Question

What is the pH of 0.304 M trimethylammonium bromide, (CH3)3NHBr? The Kb of trimethylamine, (CH3)3N, is 6.3 x 10-5.

The correct answer is 5.16. Please help me understand how to get this answer.

Answer 1

we have below equation to be used:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/6.3*10^-5

Ka = 1.587*10^-10

(CH3)3NH+ + H2O -----> (CH3)3N + H+

0.304 0 0

0.304-x x x

Ka = [H+][(CH3)3N]/[(CH3)3NH+]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.587*10^-10)*0.304) = 6.947*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.947*10^-6 M

so.[H+] = x = 6.947*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (6.947*10^-6)

= 5.16

Answer: 5.16