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The pH of $10^{-8} \mathrm{M} \mathrm{NaOH}$ is
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The correct answer is:
$7.04$
For $10^{-8} \mathrm{M} \mathrm{NaOH}$,
$\left[\mathrm{OH}^{-}\right]=10^{-8}+10^{-7} \quad$ (from water)
$=11 \times 10^{-8}$
$\therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log (11 \times 10^{-8})=8-1.04=6.96$
$\therefore \quad \mathrm{pH}=14-6.96=7.04$
$\left[\mathrm{OH}^{-}\right]=10^{-8}+10^{-7} \quad$ (from water)
$=11 \times 10^{-8}$
$\therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log (11 \times 10^{-8})=8-1.04=6.96$
$\therefore \quad \mathrm{pH}=14-6.96=7.04$
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