## Chemistry: Atoms First (2nd Edition)

$pH = 11.871$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [(C_2H_5)_2N{H_2}^+] = x$ -$[(C_2H_5)_2NH] = [(C_2H_5)_2NH]_{initial} - x = 0.05 - x$ For approximation, we consider: $[(C_2H_5)_2NH] = 0.05M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][(C_2H_5)_2N{H_2}^+]}{ [(C_2H_5)_2NH]}$ $Ka = 1.3 \times 10^{- 3}= \frac{x * x}{ 0.05}$ $Ka = 1.3 \times 10^{- 3}= \frac{x^2}{ 0.05}$ $6.5 \times 10^{- 5} = x^2$ $x = 8.062 \times 10^{- 3}$ Percent ionization: $\frac{ 8.062 \times 10^{- 3}}{ 0.05} \times 100\% = 16.12\%$ %ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 1.3 \times 10^{- 3}= \frac{x^2}{ 0.05- x}$ $6.5 \times 10^{- 5} - 1.3 \times 10^{- 3}x = x^2$ $6.5 \times 10^{- 5} - 1.3 \times 10^{- 3}x - x^2 = 0$ $\Delta = (- 1.3 \times 10^{- 3})^2 - 4 * (-1) *( 6.5 \times 10^{- 5})$ $\Delta = 1.69 \times 10^{- 6} + 2.6 \times 10^{- 4} = 2.617 \times 10^{- 4}$ $x_1 = \frac{ - (- 1.3 \times 10^{- 3})+ \sqrt { 2.617 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.3 \times 10^{- 3})- \sqrt { 2.617 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 8.738 \times 10^{- 3} (Negative)$ $x_2 = 7.438 \times 10^{- 3}$ - The concentration can't be negative, so $[OH^-]$ = $x_2$ 3. Calculate the pH value: $pOH = -log[OH^-]$ $pOH = -log( 7.438 \times 10^{- 3})$ $pOH = 2.129$ $pH + pOH = 14$ $pH + 2.129 = 14$ $pH = 11.871$