In each situation below, we are adding aqueous solutions of HNO3 to NH3, and the
ID: 924503 • Letter: I
Question
In each situation below, we are adding aqueous solutions of HNO3 to NH3, and the calculating the PH of solution formed. (For ammonia Kb=1.8x10^-5)
a)Write the reaction that describes what happens when HNO3 is added to NH3. Do not include spectator ions.
b)What is the pH if 52.0mL of 0.120 M HNO3 added to 50.0mL of 0.120 M NH3?
c)What is the pH if 50.0mL of 0.120 M HNO3 added to 50.0mL of 0.120 M NH3?
Could you please explain each step rather than simply giving me the answer, I am trying to understand the concepts.
Explanation / Answer
a) simply HNO3 + NH3 = NH4NO3 + H2O is the reaction if you wish to know about net ionic reaction then here is the details.
Ammonia is a weak base and nitric acid is a strong acid.
In a solution;
- A strong acid completely ionizes.
HNO3(aq) -----> H+(aq) + NO3-(aq)
- A weak base cannot ionize completely.
NH3(aq) + H2O(l) -----> NH4+(aq) + OH-(aq)
Although this equation looks similar to the ionization of HCl, the remarkable difference is the shapes of the arrows representing the equations.
-----> represents the complete ionization
<----> represents the partial ionization (equilibrium reaction)
The other difference is not visible, but it is a fact that the extent of the ionization cannot exceed 5%. Therefore, OH-(aq) cannot represent a weak base.
Overall: NH3(aq) + HNO3(aq) ----> NH4NO3(aq)
Total ionic:
NH3(aq) + H+(aq) + NO3-(aq) ---> NH4+(aq) + NO3-(aq)
Net ionic:
NH3(aq) + H+(aq) + ---> NH4+(aq)
b) pH = -log [ ((mmoles of acid) - (mmoles of base)) / (vol. acid + vol. base) ]
Alternativlely
n= 52*0.120 -50*0.120 = .24 moles
[H+]= no of moles / total volume
[H+] = .24 /102 = .0023
we know that pH= -log [H+]
pH= -log [.0023] = 2.63
c) in this case equal moles of acid and base are reacted together due to which neutralization of reaction takes place.
n = 0
[H+]= no of moles / total volume = 0
pH= -log [H+] = 0