Question

Calculate the pH at the equivalence point for the titration of 0.160 M methylamine (CH3NH2) with 0.160 M HCl. The Kb of methylamine is 5.0× 10–4.

Methylamine is a weak base and reacts with HCl to give the methylammonium ion.

HCl + CH3NH2 <---> CH3NH3^+ + Cl^-
0.160M 0.160M 0 0
-x -x +x +x
0.160-x 0.160-x x x

kb= x^2/(0.160-x)(0.160-x)

Have I set up the ice chart correctly? I'm not sure.

Answer 1

I apologize that the ice chart got smushed like this when i posted the problem.

Answer 2

I think you should have

CH3NH3 + H2O----> H3O+ + CH3NH2


Ka=Kw/kb= x^2/(0.160-x)

Sqrt*(Ka*(0.160))=H+

ph=-log[H+]

Answer 3

I did not get the right answer by this method.

Answer 4

Yes, because the volume doubles at the eq. pt, and the the concentration would change (i.e., the M would be cut in half). I didn't put in the right concentration when I gave you the set up.


Notice what I changed.

CH3NH3 + H2O----> H3O+ + CH3NH2


Ka=Kw/kb= x^2/(0.08-x)

Sqrt*(Ka*(0.08))=H+

ph=-log[H+]

I apologize about that!!!!

Answer 5

its alright. I figured out the answer. thanks for your help! :)