## Chemistry: Principles and Practice (3rd Edition)

$pH \approx 9.602$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: ** The image is in the end of this answer. -$[OH^-] = [Conj. Acid] = x$ -$[Base] = [Base]_{initial} - x = 1 \times 10^{- 3} - x$ For approximation, we consider: $[Base] = 1 \times 10^{- 3}M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][Conj. Acid]}{ [Base]}$ $Kb = 1.6 \times 10^{- 6}= \frac{x * x}{ 1\times 10^{- 3}}$ $Kb = 1.6 \times 10^{- 6}= \frac{x^2}{ 1\times 10^{- 3}}$ $1.6 \times 10^{- 9} = x^2$ $x = 4 \times 10^{- 5}$ Percent ionization: $\frac{ 4 \times 10^{- 5}}{ 1\times 10^{- 3}} \times 100\% = 4\%$ Since the percent ionization is less than 5 percent, this is a good approximation. Therefore: $[OH^-] = [Conj. Acid] = x = 4 \times 10^{- 5}M$ $[Base] \approx 0.001M$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 4 \times 10^{- 5})$ $pOH = 4.398$ $pH + pOH = 14$ $pH + 4.398 = 14$ $pH = 9.602$