Chem
posted by Rich on .
In a 1.0× 10–2 M solution of CH3NH3Br(aq), identify the relative molar amounts from highest to lowest for :
Br
OH
CH3NH3+
HBr
CH3NH2
H3O+
H2O

........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.....0.01x..............x........x
Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01x)
Solve for x which gives you H3O^+ and CH3NH2 as well as CH3NH3^+.
Finally, (H^+)(OH^) = Kw = 1E14 which allows you to calculate OH^
Now you can arrange them in any order you wish.
Post your work if you get stuck. 
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...