3.11    Conductivity of 0.00241 M acetic acid is 7.896\times 10^{-5} \: S cm^{-1}. Calculate its molar conductivity. If  \Delta _{m}^{0} for acetic acid is 390.5 \: S cm^{2} mol^{-1}, what is its dissociation constant?

Answers (1)

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       \Lambda _m^{\circ}= 390.5\ Scm^2\ mol^{-1}.

                                                                     \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}

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