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  4. The pH of 10−8 M NaOH aqueous solution at 25°C , is

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Q. The $ pH $ of $ 10^{−8}\, M \,NaOH $ aqueous solution at $ 25^{\circ}C $ , is

UPSEEUPSEE 2009

Solution:

$\left[ OH ^{-}\right]$ of $NaOH =10^{-8} \,M$

Since, the concentration of $OH ^{-}$ obtained from $NaOH$ is very low, the concentration of $OH ^{-}$ in water cannot be neglected. Thus,

Total, $\left[ OH ^{-}\right]=\left[ OH ^{-}\right]_{ NaOH }+\left[ OH ^{-}\right]_{ H _{2} O }$

$=10^{-8}+10^{-7}$

$=10^{-7}(1.1)=1.1 \times 10^{-7}$

$pOH =-\log \left[ OH ^{-}\right]$

$=-\log \left(1.1 \times 10^{-7}\right)=6.96$

$\therefore pH =14-6.96=7.04$