## Chemistry: The Central Science (13th Edition)

$$pH = 9.15$$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: $C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq)$ \begin{matrix} & [C_5H_5N] & [OH^-] & [C_5H_5NH^+] \\ Initial & 0.120 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.120 -x & x & x \end{matrix} For approximation, we are going to consider $0.120 - x \approx 0.120$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$ $Kb = 1.7 \times 10^{- 9}= \frac{x * x}{ 0.120}$ $Kb = 1.7 \times 10^{- 9}= \frac{x^2}{ 0.120}$ $x^2 = 1.7 \times 10^{-9} \times 0.120$ $x = \sqrt { 1.7 \times 10^{-9} \times 0.120} = 1.4 \times 10^{-5}$ Percent ionization: $\frac{ 1.4 \times 10^{- 5}}{ 0.120} \times 100\% = 0.012\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [C_5H_5NH^+] = x = 1.4 \times 10^{- 5}M$ $[C_5H_5N] \approx 0.120M$ 3. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 1.4 \times 10^{- 5})$ $pOH = 4.85$ 4. Find the pH: $pH + pOH = 14$ $pH + 4.85 = 14$ $pH = 9.15$