Question

Consider the titration of 40.0 mL 0.250 M ethylamine, C2H5NH2, with 0.350 M HCl.
Determine each of the following and sketch the titration curve.

Kb of ethylamine = 5.6x 10-4
a.
The volume of added acid required to reach the equivalence point.
b.
The initial pH.
c.
The pH when 5.0 mL of acid has been added.
d.
The pH at the half-way point.
e.
The pH at the equivalence point.
f.
The pH after adding 5.0 mL of acid beyond the equivalence point

Answer 1

You have a nightmare of a problem (actually 6 problems) here. It is important that you know how to do these. I don't mind helping you through it but first tell me what you know and what you don't understand. We can go from there. Please show your work.

Answer 2

confused on how to use the henderson-hasselbach equation. I am having trouble with part C. I got a pH of 10.75 for part C. Not sure if i'm doing it correctly

Answer 3

Thanks you. That helps a lot.
Let's call ethylamine just BNH2.

BNH2 + HCl ==> BNH3^+ + Cl^-
initial mols BNH2 = M x L = approx 0.01
mols HCl when adding 5 mL is M x L = 0.00175
mols. Put that on an ICE chart but I rather use millimols.. All of those zeros confuse me.
....BNH2 + HCl ==> BNH3^+ + Cl^-
I...10....0.......0
add.......1.75......
C.-1.75..-1.75....1.75
E...?.....0......1.75

Then substitute the E line into the HH equation and solve for pH.
Note: The HH equation uses CONCENTRATIONS (MILLIMOLS/ML) BUT SInce mL is the same for both, the you may use millimols by itself since the volume cancels.
Please follow up is this is not sufficient.

Answer 4

Check that post I made. I'm in a big hurry and I may have hit the wrong key on the calculator or misread an answer from the display. That is, confirm what I've written.