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  4. The pH of 0.01 M NaOH (aq) solution will be

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Q. The pH of 0.01 M NaOH (aq) solution will be

NEETNEET 2019Equilibrium

Solution:

$NaOH(aq)$ is strong base solution
So, $[OH^-] = N = 10^{-2}N$
$pOH = -log[OH^-] = -log10^{-2} = 2$
$pH = 14 - pOH = 14 - 2 $
$pH = 12$