Answer
pH = 10.03
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.
$Kb = \frac{[OH^-][B^+]}{ [Base]}$
$Kb = 1.3 \times 10^{- 6}= \frac{x * x}{ 8.5 \times 10^{- 3}}$
$Kb = 1.3 \times 10^{- 6}= \frac{x^2}{ 8.5 \times 10^{- 3}}$
$ 1.1 \times 10^{- 8} = x^2$
$x = 1.05 \times 10^{- 4}$
5% test: $\frac{ 1.05 \times 10^{- 4}}{ 8.5 \times 10^{- 3}} \times 100\% = 1.23\%$
%ionization < 5% : Right approximation.
$[OH^-] = x = 1.05 \times 10^{-4}$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.05 \times 10^{- 4})$
$pOH = 3.97$
$pH + pOH = 14$
$pH + 3.97 = 14$
$pH = 10.03$
