JEE Main · ChemistryHard

$The molar conductivity of a 0.025 M solution of a weak acid HA is 20 S cm^{2} / mol . If the limiting molar conductivity of HA is 400 S cm^{2} / mol, calculate the dissociation constant K_{a} of the acid.$

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Correct answer: B — $6.25 \times 1 0^{- 5}$

Solution

$Degree of dissociation α = \frac{Λ _{m}}{Λ _{m}^{0}} = \frac{20}{400} = 0.05 .$
$Dissociation constant K_{a} = \frac{c α ^{2}}{1 - α} .$
$K_{a} = \frac{0.025 \cdot ( 0.05 ) ^{2}}{1 - 0.05} = \frac{0.025 \cdot 0.0025}{0.95} .$
$K_{a} = \frac{6.25 \times 1 0 ^{- 5}}{0.95} \approx 6.57 \times 1 0^{- 5} .$
$Given the options, the closest value is 6.25 \times 1 0^{- 5} . Hence the answer is (B).$