Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: cleangrapes on January 20, 2019, 07:09:33 PM
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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+, Ka = 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) ::equil:: NH3 + H3O+. Why is the presence of NO3- in the solution not taken into consideration?
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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+.
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Why does NH4+ react in greater proportion with water, than with NO3-? Is it because water is a stronger base than NO3-?
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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