Answer
$pH = 2.093$
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.
-$[H_3O^+] = [N{O_2}^-] = x$
-$[HNO_2] = [HNO_2]_{initial} - x = 0.15 - x$
For approximation, we consider: $[HNO_2] = 0.150M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][N{O_2}^-]}{ [HNO_2]}$
$Ka = 4.6 \times 10^{- 4}= \frac{x * x}{ 0.15}$
$Ka = 4.6 \times 10^{- 4}= \frac{x^2}{ 0.15}$
$ 6.9 \times 10^{- 5} = x^2$
$x = 8.307 \times 10^{- 3}$
Percent dissociation: $\frac{ 8.307 \times 10^{- 3}}{ 0.15} \times 100\% = 5.538\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 4.6 \times 10^{- 4}= \frac{x^2}{ 0.15- x}$
$ 6.9 \times 10^{- 5} - 4.6 \times 10^{- 4}x = x^2$
$ 6.9 \times 10^{- 5} - 4.6 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 4.6 \times 10^{- 4})^2 - 4 * (-1) *( 6.9 \times 10^{- 5})$
$\Delta = 2.116 \times 10^{- 7} + 2.76 \times 10^{- 4} = 2.762 \times 10^{- 4}$
$x_1 = \frac{ - (- 4.6 \times 10^{- 4})+ \sqrt { 2.762 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 4.6 \times 10^{- 4})- \sqrt { 2.762 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 8.54 \times 10^{- 3} (Negative)$
$x_2 = 8.08 \times 10^{- 3}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 8.08 \times 10^{- 3})$
$pH = 2.093$
