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At $25^{\circ} \mathrm{C}, \mathrm{pH}$ of a $10^{-8} \mathrm{M}$ aqueous $\mathrm{KOH}$
solution will be
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solution will be
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Verified Answer
The correct answer is:
7.02
$\mathrm{NaOH} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=10^{-8} \mathrm{M}$
$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$ $\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}$
$$
\begin{aligned}
\therefore\left[\mathrm{OH}^{-}\right]_{\text {total }} &=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-7}(1.1) \\
&=1.1 \times 10^{-7} \\
\therefore \quad \mathrm{pOH} &=\log 1.1 \times 10^{-7}=6.98 \\
\therefore \quad \mathrm{pH} &=14-6.98 \\
&=7.02
\end{aligned}
$$
$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$ $\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}$
$$
\begin{aligned}
\therefore\left[\mathrm{OH}^{-}\right]_{\text {total }} &=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-7}(1.1) \\
&=1.1 \times 10^{-7} \\
\therefore \quad \mathrm{pOH} &=\log 1.1 \times 10^{-7}=6.98 \\
\therefore \quad \mathrm{pH} &=14-6.98 \\
&=7.02
\end{aligned}
$$
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