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$\mathrm{pH}$ of $10^{-3} \mathrm{M}$ solution of $\mathrm{KOH}$ is.
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Verified Answer
The correct answer is:
11
As $\mathrm{KOH}$ is a base.
$$
\begin{aligned}
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\
& \mathrm{pOH}=-\log \left[10^{-3}\right]=3 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{p}_{\mathrm{H}}=14-3=11
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\
& \mathrm{pOH}=-\log \left[10^{-3}\right]=3 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{p}_{\mathrm{H}}=14-3=11
\end{aligned}
$$
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