[slayer]

I only got half credit for this problem and I want to know what concept I overlooked

Aniline = H₂NC₆H₅ has a resonance circle

NH₃ is polar and has one set of lone pairs

this was my explanation:
Although both N's are = in electronegativity, (specifically the N) the N in aniline will keep trying to attach itself to the resonance circle. (this part was underlined by the professor to reword). In contrast, the NH₃ is more accepting. (here she asked "accepting what?")

more attraction = stronger base, therefore ammonia is a stronger base.

what did i miss?

[Ψ×Ψ]

In order to be a base, a compound has to be able to take a proton off something else.  The proton-accepting ability of a base is related to the stability of its conjugate acid, because if the conjugate acid isn't decently stable, the base can't steal protons very effectively.  Ammonia can accept a proton without changing too much of anything.  The lone pair on the nitrogen in aniline is part of the pi system (which is what I think your explanation was getting at).  If it takes on a proton, that lone pair is no longer available for resonance use.  Does that make sense?

[slayer]

so the resonance will break apart if the aniline accepts a proton?

[Yggdrasil]

You will lose any resonance structures where there is a double bond to the nitrogen.

[slayer]

The proton-accepting ability of a base is related to the stability of its conjugate acid, because if the conjugate acid isn't decently stable, the base can't steal protons very effectively.

how is a solution's ability to create a stable conjugate acid relate to a resonance structure? Are resonance structures incapable of creating stable conjugate acids? and if so, why?

[Yggdrasil]

In general, conjugation/delocalization of electrons lowers their potential energies, making them more stable.  If you remove an p-orbital from the conjugated pi system, you are lowering the amount of delocalization and thus raising the energies of the pi-system orbitals, which is unfavorable.

In this case it isn't really the stability of the conjugate acid that matters, but the stability of the acid relative to the base.  Because the base has more delocalization than its conjugate acid, the conversion from base -> acid will be disfavored relative to say aminocyclohexane where no such loss of delocalization occurs.

[slayer]

So, resonance circles, such as the one in aniline become exceedingly more unstable when protons are accepted as opposed to ammonia. The N is the same in electronegativity but in aniline, the N does not want to accept protons because it keeps trying to double bond itself to the resonance.

Does this expanation cover it?

[Yggdrasil]

That sounds like a reasonable explanation to me.