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.
I apologize that the ice chart got smushed like this when i posted the problem.
I did not get the right answer by this method.
its alright. I figured out the answer. thanks for your help! :)
Yes, you have set up the ICE (Initial, Change, Equilibrium) chart correctly. The ICE chart helps you track the changes in concentration during a reaction.
In this case, you are trying to find the pH at the equivalence point of the titration of methylamine with HCl. The reaction is as follows:
HCl + CH3NH2 ⇌ CH3NH3+ + Cl^-
Initially, you have 0.160 M HCl and 0.160 M methylamine. Both the reactants will undergo a change during the reaction, and the products will form. Let's assume that the change is "x" for the formation of CH3NH3+.
So, at equilibrium, the concentration of HCl will be 0.160 - x, the concentration of methylamine will be 0.160 - x, the concentration of CH3NH3+ will be x, and the concentration of Cl- will also be x.
Next, you can set up the expression for the base dissociation constant (Kb) of methylamine:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since methylamine is a weak base, we can assume that the concentration of OH- ions is negligible compared to the concentration of CH3NH3+ ions. Therefore, we can consider [OH-] to be approximately equal to zero.
So, Kb = x^2 / (0.160 - x)(0.160 - x)
From here, you can solve the equation using the given Kb value of methylamine (5.0× 10–4) to determine the value of x, and then calculate the pH at the equivalence point.
Remember to check the validity of your assumption that x is small compared to the initial concentrations before calculating the pH.
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+]
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!!!!