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Q. A solution of $NaOH$ contain $0.04\, gm$ of $NaOH$ per litre. Its $pH$ is

Equilibrium

Solution:

$0.04\, g$ of $NaOH$
$\Rightarrow \frac{0.04\,g}{40\,g \, mol^{-1}}=10^{-3}$ moles of $NaOH$
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$\therefore \left[ OH ^{-}\right]=10^{-3}$
$\Rightarrow p [ OH ]=3$
$\Rightarrow pH =14-3=11$