1.

The degree of dissociation of acetic acid in a 0.1 M solution is `1.32xx10^(-2)`, find out the pKa :-A. 5.75B. 3.75C. 4. 00D. 4. 75

Answer» Correct Answer - D
`Ka=Calpha^(2)0.1xx(1.32xx10^(-2))^(2)=1.74xx10^(-5)`
`pKa=5-log1.74=5-0.26`
=4.74


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