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Question: Molar conductivity of \(0.025molL^{- 1}\) methanoic acid is \(46.1Scm^{2}mol^{- 1},\) the degree of ...

Molar conductivity of 0.025molL−10.025molL^{- 1} methanoic acid is 46.1Scm2mol−1,46.1Scm^{2}mol^{- 1}, the degree of dissociation and dissociation constant will be

(Given λH+0=349.6Scm2mol−1\lambda_{H^{+}}^{0} = 349.6Scm^{2}mol^{- 1} and

λHCOO0=54.6Scm2mol−1\lambda_{HCOO}^{0} = 54.6Scm^{2}mol^{- 1})

A

11.4%,3.67×10−4molL−111.4\%,3.67 \times 10^{- 4}molL^{- 1}

B

22.8%,1.83×10−4molL−122.8\%,1.83 \times 10^{- 4}molL^{- 1}

C

52.2%,4.25×10−4molL−152.2\%,4.25 \times 10^{- 4}molL^{- 1}

D

1.14%,3.67×10−6molL−11.14\%,3.67 \times 10^{- 6}molL^{- 1}

Answer

11.4%,3.67×10−4molL−111.4\%,3.67 \times 10^{- 4}molL^{- 1}

Explanation

Solution

λºHCOOH=λºH++λºHCOO−\lambda º_{HCOOH} = \lambda º_{H^{+}} + \lambda º_{HCOO^{-}}

=349.6+54.6=404.2Scm2mol−1= 349.6 + 54.6 = 404.2Scm^{2}mol^{- 1}

α=ΔmΔºm=46.1404.2=11.4%\alpha = \frac{\Delta_{m}}{\Delta º_{m}} = \frac{46.1}{404.2} = 11.4\%

Ka=Cα21−α=0.025×(0.114)21−0.114K_{a} = \frac{C\alpha^{2}}{1 - \alpha} = \frac{0.025 \times (0.114)^{2}}{1 - 0.114}

=0.025×0.114×0.1140.886=3.67×10−4molL−1= \frac{0.025 \times 0.114 \times 0.114}{0.886} = 3.67 \times 10^{- 4}molL^{- 1}