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