## Chemistry: The Molecular Science (5th Edition)

$[OH^-] = 3.223 \times 10^{- 3}M$ $pH = 11.508$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [Conj. Acid] = x$ -$[Methylamine] = [Methylamine]_{initial} - x = 0.024 - x$ For approximation, we consider: $[Methylamine] = 0.024M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][Conj. Acid]}{ [Methylamine]}$ $Kb = 5 \times 10^{- 4}= \frac{x * x}{ 0.024}$ $Kb = 5 \times 10^{- 4}= \frac{x^2}{ 0.024}$ $1.2 \times 10^{- 5} = x^2$ $x = 3.464 \times 10^{- 3}$ Percent ionization: $\frac{ 3.464 \times 10^{- 3}}{ 0.024} \times 100\% = 14.43\%$ %ionization > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the base concentration: $Kb = 5 \times 10^{- 4}= \frac{x^2}{ 0.024- x}$ $1.2 \times 10^{- 5} - 5 \times 10^{- 4}x = x^2$ $1.2 \times 10^{- 5} - 5 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 5 \times 10^{- 4})^2 - 4 * (-1) *( 1.2 \times 10^{- 5})$ $\Delta = 2.5 \times 10^{- 7} + 4.8 \times 10^{- 5} = 4.825 \times 10^{- 5}$ $x_1 = \frac{ - (- 5 \times 10^{- 4})+ \sqrt { 4.825 \times 10^{- 5}}}{2*(-1)}$ or $x_2 = \frac{ - (- 5 \times 10^{- 4})- \sqrt { 4.825 \times 10^{- 5}}}{2*(-1)}$ $x_1 = - 3.723 \times 10^{- 3} (Negative)$ $x_2 = 3.223 \times 10^{- 3}$ - The concentration can't be negative, so $[OH^-]$ = $x_2 = 3.223 \times 10^{- 3}M$ 3. Find the pH value. $pOH = -log[OH^-]$ $pOH = -log( 3.223 \times 10^{- 3})$ $pOH = 2.492$ $pH + pOH = 14$ $pH + 2.492 = 14$ $pH = 11.508$