Question

What is the pH of a 0.29M solution of (CH3)3NHNO3? Kb for (CH3)3N = 6.4 x 10-5

Answer 1

use:
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/6.4*10^-5
Ka = 1.563*10^-10

(CH3)3NH+      + H2O ----->     (CH3)3N   +   H+
0.29                    0         0
0.29-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 ((1.563*10^-10)*0.29) = 6.731*10^-6

since c is much greater than x, our assumption is correct
so, x = 6.731*10^-6 M



so.[H+] = x = 6.731*10^-6 M


use:
pH = -log [H+]
= -log (6.731*10^-6)
= 5.17

Answer: 5.17