Here is a $0.400\;M$ solution of $HN_3$
(hydrazoic acid, $K_a = 1.9\cdot 10^{-5}$)
Equilibrium of hydrolysis:
$HN_3$ $+$ $H_2O$ 
$H_3O^+$ $+$ $N_3^-$
Which are the equilibrium concentrations of each species?
a) $(0,400-x)\;M$, $x\;M$, $x\;M$ b) $(0,200-x)\;M$, $x\;M$, $(0,200+x)\;M$ c) $x\;M$, $0\;M$, $(0,400-x)\;M$ d) $(0,400-x)\;M$, $(0,400-x)\;M$, $(0.400-x)\;M$
The answer a) is correct, $(0,400$ $-$ $ x)\;M$, $0$ $ +$ $ x$ $ =$ $ x\;M$, $0$ $+$ $x$ $=$ $ x\;M$.