Equilibrium Question 41
Question: $ pK_{a} $ of a weak acid is defined as [JIPMER 1999]
Options:
A) log $ _{10}K _{a} $
B) $ \frac{1}{log _{10}K _{a}} $
C) log $ _{10}\frac{1}{K _{a}} $
D) -log $ _{10}\frac{1}{K _{a}} $
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Answer:
Correct Answer: C
Solution:
$ pK_{a}={\log_{10}}\frac{1}{K_{a}} $