Calculate the $pH$ of a 0.894 $\frac{g}{L}$ potassium methanolate solution
$CH_3O^-$ (treated as) strong base
The number of moles of $CH_3O^-$ is equal to the number of moles of potassium methanolate !
Number of moles potassium methanolate in $1\;L$:
$n$ $=$ $\frac{m}{M}$ =
(with $m$ $ =$ 0.894 $\; g$
and $M$ = 63.01)
0.0142$ \;mol$
$c_{CH_3O^-}$ $=$ $\frac{n}{V}$ =
0.0142 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;c_{CH_3O^-}$ =
$14$ $+$ $log \;$0.0142 =
12.15
