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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 |
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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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