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Q.

The dissociation constant of a substituted benzoic acid at 25° C is $1.0 \times 10^{- 4}$ The pH of0.01 M solution of its salt is :

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answer is 8.

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Detailed Solution

$pH = \frac{1}{2} [14 + 4 + \log 0.01] = 8.0$

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