Calculate the $pH$ of a 0.193 $\frac{g}{L}$ sodium ethanolate solution
$C_2H_5O^-$ (treated as) strong base
The number of moles of $C_2H_5O^-$ is equal to the number of moles of sodium ethanolate !
Number of moles sodium ethanolate in $1\;L$:
$n$ $=$ $\frac{m}{M}$ =
(with $m$ $ =$ 0.193 $\; g$
and $M$ = 70.13)
0.0028$ \;mol$
$c_{C_2H_5O^-}$ $=$ $\frac{n}{V}$ =
0.0028 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;c_{C_2H_5O^-}$ =
$14$ $+$ $log \;$0.0028 =
11.44
