Answer
pH = $1.819$
The assumption that, x is small, is valid in this case, since the percent ionization is > 5%.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [N{O_2}^-] = x$
-$[HNO_2] = [HNO_2]_{initial} - x = 0.5 - x$
For approximation, we consider: $[HNO_2] = 0.5M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][N{O_2}^-]}{ [HNO_2]}$
$Ka = 4.6 \times 10^{- 4}= \frac{x * x}{ 0.5}$
$Ka = 4.6 \times 10^{- 4}= \frac{x^2}{ 0.5}$
$ 2.3 \times 10^{- 4} = x^2$
$x = 1.517 \times 10^{- 2}$
Percent ionization: $\frac{ 1.517 \times 10^{- 2}}{ 0.5} \times 100\% = 3.033\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [N{O_2}^-] = x = 1.517 \times 10^{- 2}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HNO_2] \approx 0.5M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.01517)$
$pH = 1.819$
