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.
5 answers
I apologize that the ice chart got smushed like this when i posted the problem.
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+]
CH3NH3 + H2O----> H3O+ + CH3NH2
Ka=Kw/kb= x^2/(0.160-x)
Sqrt*(Ka*(0.160))=H+
ph=-log[H+]
I did not get the right answer by this method.
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!!!!
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!!!!
its alright. I figured out the answer. thanks for your help! :)