Question

At ${25}^{o}C$, the $pH$ of a ${10}^{-8}M$ aqueous $KOH$ solution will be:

• A. $6.0$
• B. $7.02$
• C. $8.02$
• D. $9.02$

Answer 1

The correct option is B $7.02$

$KOH\rightleftharpoons K^{+}+{OH}^{-};\left[{OH}^{-}\right]={10}^{-8}M$
$H_{2}O\rightleftharpoons H^{+}+{OH}^{-};\left[{OH}^{-}\right]={10}^{-7}M$
$\therefore {\left[{OH}^{-}\right]}_{total}=({10}^{-8}+{10}^{-7})M$
$={10}^{-7}\left(1.1\right)=1.1\times {10}^{-7}$
$\therefore pOH=\log (1.1\times {10}^{-7})\simeq 6.98$
$\therefore pH=14-6.98=7.02$