Calculate the $pH$ of a 0.842 $\frac{g}{L}$ sodium hydroxide solution
$OH^-$ (treated as) strong base
The number of moles of $OH^-$ is equal to the number of moles of sodium hydroxide !
Number of moles sodium hydroxide in $1\;L$:
$n$ $=$ $\frac{m}{M}$ =
(with $m$ $ =$ 0.842 $\; g$
and $M$ = 40)
0.0211$ \;mol$
$c_{OH^-}$ $=$ $\frac{n}{V}$ =
0.0211 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;c_{OH^-}$ =
$14$ $+$ $log \;$0.0211 =
12.32
