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At 25⁰C, the ionic mobility of $C H_{3} C O O^{-}, H^{+}$ are respectively $4.1 \times 10^{- 4}, 3.63 \times 10^{- 3}$ cm/sec. The conductivity of 0.001 M $C H_{3} C O O H$ is $5 \times 10^{- 5} S . c m^{- 1}$ . Dissociation constant of $C H_{3} C O O H$

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$3 \times 10^{- 6}$

$3 \times 10^{- 4}$

$1.64 \times 10^{- 5}$

$3 \times 10^{- 5}$

answer is A.

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Detailed Solution

$α = \frac{\land_{M}}{\land_{M}^{0}} =$

$\land_{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}}$

$= 1000 \times 5 \times 10^{- 5} \times 10^{3}$

$50 Ω^{- 1} c m^{2} m o l^{- 1}$

$\land_{M}^{0} = 96500 (4.1 \times 10^{- 4} + 3.63 \times 10^{- 3})$

=386

$α = \frac{50}{386} = 0.129$

$α = \sqrt{\frac{k_{a}}{c}}$

${(0.129)}^{2} = \frac{k_{a}}{10^{- 6}}$

$k_{a} = 1.64 \times 10^{- 5}$

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