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The molar conductivities of $H^{+} and {HCOO}^{-}$ ions at infinite dilution are 34.6 and 5.4 ${mSm}^{2} {mol}^{- 1}$ respectively. The molar conductivity of 0.025M HCOOH is 4.0 ${mSm}^{2} {mol}^{- 1}$ . Then pK of formic acid is

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4.2

4.8

3.6

1.8

answer is A.

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Detailed Solution

$μ_{HCOOH}^{\infty} = μ_{H^{+}}^{\infty} + μ_{{HCOO}^{-}}^{\infty}$

=34.6+5.4

=40

$μ_{HCCOH} = 4$

$\propto = \frac{μ}{μ_{2}} = \frac{4}{40} = 0.1$

$K_{a} = C \propto^{2}$

$= 0.025 x 10^{- 2}$

$= 2.5 x 10^{- 4}$

$pK = 4 - \log 2.5$

$= 3.6$

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