Answer
pH = 9.63
Work Step by Step
1. Drawing the ICE table we get:
-$[OH^-] = [B^+] = x$
-$[Base] = [Base]_{initial} - x$
For approximation, we consider: $[Base] = [Base]_{initial}$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][B^+]}{ [Base]}$
$Kb = 1.1 \times 10^{- 8}= \frac{x * x}{ 0.165}$
$Kb = 1.1 \times 10^{- 8}= \frac{x^2}{ 0.165}$
$ 1.81 \times 10^{- 9} = x^2$
$x = 4.26 \times 10^{- 5}$
5% test: $\frac{ 4.26 \times 10^{- 5}}{0.165} \times 100\% = 0.0258\%$
%ionization < 5% : Right approximation.
$[OH^-] = x = 4.26 \times 10^{-5}$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 4.26 \times 10^{- 5})$
$pOH = 4.37$
$pH + pOH = 14$
$pH + 4.37 = 14$
$pH = 9.63$
