Answer
pKb = 8.839
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [Conj. Acid] = x$
-$[Base] = [Base]_{initial} - x$
2. Now, use the percent ionization to find the "x" value.
%ionization = $\frac{x}{[Initial Base]} \times 100$
0.053= $\frac{x}{ 5\times 10^{- 3}} \times 100$
0.00053= $\frac{x}{ 5\times 10^{- 3}}$
$ 2.65\times 10^{- 6}= x$
3. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][Conj. Acid]}{ [Base]}$
$Kb = \frac{x^2}{[Initial Base] - x}$
$Kb = \frac{( 2.65\times 10^{- 6})^2}{ 0.005- 2.65\times 10^{- 6}}$
$Kb = \frac{ 7.022\times 10^{- 12}}{ 4.997\times 10^{- 3}}$
$Kb = 1.405\times 10^{- 9}$
Calculate the pKb Value
$pKb = -log(Kb)$
$pKb = -log( 1.45 \times 10^{- 9})$
$pKb = 8.839$
