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The pH of $10^{- 3} M 5$ mono acidic base, if it is 1% ionized is

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answer is D.

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Detailed Solution

$C = 10^{- 3} M$

$α = 1 % = \frac{1}{100}$

$[{OH}^{-}] = C_{α} = 10^{- 3} \times \frac{1}{100} = 10^{- 5}$

$∴ P^{OH} = - \log [{OH}^{-}] = - \log 10^{5} = 5$

$P^{H} = 14 - P^{OH}$

=14-5=9

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