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The $\mathrm{pH}$ of $0.01 \mathrm{M} \mathrm{NaOH}(a q)$ solution will be
Options:
Solution:
2049 Upvotes
Verified Answer
The correct answer is:
12
$\mathrm{NaOH}$ is a strong base, thus
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =0.0 \mathrm{M}=10^{-2} \mathrm{M} \\
\mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left(10^{-2}\right)=2
\end{aligned}
$$
We know that, $\mathrm{pH}+\mathrm{pOH}=14$
$\therefore \quad \mathrm{pH}=14-2=12$
Thus, option (3) is correct.
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =0.0 \mathrm{M}=10^{-2} \mathrm{M} \\
\mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left(10^{-2}\right)=2
\end{aligned}
$$
We know that, $\mathrm{pH}+\mathrm{pOH}=14$
$\therefore \quad \mathrm{pH}=14-2=12$
Thus, option (3) is correct.
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