Question

Consider a solution having $\left[H^{+}\right]={10}^{-5}M$ at ${25}^{\circ }C$, then $O{H}^{-}$ ion concentration of this solution at the given temperature is :

• A. ${10}^{-5}M$
• B. ${10}^{-9}M$
• C. ${10}^{-10}M$
• D. ${10}^{-8}M$

Answer 1

The correct option is B ${10}^{-9}M$

$pH=-log\left[H^{+}\right]pH=-log\left({10}^{-5}\right)pH=5$
$\because pH+pOH=14$
$\therefore pOH=14-5=9$
$-log\left[O{H}^{-}\right]=9$
$\left[O{H}^{-}\right]={10}^{-9}$


Theory:
$p{K}_w$:
$p{K}_w$ at ${25}^{o}C$ is $p{K}_w=-lo{g}_{10}(1.0\times {10}^{-14})=14$
Since $pH=-lo{g}_{10}\left[H^{+}\right]$ and $pOH=-lo{g}_{10}\left[O{H}^{-}\right]$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$ $-lo{g}_{10}{K}_w=(-lo{g}_{10}[H^{+}\left]\right)+(-lo{g}_{10}[O{H}^{-}\left]\right)$
$p{K}_w=pH+pOH$,
$14=pH+pOH$