Answer
$$pH = 9.15$$
Work Step by Step
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:
$pH + pOH = 14$
$pH + 4.85 = 14$
$pH = 9.15$
