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  • Solve it, - Redox Reaction and Electrochemistry - JEE Main

The conductivity of 0.001028 mol L-1 acetic acid is 4.95 x 10-5 S cm-1. Calculate its dissociation constant if E_m^o for acetic acid is 390.5 S cm2 mol-1.

  • Option 1)

    1.85 \times 10^{-5} M

  • Option 2)

    1.55\times 10^{-5} M

  • Option 3)

    1.78\times 10^{-5}

  • Option 4)

    1.73\times 10^{-5} M

 

Answers (1)

As we have learnt,

 

Application of Kohlrausch's law -

Calculation of molar conductivities of weak electrolyte at infinite dilution.

-

 

 \\*\Lambda_m = \frac{K}{C} = \frac{4.95\times 10^{-5}}{0.001028}\times \frac{1000}{L} \\* \alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{48.15}{390.5} = 0.1233 \\*K = \frac{c\alpha^2}{1-\alpha} = \frac{0.001028\times (0.1233)^2}{1-0.1233} = 1.78\times 10^{-5} M

 


Option 1)

1.85 \times 10^{-5} M

Option 2)

1.55\times 10^{-5} M

Option 3)

1.78\times 10^{-5}

Option 4)

1.73\times 10^{-5} M

Posted by

Vakul

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