Calculating concentration of H+ Ions using Ka Value I have this homework question that I am a bit stuck on for an answer. If anyone could help that would be great. 

Question: 


  For a $\pu{0.2M}$ $\ce{HNO2}$ solution ($\pu{K_a= 4.5 \times 10^{-4}}$) calculate the concentration of the $\ce{H+}$ and $\ce{NO2-}$ ions, $\ce{HNO2}$ molecules and the $\ce{pH}$ of the solution.


If anyone could give any hints or answers that would be great.
Answer:

$K_a = 4.5*10^-4$
Concentration (From ICE Table) of products/reactants:

HNO2 = 0.2 - x
H+ = x
NO2 = x

Therefore:

$$4.5*10^-4 = x^2/(0.2-x)$$
Rearrange:
$$x^2 + x*(4.5*10^-4) - (0.2(4.5*10^-4)) = 0$$
Using quadratic formula:
$x \approx 0.009$

$$pH = -log(10)$$
$$pH \approx 2.05$$.

That should be correct!