Question

What is the $pOH$ of a NaOH solution having a concentration of $1\times {10}^{-9}M$ at ${25}^{o}C$ ?

• A. 9.69
• B. 4.69
• C. 6.99
• D. 5.99

Answer 1

The correct option is C 6.99

The solution of NaOH is very dilute, where water dissociation needs to be considered to calculate the total$\left[O{H}^{-}\right]$ ions present in the solution.
Given, $\left[O{H}^{-}\right]$ from $NaOH$ = ${10}^{-9}M$

$H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
From $H_{2}O$ ,
$\left[O{H}^{-}\right]=\left[H^{+}\right]$ = $x$
$\left[O{H}^{-}\right]$ from $H_{2}O$ = $({10}^{-9}+x)M$
We know,
$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}x({10}^{-9}+x)={10}^{-14}{x}^2+{10}^{-9}x-{10}^{-14}=0x=\frac{-{10}^{-9}\pm \sqrt{{10}^{-18}+4\times {10}^{-14}}}{2}x=9.95\times {10}^{-8}$
For calculating $pH$ :
$RememberpH=-{\log }_{10}\left[H^{+}\right]$
$pH=-{\log }_{10}[9.95\times {10}^{-8}]=7.002pH+pOH=14pOH=14-7.002=6.99$