CHEMISTRY

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Ethylamine, CH3CH2NH2, is an organic base with pKb = 3.367 at 298 K. In an experiment, a 40.0 mL sample of 0.105 mol L-1 CH3CH2NH2 (aq) is titrated with 0.150 mol L-1 HI(aq) solution at 298 K. (1a) Write a balanced chemical equation for the neutralization reaction upon which this titration is based, and indicate clearly which atom is being protonated. Then, calculate the equilibrium constant for the neutralization reaction. (Hint: To calculate the equilibrium constant, you may find it helpful to represent the neutralization reaction as the sum of two separate reactions.) (1b) Calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at the following stages of the titration. i) before the addition of any HI solution. ii) after the addition of 20.0 mL of HI solution. iii) at the equivalence point. iv) after the addition of 60.0 mL of HI solution

• CHEMISTRY -

After I went to bed last night and before I went to sleep, I worked the Keq in my head BUT you know how difficult that can be without pen and paper; therefore PLEASE double check this part to make sure I didn't slip up somewhere with something extra or something missing. If I call ethylamine, BNH2, then
if pKb = 3.367 the Kb = 4.3E-4
I followed the hint in the problem. The first equation is the ionization of the BNH2 in water.
BNH2 + HOH ==> BNH3^+ + OH^-
The second equation is the titration of the OH^- with H^+. It is
H^+ + OH^- ==> H2O
Now add those equations together and set up the Keq expression.
Now if you look at the expression you will (I think) a (BNH3^+)(OH^-)/(BNH2) which of course is Kb.
Check those off and look further. You should find a (H^+)(OH^-) in the denominator which means Kw. Further you should see (H2O) in the numerator and (H2O) in the denominator) so they cancel. If I wasn't too sleepy all that ends up with Keq = Kb/Kw but check it out.

For the titration itself, do this.
1. Determine the volume HCl to reach the equivalence point, approximately 28 mL.
2. For the beginning (zero mL HCl), set up an ICE chart for
.........BNH2 + HOH ===> BNH3^+ + OH^-
I........0.105............0........0
C..........-x..............x.......x
E........0.105-x............x......x
Set up the Kb expression substitute the equilibrium values and solve for x = (OH^-); convert that to pH.
3. For all points between zero and the eq. point, use the Henderson-Hasselbalch equation.
4. For the eq point, use the hydrolysis of the salt. If you have 0.004 mols of the salt produced (that's approximate remember), then mols/L will give you about 0.004 mols/(0.040+0.028) = about 0.06M
........BNH3^+ + HOH ==> H3O^+ + BNH2
I.......0.06M.............0........0
C.......-x................x........x
E......0.06-x.............x........x

Ka for BNH3^+ = Kw/Kb for BNH2) = (x)(x)/0.06-x and solve for x = (H3O^+) and convert to pH.
5. For all point after the eq point, that is simply the excess H^+ that has been added. Don't forget to take into account the dilution of the excess HCl.

If you have questions post your work and explain what you don't understand.

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