Calculate the $pH$ of a 0.483 $\frac{g}{L}$ sodium amide solution
$NH_2^-$ (treated as) strong base
The number of moles of $NH_2^-$ is equal to the number of moles of sodium amide !
Number of moles sodium amide in $1\;L$:
$n$ $=$ $\frac{m}{M}$ =
(with $m$ $ =$ 0.483 $\; g$
and $M$ = 39.01)
0.0124$ \;mol$
$c_{NH_2^-}$ $=$ $\frac{n}{V}$ =
0.0124 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;c_{NH_2^-}$ =
$14$ $+$ $log \;$0.0124 =
12.09
