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Q. The pH of $10^{-8}M\,NaOH$ is:

JIPMERJIPMER 2004

Solution:

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