Answer
$[H_3O^+] = 9.79 \times 10^{-3}$
$pH = 2.01$
Work Step by Step
1. Using the ICE table, we got:
$[H_3O^+] and [NO_2^-] = x$
and
$[HNO_2] = 0.143 - x$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 1.43 \times 10^{- 1}- x}$
$Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 1.43 \times 10^{- 1}- x}$
$ 1.02 \times 10^{- 4} - 7.2 \times 10^{- 4}x = x^2$
$ 1.02 \times 10^{- 4} - 7.2 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 7.2 \times 10^{- 4})^2 - 4 * (-1) *( 1.02 \times 10^{- 4})$
$\Delta = 5.18 \times 10^{- 7} + 4.11 \times 10^{- 4} = 4.12 \times 10^{- 4}$
$x_1 = \frac{ - (- 7.2 \times 10^{- 4})+ \sqrt { 4.12 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 7.2 \times 10^{- 4})- \sqrt { 4.12 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.05 \times 10^{- 2} (Negative)$
$x_2 = 9.79 \times 10^{- 3}$
Therefore: $[H_3O^+] = 9.79 \times 10^{-3}$
3. Find the pH value:
$pH = -log[H^+]$
$pH = -log( 9.79 \times 10^{- 3})$
$pH = 2.01$
