Here is a $0.400\;M$ solution of $HN_3$
(hydrazoic acid, $K_a$ $=$ $1.9\cdot 10^{-5}$)
Equilibrium of hydrolysis:
$HN_3$ $+$ $H_2O$ 
$H_3O^+$ $+$ $N_3^-$
Let's substitute the concentrations of the different species in $K_a$. Which of the following is the correct substitution ?
a) $1.9\cdot10^{-5}$ $=$ $\frac{0.400-x}{x^2}$ b) $0,400-x$ $=$ $\frac{x^2}{ 1.9 \cdot 10^{-5}}$ c) $1,9\cdot10^{-5}$ $=$ $\frac{x^2 }{0.400-x }$ d) $x^2$ $=$ $\frac{0.400-x }{ 1.9 \cdot 10^{-5}}$
The answer c) is correct, $K_a$ $=$ $1.9 \cdot 10^{-5}$ $=$ $\frac{x^2 }{0.400-x }$