CH3NH3Br ==> CH3NH3^+ + Br^-
Then the CH3NH3^+ is hydrolyzed
..CH3NH3^+ + H2O ==> CH3NH2 + H3O^+
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 = (CH3NH2) = (H3O)^+.
That will give you CH3NH3^+, CH3NH2, H3O^+; (Br^-) is 0.01, Calculate OH from (H^+)(OH^-) = Kw = 1E-14
You will need to convert these values to the RELATIVE number; I'm not sure what your prof means by that.
In a 1.0× 10–2 M solution of CH3NH3Br(aq), identify the relative molar amounts of these species.
H2O, OH-, CH3NH3+, CH3NH2, H3O+, Br-, HBr
1 answer