What is the pKa of the second ionizable group?
Since you're given that the maximum $\mathrm{p}K_\mathrm{a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons.  (Perhaps this is what you meant by &quot;at that point there should be no free $\ce{H+}$&quot;?)
Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{p}K_\mathrm{a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)
It's straightforward to then calculate:

The total number of moles of acid added
The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid.  This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{p}K_\mathrm{a}$ value since you are told the value of the final $\mathrm{pH}$.
$^\ddagger\,\mathrm{pH} = \mathrm p K_\mathrm a +\log\left(\frac{\ce{[A-]}}{[\ce{HA}]}\right) = \mathrm pK_\mathrm a + \log \left(\frac{[\mathrm{A_{total}} ] - [\ce{HA}]}{\ce{[HA]}}\right) = \mathrm p K_\mathrm a + \log\left(\frac{1}{f_\mathrm{protonated}}-1\right)$