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 :`k = 7.896 xx 10^(-5) S cm^(-1)`
Molar conductivity can be OBTAINED as under :
`Lambda_(m) = (1000 k)/M = (1000 xx 7.896 xx 10^(-5))/0.00241 = (7.896 xx 10^(3))/241 =7896/241 = = 32.76 S cm^(2) mol^(-1)`
`ALPHA = (Lambda_(m))/(Lambda_(m)^(@)) =32.76/390.5= 0.0832 = 8.38 %`
Use the FOLLOWING relation to calculate DISSOCIATION constant, `K_(a)`
`K_(a) = (Calpha^(2))/(1- alpha) =(0.00241 xx (0.084)^(2))/(1-0.084) = 1.86 xx 10^(-5)`


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