Answer
$pH = 2.183$
The assumption that, x is a small number, is not valid in this case, because 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.
-$[H_3O^+] = [N{O_2}^-] = x$
-$[HNO_2] = [HNO_2]_{initial} - x = 0.1 - x$
For approximation, we consider: $[HNO_2] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][N{O_2}^-]}{ [HNO_2]}$
$Ka = 4.6 \times 10^{- 4}= \frac{x * x}{ 0.1}$
$Ka = 4.6 \times 10^{- 4}= \frac{x^2}{ 0.1}$
$ 4.6 \times 10^{- 5} = x^2$
$x = 6.782 \times 10^{- 3}$
Percent ionization: $\frac{ 6.782 \times 10^{- 3}}{ 0.1} \times 100\% = 6.782\%$
%ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 4.6 \times 10^{- 4}= \frac{x^2}{ 0.1- x}$
$ 4.6 \times 10^{- 5} - 4.6 \times 10^{- 4}x = x^2$
$ 4.6 \times 10^{- 5} - 4.6 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 4.6 \times 10^{- 4})^2 - 4 * (-1) *( 4.6 \times 10^{- 5})$
$\Delta = 2.116 \times 10^{- 7} + 1.84 \times 10^{- 4} = 1.842 \times 10^{- 4}$
$x_1 = \frac{ - (- 4.6 \times 10^{- 4})+ \sqrt { 1.842 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 4.6 \times 10^{- 4})- \sqrt { 1.842 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 7.016 \times 10^{- 3} (Negative)$
$x_2 = 6.556 \times 10^{- 3}$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 6.556 \times 10^{- 3})$
$pH = 2.183$
