Answer
$Kb (Morphine) = 6.7\times 10^{- 7}$
Work Step by Step
1. We have these concentrations at the equilibrium:
-$[OH^-] = [Conj. Acid] = 0 + x = x$
-$[Morphine] = [Morphine]_{initial} - x$
2. Calculate the [OH^-]
pH + pOH = 14
10.5 + pOH = 14
pOH = 3.5
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 3.5}$
$[OH^-] = 3.2 \times 10^{- 4}$
Therefore : $x = 3.2 \times 10^{- 4}$
3. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][Conj. Acid]}{ [Morphine]}$
$Kb = \frac{x^2}{[Initial Morphine] - x}$
$Kb = \frac{( 3.2 \times 10^{- 4})^2}{ 0.15- 3.2\times 10^{- 4}}$
$Kb = \frac{ 1.0 \times 10^{- 7}}{ 0.15}$
$Kb = 6.7\times 10^{- 7}$
