Question
Calculate pH of $10^{-8} N KOH$ solution.
✓
Answer
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}
$
$
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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