Question

The Kb for trimethylamine, (CH3)3N is 6.40x10-5. If 100 mL of 0.655 M trimethylamine
is titrated with 0.500 M HCl calculate the pH of the solution…
a) after 30.0 mL of HCl has been added to the solution. [10.3]
b) at the equivalence point. [5.18]

Answer 1

The [10.3] and [5.18] are the answers to the respective questions, but I am not sure how to get to this answer.

Answer 2

Follow the same steps as the NH4^+ earlier. It's worked the same way.The equation is (CH3)3N + HCl --> (CH3)3NH^+ + Cl^- For the equivalence point, the pH is determined by the hydrolysis of the salt. And this is worked the same way as the post you added to the NH4^+ earlier. I didn't show you how to do that but the equivalence point for this one is done this way.
At the equivalence point the hydrolysis of the salt, which is (CH3)3NH^+, is
.................(CH3)3NH^+ + HOH ==> (CH3)3N + H3O^+
You had 65.5 millimoles trimethyl amine to start and you added 131 mL of the 0.500 M HCl in the titration to the equivalence point so the total volume is 131 + 100 = 231 mL so the concentration of the salt is 65.5millimoles/231 mL = 0.284 M. Plug that into the hydrolysis equation like this and proceed.
.................(CH3)3NH^+ + HOH ==> (CH3)3N + H3O^+
I.................0.284..................................0................0
C..................-x......................................x................x
E.................0.284.................................x.................x
Ka for (CH3)3NH^+= (Kw/Kb for (CH3)3NH2) = [(CH3)3N][H3O^+]/[(CH3)3NH^+]. Plug in the numbers to get
(1E-14/6.40E-5) = (x)(x)/(0.284 - x). Solve for x and convert to pH.
I worked both problems and the answers you cited are correct.