Consider the titration of 10.00 mL of a 0.250 M aqueous solution of ethylenediamine (H2NCH2CH2NH2) with 0.250 M HCl(aq). pKb1 = 4.07 and pKb2 = 7.15.
What is the pH of the solution after 7.50 mL of HCl are added?
Find mols of the diamine. mols = M x L
Find mols of HCl
Subtract to see which is in excess and how much of the salt has been formed.
Then use the Henderson-Hasselbalch equation to find the pH. I assume you have several of these additions of HCl. Each is done the same way UNTIL the first equivalence point is reached. The pH at the first eq. point is 1/2(pKa + pKb)
After the first eq point and until the second use the same procedure as in the first part except it's pK2 in the HH equation.
0.00250 mol H2NCH2CH2NH2
0.001875 mol HCl
=0.000625 mol H2NCH2CH2 excess
now what?
What do I plug into the HH equation?
............H2NCH2CH2NH2 + HCl ==> H2NCH2CH2NH3^+ + Cl^-
I............0.00250.....................0...............0................................0
Add....................................0.001875..................
C..........-0,001875............-0.001875......+0.001875
E..........0.000625..................0................+0.001875
Convert to mols/L. You now have 10 mL + 7.5 mL = 17.5 mL
(NH2CH2CH2NH2) = 0.000625/0.0175 = ?
(NH2CH2CH2NH3^+) = 0.001875/0.0175 = ?
(Cl^-) = 0.001875/0.0175 = ?
HH equation is
pH = pKa1 + log (base)/(acid)
Remember you can convert pKb1 to pKa1 by pKa + pKb = pKw = 14 so pKa = 14-pKb and you have pKb given in the problem. You use the first pKb (4.07 or whatever) until the first eq pt is reached.
(base) = NH2CH2CH2NH2 from above.
(acid) = NH2CH2CH2NH3^+ from above.
Substitute into the HH equation and solve for pH of the solution.