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Chemistry Question on Electrochemistry

Molar conductivity of 0.025 mol L−10.025\, mol\, L^{-1} methanoic acid is 46.1 S cm2 mol−146.1\, S\, cm^2\, mol^{-1}, the degree of dissociation and dissociation constant will be (Given : λH+∘=349.6 S cm2 mol−1\lambda^{\circ}_{H^{+}}=349.6\,S\,cm^{2}\,mol^{-1} and λHCOO−∘=54.6 S cm2 mol−1\lambda^{\circ}_{HCOO^{-}}=54.6\,S\,cm^{2}\,mol^{-1})

A

11.4%11.4\%, 3.67×10−4 mol L−13.67\times10^{-4}\,mol\,L^{-1}

B

22.8%22.8\%, 1.83×10−4 mol L−11.83\times10^{-4}\,mol\,L^{-1}

C

52.2%52.2\%, 4.25×10−4 mol L−14.25\times10^{-4}\,mol\,L^{-1}

D

1.14%1.14\%, 3.67×10−6 mol L−13.67\times10^{-6}\,mol\,L^{-1}

Answer

11.4%11.4\%, 3.67×10−4 mol L−13.67\times10^{-4}\,mol\,L^{-1}

Explanation

Solution

λHCOOH∘=λH+∘+λHCOO−∘\lambda^{\circ}_{HCOOH}=\lambda^{\circ}_{H^{+}}+\lambda^{\circ}_{HCOO^{-}} =349.6+54.6=349.6+54.6 =404.2 S cm2 mol−1=404.2\,S\,cm^{2}\,mol^{-1} α=ΛmΛm∘=46.1404.2=11.4%\alpha=\frac{\Lambda_{m}}{\Lambda^{\circ}_{m}}=\frac{46.1}{404.2}=11.4\% Ka=Cα21−αK_{a}=\frac{C\alpha^{2}}{1-\alpha} =0.025×(0.114)21−0.114=\frac{0.025\times\left(0.114\right)^{2}}{1-0.114} =0.025×0.114×0.1140.886=\frac{0.025\times0.114\times0.114}{0.886} =3.67×10−4 mol L−1=3.67\times10^{-4}\,mol\,L^{-1}