What is the ph of a 0.17 m solution of methylammonium nitrate (ch3nh3no3)? kb for methylamine (ch3nh2) is 5.0 × 10−4 . 1. 7.00 2. 5.86 3. 5.73 4. 2.04 5. 12.0 6. 5.58 7. 6.35 8. 5.49?

When we have Kb = 5 x 10^-4 & Kw= 1 x10^-14

∴ Ka = Kw / Kb 

by substitution:

∴ Ka = (1 x 10^-14) /(5 x 10^-4)

        = 2 x 10^-11

by using ICE table: 

             CH3NH3+  + H2O ↔ H3O+  + CH3NH2

initial     0.17                               0             0

change  -X                                 +X            +X

Equ       (0.17 -X)                           X             X

∴Ka = [H3O+][CH3NH2]/[CH3NH3+]

by substitution:

2 x 10^-11 = X^2 / (0.17 -X)
we can assume that [CH3NH3+] = 0.17
2 x 10^-11 = X^2 / 0.17    by solving for X 

∴X = 1.8 x 10^-6

∴[H3O+] = X = 1.8 x 10^-6 

∴ PH = - ㏒(1.8 x 10^-6)

         = 5.7