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

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

The pH of the solution is calculated to be 11.9.

First, we must set up the ICE table of the reaction;

       CH3NH2(aq) + H2O(l) ⇄   CH3NH3^+ (aq) + OH^-(aq)

I         0.175                                  0                          0

C       -x                                         +x                        +x

E       0.175 - x                               x                           x

We have to find the Ka as follows;

Ka = [ CH3NH3^+] [OH^-]/[CH3NH2]

Ka = x^2/0.175 - x  

Substituting values;

4.4 × 10−4 =  x^2/0.175 - x

4.4 × 10−4 (0.175 - x ) =  x^2

7.7 × 10−5 - 4.4 × 10^−4x = x^2

x^2 + 4.4 × 10^−4x - 7.7 × 10^−5 = 0

x = 0.0086 M

Now, pOH = -log(0.0086 M) = 2.1

pH = 14 - 2.1 = 11.9

Hence, the pH of the solution is calculated to be 11.9.

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Tolerance to high levels of salinity

Ithink 2 is the correct