[NEET Odisha 2019]
(a) $9$
(b) $7.01$
(c) $2$
(d) $12$
Correct Answer: 30.(d)
Solution:
$ \begin{aligned} & \text{ pOH }=-\log [OH^{-}]=-\log (10^{-2})=2 \\ & \text{ pH }=14-\text{ pOH }=12 \end{aligned} $
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