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The pH of $0.005 \mathrm{M} \mathrm{KOH}$ is 9.95. Calculate the $\left[\mathrm{OH}^{-}\right]$
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The correct answer is:
$8.91 \times 10^{-5} \mathrm{M}$
$\begin{aligned} & \mathrm{P}^{\mathrm{H}}+\mathrm{P}^{\mathrm{OH}}=14 \\ & \mathrm{P}^{\mathrm{OH}}=14-.9 .95 \\ & =4.05 \\ & \mathrm{P}^{\mathrm{OH}}=-\log _{10}\left[\mathrm{OH}^{-}\right] \\ & {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{P}^{\mathrm{OH}}}=10^{-4.05}} \\ & =10^{-5+0.95}=10^{-5} \times 10^{0.95}=8.91 \times 10^{-5} \mathrm{M}\end{aligned}$
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