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3.11 Conductivity of 0.00241 M acetic acid is $7.896\times 10^{-5} \: S cm^{-1}$ . Calculate its molar conductivity. If for acetic acid is $390.5 \: S cm^{2} mol^{-1}$ , what is its dissociation constant?

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Molar conductivity of a solution is given by :-

$\Lambda _M = \frac{\kappa }{C}$

So, $= \frac{7.896\times10^{-5}}{0.00241}\times1000$

or $= 32.76\ Scm^2\ mol^{-1}$

Also, it is given that _.

$\alpha = \frac{\Lambda _m}{\Lambda _m^{\circ}}$

or $\alpha = \frac{32.76}{390.5}$

$\alpha = 0.084$

For dissociation constant we have,

$K_d\ = \frac{c\alpha ^2}{(1-\alpha )}$

so, $= \frac{0.00241\times0.084^2}{(1-0.084 )}$

or $= 1.86\times 10^{-5}\ mol\ L^{-1}$

Posted by

Devendra Khairwa

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