A 0.10 mol dm–3 solution of NH2CH2CH2NH2 has a pH of 11.3. When
30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3
solution of NH2CH2CH2NH2, the final pH is 1.6.
Using the following axes, sketch the pH changes that occur
during this addition of
HCl(aq).
ANS: Since it is a diprotic base there will be two curves.
But how to calculate the end-point?
Data given only allows the student to calculate Kb1, but not Kb2.
If Kb2 is not given, the pH for the 1st and 2nd equivalence points
cannot be determined.
The final pH of 1.6 is due primarily to the excess HCl, and not
due to the acidity of N2H6 2+, since the former acid is much
stronger than the latter.
Since the question says "sketch" and since Kb2 is not given, the
student will not be required to determine the exact pH of the 1st
and 2nd equivalence points.
If Kb2 was given, then we can calculate both Ka values (ie. of
NH2CH2CH2NH3+ and +H3NCH2CH2NH3+, aka protonated ethylenediamine
and diprotonated ethylenediamine).
Only if we have both Kb (or therefore, both Ka) values, then we
can calculate the pH of the 1st and 2nd equivalence points as
follows :
1st equivalence point : only 1 amphiprotic species present,
use pH = 1/2 (pKa1 + pKa2)
2nd equivalence point : only 1 acidic species present, use
Ka of diprotonated ethylenediamine.