## General Chemistry: Principles and Modern Applications (10th Edition)

$[H_3O^+] = 9.79 \times 10^{-3}$ $pH = 2.01$
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: $pH = -log[H^+]$ $pH = -log( 9.79 \times 10^{- 3})$ $pH = 2.01$