Answer: The protonation of trimethylamine can be represented by the following equilibrium reaction:
(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-
The equilibrium constant for this reaction, which is the base ionization constant (Kb) of trimethylamine, is 6.31 x 10^-5 at 25°C.
The Kb expression for this reaction is:
Kb = [ (CH3)3NH+ ][OH-] / [(CH3)3N]
At equilibrium, we can assume that [OH-] = [ (CH3)3NH+ ] since one mole of hydroxide ion is produced for every mole of trimethylamine that is protonated. Therefore, we can simplify the Kb expression to:
Kb = [ (CH3)3NH+ ]^2 / [(CH3)3N]
We can rearrange this expression to solve for [ (CH3)3NH+ ]:
[ (CH3)3NH+ ] = sqrt(Kb * [(CH3)3N])
Plugging in the given values, we get:
[ (CH3)3NH+ ] = sqrt(6.31 x 10^-5 * 0.36 M) = 0.0104 M
The concentration of hydroxide ion in the solution is also equal to [ (CH3)3NH+ ] since the reaction produces one mole of hydroxide ion for every mole of trimethylamine that is protonated.
pOH = -log[OH-] = -log[ (CH3)3NH+ ] = -log(0.0104) = 1.98
Using the relation pH + pOH = 14, we get:
pH = 14 - pOH = 14 - 1.98 = 12.02
Therefore, the pH of the 0.36 M solution of trimethylamine is 12.0 (rounded to 1 decimal place).