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

What is the pH of 0.1 M trimethylammonium bromide,
(CH3)3NHBr?
The Kb of trimethylamine, (CH3)3N,
is 6.3 x 10-5.

What must be the molarity of an aqueous solution of
trimethylamine, (CH3)3N, if it has a pH = 11.04?
(CH3)3N+H2O⇌(CH3)3NH++OH−Kb=6.3×10−5

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HNO3(aq) after 19.5 mL of the acid have been added.
Kb of trimethylamine = 6.5 x 10-5.
2)Determine the volume in mL of 0.13 M KOH(aq) needed to reach
the half-equivalence (stoichiometric) point in the titration of 42
mL of 0.23 M C6H5OH(aq). The Ka of
phenol is 1.0 x 10-10.
Please explain

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