Calculating concentration of H+ Ions using Ka Value
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!