Question

A solution of sodium hydroxide (NaOH) has a pH of 10. What will be the concentration of $O{H}^{-}$ ions in the given solution?

• A. ${10}^{-2}M$
• B. ${10}^{-7}M$
• C. ${10}^{-4}M$
• D. ${10}^{-5}M$

Answer 1

The correct option is C ${10}^{-4}M$

$WeknowpH+pOH=1410+pOH=14pOH=14-10=4pOH=-lo{g}_{10}\left[O{H}^{-}\right]=4\left[O{H}^{-}\right]={10}^{-4}M$
Hence concentration of $O{H}^{-}$ is ${10}^{-4}M$

Theory :
pH of a solution is the negative logarithm to the base 10 of the activity ($a_{H^{+}}$) of $H^{+}$ ions.
pH and pOH :
$pH=-lo{g}_{10}\left[H^{+}\right]=lo{g}_{10}\frac{1}{\left[H^{+}\right]}$,$pOH=-lo{g}_{10}\left[O{H}^{-}\right]=lo{g}_{10}\frac{1}{\left[O{H}^{-}\right]}$
In general for any quantity X we can define pX as $pX=-lo{g}_{10}X$
$p{K}_w$:
$p{K}_w$ 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 $pOH=-lo{g}_{10}\left[O{H}^{-}\right]$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$ $-lo{g}_{10}{K}_w=(-lo{g}_{10}[H^{+}\left]\right)+(-lo{g}_{10}[O{H}^{-}\left]\right)$
$p{K}_w=pH+pOH$
$14=pH+pOH$