Answer
The pH value is approximately $8.77$.
Work Step by Step
1. As we calculate in 109a, the pKb of that base is 9.16, use it to find the Kb:
$Kb = 10^{-pKb}$
$Kb = 10^{-9.16}$
$Kb = 6.92 \times 10^{-10}$
2. Now, find $[OH^-]$
$Kb = \frac{[OH^-][B^+]}{[BOH]}$
$[OH^-] = [B^+] = x$
$6.92 \times 10^{-10} = \frac{x^2}{0.05-x}$
Since $x$ has a very low value: $(0.05-x) \approx (0.05)$
$6.92 \times 10^{-10} = \frac{x^2}{0.05}$
$3.46 \times 10^{-11} = x^2$
$x = 5.88 \times 10^{-6}M = [OH^-]$
3. Find the pH:
$pOH = -log(5.88 \times 10^{-6})$
$pOH = 5.23$
$pH = 14 - pOH $
$pH = 8.77$
