## Chemistry: Atoms First (2nd Edition)

$pH = 12.013$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [C_2H_5N{H_3}^+] = x$ -$[C_2H_5NH_2] = [C_2H_5NH_2]_{initial} - x = 0.2 - x$ For approximation, we consider: $[C_2H_5NH_2] = 0.2M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][C_2H_5N{H_3}^+]}{ [C_2H_5NH_2]}$ $Ka = 5.6 \times 10^{- 4}= \frac{x * x}{ 0.2}$ $Ka = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.2}$ $1.12 \times 10^{- 4} = x^2$ $x = 1.058 \times 10^{- 2}$ Percent ionization: $\frac{ 1.058 \times 10^{- 2}}{ 0.2} \times 100\% = 5.292\%$ %ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.2- x}$ $1.12 \times 10^{- 4} - 5.6 \times 10^{- 4}x = x^2$ $1.12 \times 10^{- 4} - 5.6 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 5.6 \times 10^{- 4})^2 - 4 * (-1) *( 1.12 \times 10^{- 4})$ $\Delta = 3.136 \times 10^{- 7} + 4.48 \times 10^{- 4} = 4.483 \times 10^{- 4}$ $x_1 = \frac{ - (- 5.6 \times 10^{- 4})+ \sqrt { 4.483 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 5.6 \times 10^{- 4})- \sqrt { 4.483 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 1.087 \times 10^{- 2} (Negative)$ $x_2 = 1.031 \times 10^{- 2}$ - The concentration can't be negative, so $[OH^-]$ = $x_2$ 3. Calculate the pH value: $pOH = -log[OH^-]$ $pOH = -log( 0.01031)$ $pOH = 1.987$ $pH + pOH = 14$ $pH + 1.987 = 14$ $pH = 12.013$