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

$[H^+] = 7.58 \times 10^{- 13}M$ $pH = 12.12$ $pOH = 1.88$ $[OH^-] = 1.3 \times 10^{-2}M$
$Kb (CH_3NH_2): 4.4 \times 10^{-4}$ 1. Drawing the ICE table, we get: -$[OH^-] = [B^+] = x$ -$[BOH] = [BOH]_{initial} - x$ For approximation, we consider: $[BOH] = [BOH]_{initial}$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][B^+]}{ [BOH]}$ $Kb = 4.4 \times 10^{- 4}= \frac{x * x}{ 3.86 \times 10^{- 1}}$ $Kb = 4.4 \times 10^{- 4}= \frac{x^2}{ 3.86 \times 10^{- 1}}$ $1.69 \times 10^{- 4} = x^2$ $x = 1.3 \times 10^{- 2}$ 5% test: $\frac{ 1.3 \times 10^{- 2}}{ 3.86 \times 10^{- 1}} \times 100\% = 3.37\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = 1.3 \times 10^{-2}M$ 3. Find $[H_3O^+]$, pH and pOH: $pOH = -log[OH^-]$ $pOH = -log( 1.3 \times 10^{- 2})$ $pOH = 1.88$ $pH + pOH = 14$ $pH + 1.88 = 14$ $pH = 12.12$ $[H^+] = 10^{-pH}$ $[H^+] = 10^{- 12.12}$ $[H^+] = 7.58 \times 10^{- 13}$