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$10^{-6} \mathrm{M} \mathrm{NaOH}$ is diluted 100 times. The $\mathrm{pH}$ of the diluted base is
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2070 Upvotes
Verified Answer
The correct answer is:
between 7 and 8
$\left[\mathrm{OH}^{-}\right]$in the diluted base $=\frac{10^{-6}}{10^{2}}=10^{-8}$
Total $\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]$of water
$$
\begin{aligned}
&=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-8}[1+10] \mathrm{M}
\end{aligned}
$$
$\begin{aligned} &=11 \times 10^{-8} \mathrm{M} \\ \mathrm{pOH} &=-\log 11 \times 10^{-8} \\ &=-\log 11+8 \log 10 \\ &=6.9586 \\ \mathrm{pH} &=14-6.9586 \\ &=7.0414 \end{aligned}$
Total $\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]$of water
$$
\begin{aligned}
&=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-8}[1+10] \mathrm{M}
\end{aligned}
$$
$\begin{aligned} &=11 \times 10^{-8} \mathrm{M} \\ \mathrm{pOH} &=-\log 11 \times 10^{-8} \\ &=-\log 11+8 \log 10 \\ &=6.9586 \\ \mathrm{pH} &=14-6.9586 \\ &=7.0414 \end{aligned}$
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