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NCERT Class XII Chemistry
Chapter - Electrochemistry
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Question : 27 of 32
Marks: +1, -0
The molar conductivity of 0.025 mol L−1^{-1} methanoic acid is 46.1 S cm2^{2} mol−1^{-1}. Calculate its degree of dissociation and dissociation constant. Given λ\lambda(H+^{+} ) = 349.6 S cm2^{2} mol−1^{-1} and l(HCOO−^{-} ) = 54.6 S cm2^{2} mol−1^{-1}
Solution:  
Λm0(HCOOH)=λ0(H+)+λ0(HCOO−)\Lambda_{m}^{0}(\mathrm{HCOOH})=\lambda^{0}(\mathrm{H}^{+})+\lambda^{0}(\mathrm{HCOO}^{-})
=349.6+54.6=404.2 S cm2 mol−1=349.6+54.6=404.2\ \mathrm{S}\ \mathrm{cm}^{2}\ \mathrm{mol}^{-1}
α=λmλm0\alpha=\frac{\lambda_{m}}{\lambda_{m}^{0}}
=46.1404.2=0.114=\frac{46.1}{404.2}=0.114
⇒α=11.4%\Rightarrow\alpha=11.4\%
Ka=Cα21−αK_{a}=\frac{C\alpha^{2}}{1-\alpha}
=0.025×(0.114)21−0.114=\frac{0.025\times(0.114)^{2}}{1-0.114}
=0.025×0.114×0.1140.886=\frac{0.025\times0.114\times0.114}{0.886}
=3.67×10−4=3.67\times10^{-4}
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