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}} $

Show Answer

Answer:

Correct Answer: C

Solution:

$ pK_{a}={\log_{10}}\frac{1}{K_{a}} $



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