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At 25 ^(@)C , the molar conductance of 0.007 M hydrofluoric acid is 150 mho cm^(2) mol^(-1) and its Lambda_(m)^(@) = 500 mho cm^(2) mol^(-1) . The value of the dissociation constant of the acid at the given concentration at 25^(@)C is |
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Answer» `7 xx 10^(-4)` M `Lambda_(m)^(@) = 500 mho cm^(2) mol^(-1)` Degree of DISSOCIATION `(alpha) = (Lambda_m)/(Lambda_m^@) = (150)/(500) =0.3` `""HF hArr H^(+) + F^(-)` C "" 0 "" 0` `Final : C - Calpha""C alpha ""Calpha` `therefore K_(a)= (C_(alpha).C_(alpha))/(C-C_(alpha)) = (Calpha^(2))/(1 - alpha)` `K_(a) = (7 xx10^(-3) xx (0.3)^(2))/(1-0.3)=(7xx 10^(-3)xx9 xx 10^(-2))/(0.7)` `rArr K_(a) = (7xx9xx 10^(-5))/(7XX10^(-1)) = 9 xx10^(-4) M` |
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