Question

What is the ph of 0.175 m methylammonium bromide, ch3nh3br? (kb of ch3nh2 = 4.4 × 10−4.)?

Answer 1

By using ICE table:

              CH3NH3+  +  H2O  → CH3NH4 2+   +  OH-

 initial       0.175                                   0                      0

change     -X                                     +X                     +X

Equ      (0.175-X)                                 X                      X

when: Ka = Kw / Kb 

     = (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11

when Ka = [CH3NH42+][OH-] / [CH3NH3+]

by substitution:

2.3 x 10^-11 = X^2 / (0.175 - X )    by solving for X

∴ X = 2 x 10^-6

∴[OH-] = 2 x 10^-6 

∴POH = -㏒[OH-]

           = -㏒(2 x 10^-6)

           = 5.7

when PH + POH = 14 

∴PH = 14 - 5.7 = 8.3