1000 M = 5 × 1 0 - 5 × 1000 1 × 1 0 - 3 = 50 S c m 2 m o l - 1 So α = Λ v Λ ∞ = 50 390.5 = 0.13 Now, κ a = cα 2 1 − α = 10 − 3 × ( 0.13 ) 2 0.87 = 1.9 × 10 − 5 ≃ 19 × 10 − 6

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumpH of Strong and Weak Acids and BasesEasy2 minPYQ_2020

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The conductivity of $1 \times 1 0^{- 3}$ M acetic acid is $5 \times 1 0^{- 5} S c m^{- 1}$ and $Λ °$ is 390.5 S $c m^{2} m o l^{- 1}$ . The dissociation constant of acetic acid is

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$Λ_{v} = κ_{v} \times \frac{1000}{M} = \frac{5 \times 1 0^{- 5} \times 1000}{1 \times 1 0^{- 3}} = 50 S c m^{2} m o l^{- 1}$
So $α = \frac{Λ_{v}}{Λ \infty} = \frac{50}{390.5} = 0.13$
Now, $κ_{a} = \frac{{cα}^{2}}{1 - α} = \frac{10^{- 3} \times {(0.13)}^{2}}{0.87} = 1.9 \times 10^{- 5} ≃ 19 \times 10^{- 6}$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Strong and Weak Acids and Bases

⏱ 2mℹ️ Source: PYQ_2020

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