KOH is a strong acid. For this $ N=\left[OH^{-}\right]=10^{-8} $ Hence $ \begin{array}{l} pOH=-\log \left[OH^{-}\right] \\ \text {pOH }=-\log 10^{-8} \\ \text { pOH }=8 \text { hence } pH=6 \end{array} $ This is not possible because pH of base can't be less than 7. As it is very dilute solution, hence in it $\left[ OH ^{-}\right]$in water must be added which is $10^{-7}$. $ \begin{array}{ll} \text { Hence } & {\left[OH^{-}\right]=10^{-7}+10^{-8}} \\ & {\left[OH^{-}\right]=10^{-7}(1+0.1)} \\ & {\left[OH^{-}\right]=10^{-7}(1.1)} \\ & pOH=-\log 10^{-7}-\log 1.1 \\ & pOH=7-(0.0414)(\log 1.1=0.0414) \\ & pOH=7-0.0414 \\ & pOH=6.95 \\ & pOH e \\ & pH=14-pOH \\ & pH=14-6.95=7.05 \end{array} $
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