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

$[H_3O^+] = x= 7.24 \times 10^{-2}$ $[OH^-] = 1.38 \times 10^{-13}$ $pH = 1.14$ $pOH = 12.86$
1. Drawing the ICE table we get: - $[H_3O^+] = [A^-] = x$ - $[HA] = [HA]_{initial} - x$ approximation: $[HA] \approx [HA]_{initial}$ 2. Use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{[HA]}$ $Ka = 1.1 \times 10^{- 2}= \frac{x * x}{ 5.5 \times 10^{- 1}}$ $Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 5.5 \times 10^{- 1}}$ $6.05 \times 10^{- 3} = x^2$ $x = 7.77 \times 10^{- 2}$ 5% test: $\frac{ 7.77 \times 10^{- 2}}{ 5.5 \times 10^{- 1}} \times 100\% = 14.1\%$ %ionization > 5% : Not accurate approximation. - So we need to consider $[HA] = [HA]_{initial} - x$: $Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 5.5 \times 10^{- 1}- x}$ $6.05 \times 10^{- 3} - 1.1 \times 10^{- 2}x = x^2$ $6.05 \times 10^{- 3} - 1.1 \times 10^{- 2}x - x^2 = 0$ $\Delta = (- 1.1 \times 10^{- 2})^2 - 4 * (-1) *( 6.05 \times 10^{- 3})$ $\Delta = 1.21 \times 10^{- 4} + 2.42 \times 10^{- 2} = 2.43 \times 10^{- 2}$ $x_1 = \frac{ - (- 1.1 \times 10^{- 2})+ \sqrt { 2.43 \times 10^{- 2}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.1 \times 10^{- 2})- \sqrt { 2.43 \times 10^{- 2}}}{2*(-1)}$ $x_1 = - 8.34 \times 10^{- 2} (Negative)$ $x_2 = 7.24 \times 10^{- 2}$ - Concentrations can't be negative. Therefore: $[H_3O^+] = x= 7.24 \times 10^{-2}$ 3. Find the pH, pOH and OH. $pH = -log[H^+]$ $pH = -log( 7.24 \times 10^{- 2})$ $pH = 1.14$ $pOH + pH = 14$ $pOH = 14 - pH = 14 - 1.14$ $pOH = 12.86$ $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{-12.86}$ $[OH^-] = 1.38 \times 10^{-13}$