[cleangrapes]

It looks to me like ammonium nitrate (NH₄NO₃) will dissociate in water into NH₄⁺ and NO₃^-. I calculated the pH using the provided acid dissociation constant for NH₄⁺, K_a = 5.6x10^-10 and got the answer pH = 5.13. This is correct, as per the answer key, so it seems that I used the correct procedure.
Still, I am confused. It seems to me like the solution contains a stronger base than water, NO₃^-. I am trying to understand why the calculation is done without factoring in this in, as if we were just dealing with NH₄⁺(aq). The calculation, which yields the correct answer, uses the equation NH₄⁺(aq) + H₂O(l) NH₃ + H₃O⁺. Why is the presence of NO₃^- in the solution not taken into consideration?

[Alt+Tab]

In my opinion, the NO3- ion is a weak base, since it originates from a strong acid (HNO3). To answer your question, it is necessary to remember that the salt is formed by the reaction of a strong acid and a weak base. Therefore, for the same reason that NO3- is a weak base, NH4 + is a strong acid because it originates from a weak base. And finally, the dissociated NH4 + salt will react in greater proportion with H2O leading to NH3 + H3O+.

[cleangrapes]

Why does NH₄⁺ react in greater proportion with water, than with NO₃^-? Is it because water is a stronger base than NO₃^-?

[Borek]

NO₃^- is a very weak base, with pKb = 15.

You haven't told us what is the concentration, I assume 0.1 M.

Look at the exact calculations result: pH of 0.1 M ammonium nitrate solution is 5.13*, while pH of 0.1 ammonium chloride (much stronger acid) is 5.12. This 0.01 pH unit difference can be definitely attributed to the NO₃^- being a base (still, very small fraction of the NO₃^-, around 10^-7, gets protonated).

*actually around 5.28, due to ionic strength of the solution