In a 1.0× 10–2 M solution of CH3NH3Br(aq), identify the relative molar amounts from highest to lowest for :

Question

-Br-
-OH-
-CH3NH3+
-HBr
-CH3NH2
-H3O+
-H2O

Answer 1

........CH3NH3Br ==> CH3NH3^+ + Br^-
I......0.01...........0..........0
C.....-0.01..........0.01.......0.01
E........0...........0.01.......0.01

Then the hydrolysis of the cation.
......CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
I.....0.01................0........0
C......-x.................x........x
E.....0.01-x..............x........x

Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01-x)
Solve for x which gives you H3O^+ and CH3NH2 as well as CH3NH3^+.

Finally, (H^+)(OH^-) = Kw = 1E-14 which allows you to calculate OH^-

Now you can arrange them in any order you wish.
Post your work if you get stuck.

Answer 2

I can't find the value x by calculation. It ends up to give me a quadratic equation and the value of x is negative! please help me ASAP...

Answer 3

Damn Bob you left nothing to help

Answer 4

I miss 2013