The dissociation constant of a weak acid is $10^{- 6}$ . Then the $P^{H}$ of $0.01 N$ of that acid is:

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Solution

The correct option is C 4
Correct answer is D $C H_{3} C O O H \Leftrightarrow C H_{3} C O O^{-} + H^{+ 1} k_{d} = \frac{[C H_{3} C O O^{-}] [H^{+ 1}]}{[C H_{3} C O O^{-}]} k_{d} = 10^{- 6} L e t y b e t h e a m o u n t o f a c i d d i s s o c i a t e d (H^{+ 1} i o n) y^{2} = 0.01 \times 10^{- 6} = 10^{- 8} y = 10^{- 4} p H = - l o g (10^{- 4}) = 4$

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