Answer
pKb = 4.645
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.
-$[OH^-] = [Conj. Acid] = x$
-$[Benzamide] = [Benzamide]_{initial} - x$
2. Calculate the $[OH^-]$
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 2.91}$
$[OH^-] = 1.23 \times 10^{- 3}$
$x = 1.23 \times 10^{- 3}$
3. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][Conj. Acid]}{ [Benzamide]}$
$Kb = \frac{x^2}{[Initial Benzamide] - x}$
$Kb = \frac{( 1.23\times 10^{- 3})^2}{ 0.068- 1.23\times 10^{- 3}}$
$Kb = \frac{ 1.513\times 10^{- 6}}{ 0.06677}$
$Kb = 2.266\times 10^{- 5}$
4. Calculate the pKb Value
$pKb = -log(Kb)$
$pKb = -log( 2.266 \times 10^{- 5})$
$pKb = 4.645$
