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The conductivity of 0.001028M acetic aci...

The conductivity of `0.001028M` acetic acid is `4.95xx10^(-5)Scm^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(0)` for acetic acid is `390.5Scm^(2)mol^(-1)`.

Text Solution

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We know, `Lambda_(m)= k xx 1000/M`………….(i)
Given: `k = 4.95 xx 10^(-5) S cm^(-1), M = 0.001028`
`therefore` From (i) `Lambda_(m) = 4.95 xx 10^(-5) xx 1000/(0.001028)`
`=48.15 ohm^(-1) cm^(2) mol^(-1)`
Degree of dissociation,
`alpha = Lambda_(m)^( c)/Lambda_(m)^(@) =(48.15)/(390.5) = 0.1233`
`{:(CH_(3)COOH, Dissociation constant,
`K =([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH]) = (Calpha xx C alpha)/(C-Calpha) =(Calpha^(2))/(1-alpha)`
Substituting the values of .C. and `alpha` in above equation, we get,
`K = (0.001028 xx (0.1233)^(2))/(1-0.1233) = 1.78 xx 10^(-5)`
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