$p H$ of one litre solution containing 40 gm of $N a O H$ is:

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Solution

The correct option is D 14
1 L of 40 gm (=1 mol) $N a O H$ (molecular weight $=$ 40) corresponds to 1 M solution.
$[O H^{-}] = 1$
$p O H = - l o g [O H^{-}] = - l o g (1) = 0$
But $p H + p O H = 14$ .
Hence, $p H = 14 - p O H = 14 - 0 = 14$ .

Option D is correct.

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