The equivalent conductance of $\frac{M}{32}$ solution of a weak monobasic acid is 8.0 $mho {cm}^{2} . {eq}^{- 1}$ and at infinite dilution is 400 $mho {cm}^{2} . {eq}^{- 1}$ .The dissociation constant of this acid is

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$1.625 \times 10^{- 4}$

$1.25 \times 10^{- 4}$

$1.25 \times 10^{- 5}$

$1.25 \times 10^{- 6}$

answer is A.

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Detailed Solution

$α = \frac{λ_{v}}{λ_{\infty}} = \frac{8.0}{400} = 2 \times 10^{- 2}; k_{a} = {Cα}^{2} = \frac{1}{32} \times {(2 \times 10^{- 2})}^{2} = 1.25 \times 10^{- 5}$

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