1.

Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity. If Lambda_(m)^(@) for acetic acid is 390.5 S cm^(2) mol^(-1).what is its dissociation constant ?

Answer»

SOLUTION :Using the following relation and substituting the values, we get 5) S cm^(-1)) xx 1000 cm^(3) L^(-1))/(0.00241 mol L^(-1)) = 32.76 S cm^(2) mol^(-1)`
`alpha = Lambda_(m)^( C)/Lambda_(m)^(@) = (32.76)/(390.5) = 8.4 xx 10^(-2)`
DEGREE of dissociation may be obtained as under:
`K_(a) -(CALPHA^(2))/(1- alpha) = (0.00241 xx (8.4 xx 10^(-2))^(2))/(1-0.084) = 1.86 xx 10^(-5)`


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