####36. The $pH$ of $0.01 M NaOH$ (aq) solution will be [NEET Odisha 2019]
(a) 9
(b) 7.01
(c) 2
(d) 12
$ \begin{aligned} & \text{ pOH }=-\log [OH^{-}]=-\log (10^{-2})=2 \\ & \text{ pH }=14-\text{ pOH }=12 \end{aligned} $
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