Question

What is the pH of a 0.00831 M solution of NH(CH3)3Cl? The Kb of N(CH3)3 is 6.3×10–5.

Answer 1

use:

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+

8.31*10^-3 0 0

8.31*10^-3-x x x

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

Ka = x*x/(c-x)

Assuming 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)*8.31*10^-3) = 1.148*10^-6

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

so, x = 1.148*10^-6 M

So, [H+] = x = 1.148*10^-6 M

use:

pH = -log [H+]

= -log (1.148*10^-6)

= 5.9399

Answer: 5.94