What is the sum of adding $p H$ and $p O H$ ?

3

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7

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14

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0

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Solution

The correct option is C $14$
At $25^{o} C$ , $[H^{+}] \times [O H^{-}] = 10^{- 14}$ [Ionic product of water]
Taking negative logarithm on both sides,
$- log [H^{+}] + - log [O H^{-}] = - log 10^{- 14}$
$⟹ p H + p O H = 14$

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