The pH of $0.1 M$ monobasic acid with dissociation constant $10^{- 3}$ will be:

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Solution

Dissociation constant is $K_{a} = 10^{- 3}$
$p K_{a} = - l o g K_{a} = - l o g [10^{- 3}] = + 3$
From hinderson's equation,
$p H = p K_{a} + l o g \frac{[S a l t]}{[A c i d]}$
$\Rightarrow p H = 3 + l o g \frac{1}{0.1} = 3 - 1 = 2$
$p H = 2$

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